Class 11 Medical Chemistry - Thermodynamics

The standard heat of formation $(\Delta_f H^0_{298})$ of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite are $-1560, -393.5$ and $-286\ kJ/mol$ , respectively is ______.

We have to calculate the heat of formation of ethane.

The reaction of formation of ethane is,

$2C_{(s)} + 3H_{2(g)} \rightarrow C_2H_{6(g)}$

Heat of combustion of ethane is,

$-1560\ kJ/mol$

$C_2H_{6(g)} + \dfrac{7}{2} O_2(g) \rightarrow 2CO_2 + 3H_2O$ ----- (I)

$\Delta H = -1560\ kJ/mol$

Heat of combustion or hydrogen is,

$-393.5\ kJ/mol$

$H_{2(g)} + \dfrac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)}$ ------- (II)

$\Delta H = -393.5\ kJ/mol$

Heat of combustion of graphite is,

$-286\ kJ/mol$

$C(s) + O_2 (g) \rightarrow CO_{2(g)}$ ___(III)

$\Delta H = -286\ kJ/mol$

Multiply equation (III) by 2 multiply equation (II) by 3 and inverse equation (I)

$22C_{(s)} +2O_{2(g)} \rightarrow 2CO_{2(g)}$ -------- 1

$\Delta H_1 = -286 \times 2 = -572\ kJ/mol$

$3H_2 + \dfrac{3}{2} O_2(g) \rightarrow 3H_2O (g)$ ------ 2

$\Delta H_2 = -393.5 \times 3 = -1180.5\ kJ/mol$

$2CO_{2(g)} + 3H_2 O_{(g)} \rightarrow C_2 H_6 (g) + \dfrac{7}{2} O_2 (g)$ ----3

$\Delta H_3 = 1560\ kJ/mol$

Adding 1, 2 and 3 equation, we get

$2C_{(s)} + 3 H_2 (g) \rightarrow C_2 H_6 (g)$

$\Delta H =\Delta H_1+\Delta H_2+\Delta H_3= -572 - 1180.5 + 1560$

$\Delta H = -192.5\ kJ/mol$

The surface of copper gets tarnished by the formation of copper oxide. $N_{2}$ gets was passed to prevent the oxide formation during heating of copper at $1250\ K$ . However, the $N_{2}$ gas contains $1$ mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:
$2Cu(s) + H_{2}O(g) \rightarrow Cu_{2}O(s) + H_{2}(g)$
$\rho_{H_{2}}$ is the minimum partial pressure of $H_{2}$ (in bar) needed to prevent the oxidation at $1250\ K$ . The value of $ln (\rho_{H_{2}})$ is _________.

Given:
Total pressure $= 1\ bar$
$R$ (universal gas constant) $= 8\ JK^{-1} mol^{-1}$
$ln (10) = 2.3$
$Cu(s)$ and $Cu_{2}O(s)$ are mutually immiscible
At $1250\ K$ :
$2Cu(s) + 1/2O_{2}(g) \rightarrow Cu_{2}O(s); \ \Delta G^{\circ} = -78,000\ J\ mol^{-1}$ $H_{2}(g) + 1/2O_{2}(g) \rightarrow H_{2}O(g); \ \Delta G^{\circ} = -1,78,000\ J\ mol^{-1}$
[Upto one decimal point and take modulus of the answer]

$2Cu(s) + \dfrac {1}{4} O_{2}(g) \rightarrow Cu_{2}O(s) \ \Delta G^{\circ} = -78\ kJ$

$[H_{2}(g) + \dfrac {1}{2} O_{2}\rightarrow H_{2}O(g) \ \Delta G^{\circ} = -178\ kJ]\times (-1)$
Hence,

$2Cu(s) + H_{2}O(g)\rightarrow Cu_{2}O + H_{2}(g)\ \Delta G^{\circ} = +100\ kJ$

$\Delta G = \Delta G^{\circ} + RT\ ln\ Q$

$0 = + 100 + \dfrac {8}{1000}\times 1250\ ln \dfrac {\rho_{H_{2}}}{\rho_{H_{2}O}}$

$- \dfrac {100\times 1000}{8} = 1250\ ln \dfrac {\rho_{H_{2}}}{\left (\dfrac {1}{100}\times 1\right )}$

$\therefore ln\ \rho_{H_{2}} = -14.6$ .

The degree of freedom of a three atomic molecule can not be 5. true=1, false=0

A polyatomic linear molecule having N molecules has :

(i)

$3$ translation degrees of freedom .

(ii)

$2$ rotational degrees of freedom .

(i)

$3N-5$ vibrational degrees of freedom .

Hence total degrees of freedom ,

$f=3+2+3N-5=3N$

But for a triatomic linear molecule , if there is no vibrational motion then

$f=5$ , the degrees of freedom for a triatomic molecule can be 5 .

Given statement is false=0.

What is meant by an equilibrium?

When net external force acting on the object is zero, the object is said to be in equilibrium.

The specific heat of a gas is 0.125 and 0.075 cal/g. The value of its molar mass is ______.

As we know,

${ C }_{ p }-{ C }_{ V } = \dfrac{R}{M}$

$M(0.125-0.075)=2$

$M=40\\$

Molar mass of gas

$= 40$ .

Thermochemical equations can be reversed by changing the sign of $\Delta H$ .

If true enter 1, else enter 0.

A chemical equation undergoes in forward reaction when

$\triangle$ G <0 and in backward direction when

$\triangle$ G>0

$\triangle$ G =

$\triangle$ H-T

$\triangle$ S

We can change the sign of

$\triangle$ G by increasing or decreasing

$\triangle$ H.

Hence, the reaction can be reversed from forward to backward and vice-versa

$\displaystyle \Delta G=\Delta H-T\Delta S$ .

If true enter 1, else enter 0.

Gibbs Energy is the maximum (or reversible) work that a thermodynamic system can perform at a constant temperature and pressure.

We can mathematically define the Gibbs free energy by the equation:

$\triangle G=\triangle H-T \triangle S$

where

$H=heat \quad content$ ;

$S=entropy$ ; and

$T=temperature$

Which has more heat capacity: a bucket full of water or a mug full of water?

This is quite obvious that a bucket full of water contains more water than a mug full of water. Because the heat capacity of a substance depends on mass, therefore a bucket full of water will have more heat capacity than a mug of water.

Given equal masses of aluminium and copper, which would require more heat if both are heated to the same temperature. Justify your answer.
[Specific heat capacity of $Al= 899 Jkg^{-1} C^{-1}$ and $Cu= 387 Jkg^{-1} C^{-1}$ ]

Since the specific heat capacity of aluminium (899

$Jkg^{-1}C^{-1}$ ) is more than that of copper (387

$Jkg^{-1} C^{-1}$ ) so aluminium would require more amount of heat than copper to raise an equal amount of temperature.

Differentiate between heat capacity and specific heat capacity.

The heat required to raise the temperature of the whole substance by

$1^oC$ is called the heat capacity while the heat required to raise the temperature of just 1 g or 1 kg of it is called its specific heat capacity.

The unit of heat capacity is

$JK^{-1}$ and that of specific heat capacity is

$JK^{-1}g^{-1}$ .

Why is water used as a coolant in car radiations and factories to keep the engine and other parts cool?

Due to its large specific heat capacity, water absorbs large amounts of heat from the heated parts without much rise in temperature.

What is the quantity of heat released per kg of water per 1 $^o$ C fall in temperature?

By definition, amount of heat released per kg per 1

$^o$ C fall in temperature is specific heat capacity of water

$= 4200 J kg^{-1} {}^oC^{-1}$

Calculate the heat energy that will be released when 5.0 kg of steam at 100 $^o$ C condenses to form water at 100 $^o$ C. Express your answer in S.I. unit (specific heat of vapourisation of steam is 2268 kJ/kg)

Heat produced (H)

$= mL = 5kg \times 2268 k J g^{-1} = 11340 kJ$

Discuss the change in energy and arrangement of molecules on increasing the temperature of ice from -5 $^o$ C to 10 $^o$ C at 1 atm pressure.

When ice is heated at

$-5^oC$ its temperature increases up to

$0^oC$ that is the kinetic energy of the molecules increases. At

$0^oC$ , the ice starts melting. During this process, the energy supplied is utilized to increase the potential energy of the molecules keeping kinetic energy constant, and the arrangement of molecules changes. Once the process ends, the heat supplied is again used to increase the temperature of water by increasing the kinetic energy. However, from

$0^oC$ to

$4^oC$ , the molecules of water come closer and above

$4^oC$ the molecules move farther away.

Water falls from a height of 50 cm. Calculate the rise in temperature of water when it strikes the bottom. ( $g = 10 ms^{-2}$ , sp. heat capacity of water $=4200 J/kg^oC$ )

Let the rise in temperature be T

$^o$ C. Then, Heat produced by water = Loss in potential energy

$\Rightarrow msT = mgh \Rightarrow T = \displaystyle \frac{gh}{s} = \frac{10 \times 50}{4200} = \frac{5}{42} = 0.12^oC$

The rise in temperature =

$0.12^oC$

Explain why
(a) Two bodies at different temperatures T $\displaystyle _{1}$ and T $\displaystyle _{2}$ if brought in thermal contact do not necessarily settle to the mean temperature (T $\displaystyle _{1}$ +T $\displaystyle _{2}$ )/2
(b) The coolant in a chemical or a nuclear plant (i.e. the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat
(c) Air pressure in a car tyre increases during driving
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude

(a)

Law of conservation of Energy says that the energy is transferred from one body to another and from one format to another.
If we consider ideal conditions for heat transfer, when two bodies of temperature

$T_1$ and

$T_2$ come in thermal contact, heat transfer takes place from body of higher temperature to body of lower temperature.
The quantity thermal energy is the function of mass and temperature. Hence, the mean temperature of the bodies in contact will be given by the weighted average not normal average.

$\therefore T_{avg} = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}$
Hence only when

$m_1 = m_2$ , then

$T_{avg} = \frac{T_1 + T_2}{2}$

(b)

A coolant is required to have high heat capacity so that its temperature does not change much during heat transfer. Due to this, the coolant is able to hold more heat and allows quicker temperature drop in the hot body.

(c)

As the car moves, temperature of the gas inside the tyre increases.

$PV=nRT$

Since volume is almost constant, air pressure increases.

(d)

Water has a very high value of specific heat capacity. Due to this, the temperature of the water bodies do not change easily. Hence, climate near a harbour town is moderate.

Sand has a very low value of specific heat capacity. Due to this, the temperature of the sand changes rapidly. Hence, climate of a desert is extreme.

For the reaction,
2 A(g)+ B(g) $\displaystyle \rightarrow$ 2D(g), $\displaystyle \bigtriangleup U^{\ominus }=-10.5KJ$ and $\bigtriangleup S^{\ominus}=-44.1 JK^{-1}$

Calculate $\displaystyle \bigtriangleup G^{\ominus }$ for the reaction and predict whether the reaction may occur spontaneously.

The change in the number of moles of gaseous species,

$\Delta n_g= \displaystyle 2-3=-1$

$\displaystyle \Delta H^0 = \Delta U^0 + \Delta n_gRT =-10.5+(-1) \times 8.314 \times 10^{-3} \times 298=-12.98 kJ$

$\displaystyle\Delta G^0 =\Delta H^0-T\Delta S^0= -12.98-298(-44.10 \times 10^{-3}) = +0.16 kJ$

Since

$\Delta G^o$ for the reaction is positive, the reaction will not occur spontaneously.

State Hess's law of constant heat summation and explain it with an example.

Hess's law of constant heat summation:
The law states that the change in enthalpy for a reaction is the same whether the reaction takes place in one or a series of steps. The Hess's law can also be stated as the enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs.

For example, consider following two paths for the preparation of methylene chloride
Path I :

$\displaystyle CH_4(g) + 2Cl_2(g) \rightarrow CH_2Cl_2(g) + 2HCl (g) \: \: \: \Delta H^0_1=-202.3 \: kJ$

Path II :

$\displaystyle CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl (g) \: \: \: \Delta H^0_2=-98.3 \: kJ$

$\displaystyle CH_3Cl(g) + Cl_2(g) \rightarrow CH_2Cl_2(g) + HCl (g) \: \: \: \Delta H^0_3=-104.0 \: kJ$

Adding two steps

$\displaystyle CH_4(g) + 2Cl_2(g) \rightarrow CH_2Cl_2(g) + 2HCl (g) \: \: \: \Delta H^0=-202.3 \: kJ$

Thus whether we follow path I or path II, the enthalpy change of the reaction is same.

$\displaystyle \Delta H^0_1=\Delta H^0_2+\Delta H^0_3=-202.3 \: kJ$

State Hess's law.

$1$ mole of ${ Fe }_{ 2 }{ O }_{ 3 }$ and $2$ moles of $Al$ are mixed at temperature ${ 25 }^{ o }C$ and the reaction is completed to give:
${ Fe }_{ 2 }{ O }_{ 3 }(s)+2Al(s)\longrightarrow { Al }_{ 2 }{ O }_{ 3 }(s)+2Fe(l);\Delta H=-850kJ$
The liberated heat is retained within the products, whose combined specific heat over a wide temperature range is about $0.8J{ g }^{ -1 }{ K }^{ -1 }$ . The melting point of iron is ${ 1530 }^{ o }C$ . Show that the quantity of heat liberated is sufficient to raise the temperature of the product to the melting point of iron in order to get it welded.

Mass of products
=Mass of one mole of

${Al}_{2}{O}_{3}+$ Mass of two moles of

$Fe$

$=214g$

$q=ms\Delta T=214\times 0.8\times (1803-298)$

$=257656J=257.656J$
Heat required is less than heat released, hence the temperature can be easily raised to the required value.

A gas expands from a volume of $3.0\ {dm}^{3}$ to $5.0\ {dm}^{3}$ against a constant pressure of $3.0$ atm. The work done during expansion is used to heat $10.0$ mole of water of temperature $290.0K$ . Calculate the final temperature of water (specific heat of water $=4.184\ J\ { K }^{ -1 } { g }^{ -1 }$ )

Work done

$=P\times dV=3.0\times (5.0-3.0)$

$=6.0litre-atm=6.0\times 101.3J$

$=607.8J$
Let

$\Delta T$ be the change in temperature
Heat absorbed

$=m\times s\times \Delta T$

$=10.0\times 18\times 4.184\times \Delta T$
Given,

$P\times dV=m\times s\times \Delta T$
or

$\Delta T=\cfrac { P\times dV }{ m\times s } =\cfrac { 607.8 }{ 10.0\times 18.0\times 4.184 } =0.807$
Final temperature

$=290+0.807=290.807K$

A gas contained in a cylinder is filled with a frictionless piston expands against a constant pressure $1$ atmosphere from a volume of $4$ litre to a volume of $14$ litre. In doing so, it absorbs $800J$ thermal energy from surroundings. Determine $\Delta U$ for the process.

Given,

$q=800J$

$\Delta V=(14-4)=10$ litre

$w=-P\times \Delta V=-1\times 10=-10litre-atm$
But

$0.082$ litre-atm

$=1.987cal$
So,

$w=\cfrac { 10\times 1.987 }{ 0.082 } =-242.3cal$
But

$1$ calorie

$=4.184J$
So,

$w=-242.3\times 4.184=-1013.7J$
Substituting the values in the equation

$\Delta U=q+w=(800-1013.7)=-213.7J$

$5$ mole of oxygen are heated at constant volume from ${10}^{o}C$ to ${20}^{o}C$ . What will be the change in the internal energy of gas? The molar heat of oxygen at constant pressure, ${C}_{P}=7.03cal$ ${ mol }^{ -1 }{ deg }^{ -1 }$ and $R=8.31J\quad { mol }^{ -1 }{ deg }^{ -1 }$ .

$R=8.31J\quad { mol }^{ -1 }{ deg }^{ -1 }=\cfrac{8.31}{4.18}cal$

${ mol }^{ -1 }{ deg }^{ -1 }$

$=1.99cal$

${ mol }^{ -1 }{ deg }^{ -1 }$
We know that

${C}_{P}-{C}_{V}=R$
or

${C}_{V}={C}_{P}-R=7.03-1.99=5.04cal$

${ mol }^{ -1 }{ deg }^{ -1 }$
Heat absorbed by

$5$ mole of oxygen in heating from

${10}^{o}C$ to

${20}^{o}C$

$=5\times {C}_{V}\times \Delta {T}=5\times 5.04\times 10=252cal$
Since, the gas is heated at constant volume, no external work is done,
i.e.,

$w=0$
So, change in internal energy will be equal to heat absorbed,

$\Delta U=q+w=252+0=252cal$

Determine the heat of transformation of
${C}_{(diamond)}\rightarrow {C}_{(graphite)}$ from the following data
${C}_{(diamond)}+{O}_{2}(g)\rightarrow C{CO}_{2}(g);(\Delta H=-94.5kcal)$
${C}_{(graphite)}+{O}_{2}(g)\rightarrow C{CO}_{2}(g);(\Delta H=-94.0kcal)$

${C}_{(diamond)}+{O}_{2}(g)\rightarrow C{CO}_{2}(g);\Delta H=-94.5kcal)....(i)$

${C}_{(graphite)}+{O}_{2}(g)\rightarrow C{CO}_{2}(g);\Delta H=-94.0kcal).....(ii)$
Subtracting eq(ii) from (i) the required equation is obtained

$\Delta {H}_{transformation}=-94.5-(-94.0)$

$=-0.5kcal$

The equilibrium constant at ${25}^{o}C$ for the process:
${ Co }^{ 3+ }(aq.)+6{ NH }_{ 3 }(aq.)\rightleftharpoons { \left[ Co{ \left( { NH }_{ 3 } \right) }_{ 6 } \right] }^{ 3+ }(aq.)\quad$ is $\quad 2\times { 10 }^{ 7 }$
Calculate the value of $\Delta { G }^{ o }$ at ${25}^{o}C$ . $\left( R=8.314J{ K }^{ -1 }{ mol }^{ -1 } \right)$

We know that

$\Delta { G }^{ o }=-2.303RT\log { { K }_{ c } }$
Given,

${ K }_{ c }=2\times { 10 }^{ 7 },T=298K,R=8.314J{ K }^{ -1 }{ mol }^{ -1 }$
Thus, from the above equation

$\Delta { G }^{ o }=-2.303\times 8.314\times 298\log { (2\times { 10 }^{ 7 }) } =-12023.4J{ mol }^{ -1 }=-12.023kJ\quad { mol }^{ -1 }$
Since,

$\Delta { G }^{ o }$ is negative hence reaction is spontaneous forward direction.

For the equilibirium,
$P{ Cl }_{ 5 }(g)\rightleftharpoons P{ Cl }_{ 3 }(g)+{ Cl }_{ 2 }(g)$
${ K }_{ c }=1.8\times { 10 }^{ -7 }$
Calculate $\Delta { G }^{ o }$ for the reaction $\left( R=8.314J{ K }^{ -1 }{ mol }^{ -1 } \right)$

We know that

$\Delta { G }^{ o }=-2.303RT\log { { K }_{ c } }$

$-2.303\times 8.314\times 298\log { (1.8\times { 10 }^{ -7 }) } =38484J{ mol }^{ -1 }=38.484kJ\quad { mol }^{ -1 }$

For the reaction
$2NO(g)+{ O }_{ 2 }(g)\longrightarrow 2N{ O }_{ 2 }(g)$
Calculate $\Delta G$ at $700\ K$ when enthalpy and entropy changes are $-113\ kJ{ mol }^{ -1 }$ and $-145\ J{ K }^{ -1 }{ mol }^{ -1 }\quad$ respectively.

We know that,

$\Delta G=\Delta H-T\Delta S$
Given that,

$\Delta H=-113kJ{ mol }^{ -1 }=-113000J{ mol }^{ -1 }$

$\Delta S=-145J{ K }^{ -1 }{ mol }^{ -1 }\quad$

$T=700K$
Substituting these values in the above equation

$G=-113000-700\times (-145)=-11500J{ mol }^{ -1 }=-11.5kJ{ mol }^{ -1 }\quad$

Zinc reacts with dilute hydochloric acid to give hydrogen at ${17}^{o}C$ . The enthalpy of the reaction is $-12.55kJ{mol}^{-1}$ of zinc and entropy change equals $5.0J{ K }^{ -1 }{ mol }^{ -1 }$ for the reaction. Calculate the free energy change and predict whether the reaction is spontaneous or not.

Given

$\Delta H=-12.55kJ{ mol }^{ -1 }$

$\Delta S=5.0J{ K }^{ -1 }{ mol }^{ -1 }=0.005kJ{ K }^{ -1 }{ mol }^{ -1 }$

$T=17+273=290K$
Applying

$\Delta G=\Delta H-T\Delta S$

$=-12.55-0.005\times 290$

$=-14.00kJ{ mol }^{ -1 }$
since,

$\Delta G$ is negative, the reaction will be spontaneous.

In the reaction
${ A }^{ + }+B\longrightarrow A+{ B }^{ + }$ ,
there is no entropy change. If enthalpy change is $22kJ$ of ${A}^{+}$ , calculate $\Delta G$ for the reaction.

For the given reaction

$\Delta H=22kJ,\Delta S=0$

$\Delta G=\Delta H-T\Delta S$

$\Delta G=22-T\times 0=22kJ{ mol }^{ -1 }$

$\Delta H$ and $\Delta S$ for the system
${ H }_{ 2 }O(l)\rightleftharpoons { H }_{ 2 }O(g)$
at $1$ atmospheric pressure are $40.63\ kJ{ mol }^{ -1 }$ and $108.8\ J{ K }^{ -1 }{ mol }^{ -1 }$ respectively. Calculate the temperature at which the rates of forward and backward reactions will be the same. Predict the sign of free energy for this transformation above this temperature.

Given,

$\Delta H=40.63kJ{ mol }^{ -1 }$

$\Delta S=108.8J{ K }^{ -1 }{ mol }^{ -1 }=0.1088kJ{ K }^{ -1 }{ mol }^{ -1 }$

$\Delta G=0$ (when the system is in equilibrium)
Applying

$\Delta G=\Delta H-T\Delta S$
The sign of

$\Delta G$ above

$373K$ , say

$374K$ may be calculated as follows
Again applying

$\Delta G=\Delta H-T\Delta S$

$40.63-374\times 0.1088$

$=-0.06kJ$

$\Delta G$ will be negative; hence the reaction will be spontaneous.

For the reaction
$SO{ Cl }_{ 2 }+{ H }_{ 2 }O\longrightarrow S{ O }_{ 2 }+2HCl$
the enthalpy of reaction is $49.4kJ$ and the entropy of reaction is $336J{K}^{-1}$ . Calculate $\Delta G$ at $300K$ and predict the nature of the reaction.

$\Delta G=\Delta H-T\Delta S$

$=49.4-(300\times 336\times { 10 }^{ -3 })=-51.4kJ$
Since, the free energy change is negative, the given reaction is spontaneous.

$\Delta H$ and $\Delta S$ for the reaction,
${ Ag }_{ 2 }O(s)\longrightarrow 2Ag(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)$
are $30.56kJ{ mol }^{ -1 }$ and $66.0J{ K }^{ -1 }{ mol }^{ -1 }$ respectively. Calculate the temperature at which free energy change for the reaction will be zero. Predict whether the forward reaction will be forward above or below this temperature.

We know that

$\Delta G=\Delta H-T\Delta S$
At equilibrium

$\Delta G=0$
so that

$0=\Delta H-T\Delta S$
or

$T=\cfrac { \Delta H }{ \Delta S }$
Given that,

$\Delta H=30.56kJ{ mol }^{ -1 }=30560J{ mol }^{ -1 }\quad$

$\Delta S=66.0J{ K }^{ -1 }{ mol }^{ -1 }$

$T=\cfrac { 30560 }{ 66 } =463K\quad$
Above this temperature,

$\Delta G$ will be negative and the process will be spontaneous in forward direction.

Calculate the temperature at which liquid water will be in equilibrium with water vapour.
$\Delta { H }_{ vap }=40.73\ kJ{ mol }^{ -1 }$ and $\Delta { S }_{ vap }=0.109\ kJ{ mol }^{ -1 }{ K }^{ -1 }$

Given

$\quad \Delta { H }_{ vap }=40.73kJ{ mol }^{ -1 };\Delta { S }_{ vap }=0.109kJ{ mol }^{ -1 }{ K }^{ -1 }$
and

$\Delta G=0$
Applying

$\Delta G=\Delta H-T\Delta S$

$0=40.73-T\times 0.109$

$T=\cfrac { 40.73 }{ 0.109 } =373.6K$

For the reaction, ${ Ag }_{ 2 }O(s)\rightleftharpoons 2Ag(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)$
$\Delta H,\Delta S,\Delta T$ are $40.63\ kJ\ { mol }^{ -1 },1.808\ J{ K }^{ -1 }{ mol }^{ -1 }$ and $373.4\ K$ respectively. Free energy change $\Delta G$ of the reaction will be:

$AgO(s) \rightleftharpoons 2Ag (s) + O_2(g)$

$\triangle G = \triangle H - T\triangle S$

$\triangle G = 40.63 * 10^3 - 373.4 * 1.808$

$\triangle G= 675.103 J$

$\triangle _rG^o= 0$ , the reaction is at equilibrium

Therefore,

$KC=1$ , therefore, Concentration of reactants is equal to concentration of products.

For the reaction,
$4C(graphite)+5{ H }_{ 2 }\longrightarrow n{ C }_{ 4 }{ H }_{ 10 };\Delta { H }^{ o }=-124.73kJ\quad { mol }^{ -1 }$
$4C(graphite)+5{ H }_{ 2 }\longrightarrow iso-{ C }_{ 4 }{ H }_{ 10 };\Delta { S }^{ o }=-365.8J{ K }^{ -1 }{ mol }^{ -1 }\quad$
$\Delta { H }^{ o }=-131.6kJ\quad { mol }^{ -1 }$
$\Delta { S }^{ o }=-381.079J{ K }^{ -1 }{ mol }^{ -1 }$
Indicate whether normal butane can be spontaneously converted to iso-butane or not.

Yes,

$\Delta { G }^{ o }=-2.32kJ$
For

$n{ C }_{ 4 }{ H }_{ 10 }$

$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }\quad$

$=-15.72kJ$
For

$iso-{ C }_{ 4 }{ H }_{ 10 }$

$\Delta { G }^{ o }=-18.04kJ$
For conversion

$\Delta { G }^{ o }=-18.04-(-15.72)=-2.32kJ$

All the energy released from the reaction $X\rightarrow Y, \triangle_{r}G^{\circ} = -193\ kJ\ mol^{-1}$ is used for oxidising $M^{+}$ as $M^{+}\rightarrow M^{3+} + 2e^{-}, E^{\circ} = -0.25\ V$ . Under standard conditions, the number of moles of $M^{+}$ oxidised when one mole of $X$ is converted to $Y$ is: $(F = 96500\ C\ mole^{-1})$

Given

$X \rightarrow Y, \triangle_{r}G^{\circ} = -193\ kJ/ mol$

$M^{+}\rightarrow M^{3+} + 2e^{-}$

$\triangle G^{\circ}$ for above reaction is:

$\triangle G^{\circ} = -nFE^{\circ}$

$= -2\times 96500\times (-0.25)$

$= 48250\ J/mol$

$= 48.25\ kJ/mol$

$\therefore$ The number of moles of

$M^{+}$ oxidised from

$(X\rightarrow Y)$ .

$= \dfrac {193}{48.25} = 4$ .

$Zn(s) + 2AgCl(s) \rightleftharpoons ZnCl_{2} (0.555\ M) + 2Ag(s)$
$E_0 = 1.015\ volt \left (\dfrac {dE}{dT}\right )_{P} = -4.02\times 10^{-4} volt$ per degree.
Find $\triangle G, \triangle S$ .

We know that,

$\triangle H = nF \left [T \left (\dfrac {dE}{dT}\right )_{P} - E\right ]$

$= 2\times 96500 [ -273 \times 4.02\times 10^{-4} - 1.015]$

$= - 217075.98\ joule = -217.075\ kJ$

$\triangle G = -nFE = -2\times 96500 \times 1.015 = -195895\ J$

$= -195.895\ kJ$

$\triangle G = \triangle H - T\triangle S$

$-195.895 = -217.075 - 273\times \triangle S$

$\triangle S = -0.07758\ kJ = -77.58\ J = -18.55\ cal$ .

Commercial sample of $H_2O_2$ is labeled as $10\ Volume$ . Its % strength is nearly:

1410525_647517_ans_53c1351d0b3147e09cd3318bca9d873b.png

The following data is known about the melting of $KCl$ :
$\Delta H=7.25kJ\quad { mol }^{ -1 };\Delta S=+0.007KJ{ K }^{ -1 }{ mol }^{ -1 }$
Calculate its melting point.

we have,

$\Delta H = T\Delta S$

$T = \dfrac{\Delta H}{\Delta S}$

$T = \dfrac{7250}{7} K$

$T = 1035 K$

Show that the reaction,
$CO(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g)$
at $300\ K$ is spontaneous and exothermic, when the standard entropy change is $-0.094\ kJ\ { mol }^{ -1 } {K}^{-1}$ . The standard Gibbs free energies of formation for ${CO}_{2}$ and $CO$ are $-394.4$ and $-137\ kJ\ { mol }^{ -1 }$ respectively.

The given reaction is

$CO(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow C{ O }_{ 2 }(g)$

$\Delta { G }^{ o }(for\quad reaction)={ G }_{ C{ O }_{ 2 } }^{ o }-{ G }_{ CO }^{ o }-{ G }_{ { O }_{ 2 } }^{ o }=-394.4-(-137.2)-0=-257kJ\quad { mol }^{ -1 }\quad$

$\Delta { G }^{ o }=\Delta { H }^{ o }-T{ \Delta S }^{ o }\quad$

$-257.2=\Delta { H }^{ o }-298\times (-0.094)\quad$

$\Delta { H }^{ o }=-285.2kJ$

For the reaction,
$\quad { Ag }_{ 2 }O\rightleftharpoons 2Ag(s)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)$
Calculate the temperature at which free energy change is zero. At a temperature lower than this, predict whether the forward of the reverse reaction will be forward. Give reason
( $\Delta H=+30.56kJ;\Delta S==+0.066kJ{ K }^{ -1 }{ mol }^{ -1 }$ )

We have,

$\Delta G = \Delta H - T \Delta S$

if free energy change is zero then we gave

$0 = \Delta H - T \Delta S$

$0 = 30560 - T* 66$

$30560 = T* 66$

we get T = 463 K

at temperature lower than that of 463K, free energy change become positive and reaction goes in backward direction.

Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows etc.

During day the temperature increases

During night, the temperature decreases.

But

$PV$ =constant.

$PV=nRT=nT=\dfrac{PV}{R}$

Internal energy,

$U=nRT$ =constant

$\times R$ =constant.

$\therefore$ In the room, internal energy remains constant.

Calculate the standard internal energy change for the reaction
${C}_{(graphite)} + {H}_{2}O(g) \rightarrow CO(g) + {H}_{2}O(g)$
${\Delta} ^ {o}_{f}$ at $25^o$ ; ${H}_{2}O(g) = -241.8 kJ {mol}^{-1}$
$CO(g) = -110.5 kJ {mol}^{-1}$
$R = 8.314 J{K}^{-1}{mol}^{-1}$

$\begin{array}{l} \Delta H=\Delta U+\Delta ngRT & \Delta H=\Delta { H_{ f } }CO+\Delta { H_{ f } }{ H_{ 2 } }O \\ \Delta U=\Delta H-\Delta ngRT & \Rightarrow -241.8-110.5 \\ \Rightarrow -352.3-\left( { 2-1 } \right) 8.314\times 298 & \Rightarrow -352.3\, \, KJ \\ \Rightarrow -352.3\, \, KJ-\left( { 8.314\times 298 } \right) J & \\ \Rightarrow -352.3\, \, KJ-2.477\, \, KJ\Rightarrow -354.78\, \, KJ=\Delta U & \end{array}$

The Born-Haber cycle for formation of rubidium chloride ( $RbCl$ )is given below (the enthalpies are in $kCal$ ${mol}^{-1}$ ).
Find the value of $X$ .

Heat of formotion = Heat of sublimation + Dessolution energy +Ionzation + energy + (electron affnutr+)

Substituting values

$\Delta H_f =\Delta H_S +\Delta H_D +\Delta IE +\Delta E_a +\Delta U$

$-105=+20.5+28.75+96.0+X-159.5$

$\boxed {X=-90.75} \ Kcel /mol$

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Determine whether the following reaction will be spontaneous or nonspontaneous under standard conditions.
$Zn(s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu (s) ,\ \ \ \Delta H^0 = -219 kJ, \\ \ \Delta S^0 = -21 JK^{-1}$ .

$\triangle G$ =

$\triangle H$ - T

$\triangle S$

Where

$\triangle G$ is change in Gibbs free energy

$\triangle H$ is change in enthalpy

$\triangle S$ is entropy change .

Any reaction is spontaneous if

$\triangle G$ is negative for that reaction.

T is 273K (standard conditions)

$\triangle G$ = -219000 - 273 ×(- 21)

$\triangle G$ = -213267 J

Since

$\triangle G$ is negative hence reaction is spontaneous .

Calculate the heat change during the reaction $24$ g C and $128$ g S following :the change $C+S_2\rightarrow CS_2$ ; $\Delta$ H $=22$ K cal.

$C + {S}_{2} \longrightarrow C{S}_{2} \quad \Delta{H} = 22 \; kcal$

$1$ mole of

$C = 12 \; g$

$1$ mole of

${S}_{2} = 64 \; g$

Therefore,

$24 \; g$ of

$C = 2 \text{ moles}$

$128 \; g$ of

${S}_{2} = 2 \text{ moles}$

Given that heat change during the reaction of

$12 \; g$ of

$C$ and

$64 \; g$ of

$S$ is

$22 \; kcal$

Therefore

$2 \times \left[ C + {S}_{2} \longrightarrow C{S}_{2} \quad \Delta{H} = 22 \; kcal \right]$

$2C + 2{S}_{2} \longrightarrow 2 C{S}_{2} \quad \Delta{H} = 44 \; kcal$

Hence the heat change during the reaction

$24 \; g$ of

$C$ and

$128 \; g$ of

$S$ is

$44 \; kcal$ .

For a chemical reaction, $\Delta H$ and $\Delta S$ are negative. State, giving reason, under what conditions this reaction is expected to occur spontaneously.

Let us consider the

$\Delta G = \Delta H - T\Delta S$

In order to any reaction to be spontaneous

$\Delta G$ must have Positive value.

Therefore,

$\Delta H$ and

$\Delta S$ are negative then,

$T\Delta S < \Delta H$ .

For the reaction, $2A(g)+B(g)\longrightarrow 2D(g),\Delta { U }=-10.5KJ$ and $\Delta { S }=-44.10J{ K }^{ -1 }$ . Calculate $\Delta { G }$ for the reaction and reaction and predict whether the reaction may occur spontaneously.

$Let\quad T=298\\ \Delta H=\Delta U+AngRT\\ =-10500-1\times 8.314\times 298\\ =-10500-2477.572\\ \Delta H=-12977.572\\ \Delta G=\Delta H-T\Delta S\\ =-12977.572(-298\times -44.1)\\ =-12977.572+13141.8\\ \Delta G=-164.228$

For a reaction both enthalpy change and entropy change are positive. Under what conditions the reaction will be spontaneous?

Let us consider the equation for gibbs free energy

$\Delta G=\Delta H-T\Delta S$

Where,

$\delta H$ is enthalpy change

$(KJ\;mol^{-1})$

$T$ is absolute temperature

$(K)$

$\Delta S$ is entropy change

$(JK^{-1})$

For a reaction to be considered spontaneous

$\Delta G<0$

Therefore,

$\Delta H-T\Delta S<0$

$\Delta H < T \Delta S$

$T>\Delta H \Delta S$

For the reaction:

$2A(g)+B(g)\rightarrow 2D(g)$ ,

$\Delta U^{\ominus}=-10.5$ kJ and $\Delta S^{\ominus}=-44.1$ $JK^{-1}$ .Does the reaction occur spontaneously?
[Give your answer in terms of Yes or No]

Given:-

$\Delta{{U}^{0}} = -10.5 \; kJ$

$\Delta{{S}^{0}} = -44.1 \; {J}/{K}$

$2 {A}_{\left( g \right)} + {B}_{\left( g \right)} \longrightarrow 2 {D}_{\left( g \right)}$

For the above reaction-

$\Delta{{n}_{g}} = 2 - \left( 2 + 1 \right) = -1$

As we know that,

$\Delta{{H}^{0}} = \Delta{{U}^{0}} + \Delta{{n}_{g}} RT$

$\Delta{{H}^{0}} = -10.5 + \left( -1 \times 8.314 \times {10}^{-3} \right) \left( 298 \right)$

$\Rightarrow \Delta{{H}^{0}} = - 12.98 \; kJ$

Again as we know that,

$\Delta{{G}^{0}} = \Delta{{H}^{0}} - T \Delta{{S}^{0}}$

$\Rightarrow \Delta{{G}^{0}} = -12.98 - 298 \times \left( -44.1 \times 10^{-3}\right) = + 0.16 \; kJ$

Since

$\Delta{{G}^{0}}$ for the reaction is positive, the reaction will not occur spontaneously.

Hence the correct answer is No.

How do you define specific heat ? What is specific heat of water ?

It is defined as the amount of heat required to raise the temperature of 1 g of a substance through 1

$^{o}$ C.

Specific heat of water is

$4200 \ Jkg^{-1}K^{-1}$ in SI units and

$1 \ cal g^{-1 \ o}C^{-1}$ in CGS system.

Explain Third law of thermodynamics.

Third law of thermodynamics states that at absolute zero temperature, the entropy of system approaches a constant value. This value depends on some parameters like pressure and magnetic field. The value of minimum possible energy of the system at absolute zero is zero. Temperature is defined as

$T={{\left( \dfrac{\partial U}{ds} \right)}_{V,N}}$

According to third law $\underset{T\to 0}{\mathop{\lim }}\,\Delta S=0$

For oxidation of iron,
$4Fe\left( s \right) +{ 3O }_{ 2 }\left( g \right) \rightarrow 2{ Fe }_{ 2 }{ O }_{ 3 }\left( s \right)$
entropy change is - 549.4 ${ JK }^{ -1 }{ mol }^{ -1 }$ at 298 K. In spite of negative entropy change of this reaction, why is the reaction spontaneous?
( $\triangle ,{ H }^{ \circleddash }$ for this reaction is $-1648\times { 10 }^{ 3 }J\quad { mol }^{ -1 }$ )

For a spontaneous reaction, total entropy must be greater than zero i.e. positive

$4Fe(s)+3O_2(g)\longrightarrow 2Fe_2O_3(s)$

$\triangle H_f=-1648$

$KJ mol^{-1}$

reaction is exothermic and surroundings absorbed

$1648KJ$ of heat

$\triangle S_{surr}=\cfrac {\triangle H_f}{T}=\cfrac {1648}{298}=5.53$

$KJ mol^{-1}k^{-1}$

$=5530$

$J mol^{-1} k^{-1}$

$\triangle S_{total}=\triangle S_{system}+\triangle S_{surrounding}$

$=-549.4+5530$

$=4980.6$

$Jk^{-1}mol^{-1}$

Many of the photochemical changes have positive sign of $\delta G$ , yet they are spontaneous. Why?

In photochemical changes,

$h\nu$ is included and

$h\nu$ is not thermal energy, so thermodynamic conditions do not apply. We cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics. The main reason of excitation is not a result of thermal energy transfer, but rather the absorption of photon which changes the quantum state.

What is specific heat?

The amount of heat per unit mass required to raise the temperature by 1 degree Celsius is called the specific heat.

The relationship between heat and temperature is usually expressed as:

$Q=cm\Delta T$

where

$Q$ = heat in joules

$c$ = specific heat capacity

$m$ = mass in grams

$\Delta T$ = change in temperature in

$^{\circ}C$

If $1.0 kcal$ of heat is added to $1.2 L$ of ${ O }_{ 2 }$ in a cylinder at constant pressure of $1$ atm, the volume increase to $1.5 L$ . Calculate $\Delta U$ and $\Delta H$ of the process $(1 L-atm=100 J)$

As we know that,

$\Delta{U} = q - P \Delta{V}$

Here,

$q = 1 \text{ kcal} = 1000 \text{ cal.}$

$P = \text{Pressure} = 1 \text{ atm}$

$\Delta{V} =$ Change in volume

$= \left( 1.5 - 1.2 \right) L = 0.3 L$

$\therefore \; \Delta{U} = 1000 - 1 \times 0.3 = 992.7 \text{cal.} = 0.993 \text{ kcal}$

At constant pressure,

$\Delta{H} = {q}_{P} = 1 \text{ kcal}$

Hence,

$\Delta{U} = 0.993 \text{ kcal} \; \ and \; \Delta{H} = 1 \text{ kcal}$ .

Identify the correct statement regarding a spontaneous process
1) For a spontaneous process in an isolated system, the change in entropy is positive
2) Endothermic process are never spontaneous
3) Exothermic processes are always spontaneous (4) Lowering of energy in the reaction process is the only criterion for spontaneity

Answer: 3) statement.

Spontaneous process are the process there is entropy of a system increases and decrease in enthalpy .In case of exothermic process, energy of the system decreases.

State and explain Hess's law of constant heat summation? Give its application.

It states that, “The amount of heat evolved or absorbed in a chemical change is the same whether the process takes place in one step or in several steps”.

(i) Determination of transition :It helps in the determination of enthalpy of transition during allotropic modification.

(ii) Determination of enthalpy of formation :It helps in the determination of enthalpy of formation which cannot be determined experimentally e.g. it is not possible to calculate enthalpy of formation of $CO$ experimentally, but can be calculated by Hess’s law.

Derive relationship between H and U.

Let

${H}_{i}$ be the enthalpy of a system in the initial state and

${H}_{f}$ be the enthalpy of a system in the final state. Let

${U}_{i}$ and

${V}_{i}$ be the internal energy and volume in the initial state and

${U}_{f}$ and

${V}_{f}$ be the internal energy and volume in a final state.

Now, as we know that,

$H = U + PV$

Therefore,

${H}_{i} = {U}_{i} + P {V}_{i} ..... \left( 1 \right)$

${H}_{f} = {U}_{f} + P {V}_{f} ..... \left( 2 \right)$

Subtracting equation

$\left( 1 \right)$ from

$\left( 2 \right)$ , we have

${H}_{f} - {H}_{i} = \left( {U}_{f} + P {V}_{f} \right) - \left( {U}_{i} + P{V}_{f} \right)$

$\Rightarrow {H}_{f} - {H}_{i} = \left( {U}_{f} - {U}_{i} \right) + P \left( {V}_{f} - {V}_{i} \right)$

$\Rightarrow \Delta{H} = \Delta{U} + P \Delta{V} ..... \left( 3 \right)$

Here,

$\Delta{U}$ and

$P \Delta{V}$ are the change in internal energy and work energy respectively.

Hence equation

$\left( 3 \right)$ is the relationship between

$H$ and

$U$ .

Write the importance of Hess's law of constant heat summation.

Hess's law of heat summation is important because it demonstrates, chemically, the conservation of energy, and the laws of thermodynamics.

A gas cylinder contains 14.2 Kg butane gas. it is consumption of energy in a family is 10000 KJ for cooking purposes how long this cylinder will be sufficient to supply. If the enthalpy of combustion of butane is 2658 KJ mol

Solution:-

Enthalpy of combustion of butane

$= 2658 \; {kJ}/{mol} \quad \left( \text{Given} \right)$

Molecular weight of butane

$= 58 \; {g}/{mol}$

Amount of butane gas contained in a cylinder

$= 14.2 \; kg = 14200 \; g$

Therefore,

Heat released during combustion of

$58 \; g$ of butane

$= 2658 \; kJ$

Heat released during combustion of

$14200\; g$ of butane

$= \cfrac{2658}{58} \times 14200 \; kJ$

Now,

$10000 \; kJ$ of energy is consumed in one day.

Therefore,

No. of days require to consume

$\left( \cfrac{2658}{58} \times 14200 \right) kJ$ of energy

$= \cfrac{1}{10000} \times \cfrac{2658}{58} \times 14200 \approx 65 \text{days}$

Hence the cylinder will be sufficient for

$65$ days.

How it is possible to find the value of $\Delta U$ and not the value of $U$

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Calculate $\text E_{cell}$ and $\Delta G$ for the following $28 ^ { \circ } \mathrm { C } :\mathrm { Mg } _ { ( s ) } + \mathrm { Sn } ^ { 2 + } ( 0.04 \mathrm { M } ) \rightarrow \mathrm { Mg } ^ { 2 + } ( 0.06 \mathrm { M } ) + \mathrm { Sn } _ { ( s ) }$
$E _ { \text { cell } } ^ { \circ } = 2.23 \mathrm { V }$
Is the reaction spontaneous?

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The reaction of nitrogen with hydrogen to make ammonia has $\Delta H=-92 kJ$ $N_2(g)+3H_2(g)\rightarrow2NH_3(g)$ . What is the value of $\Delta U$ (in kJ) if the reaction is carried out at a constant pressure of $40\ bar$ and the volume change is $-1.25\ litre$ ?

$\Delta{H}=\Delta{U}+P\Delta{V}$

$\Delta{U}=\Delta{H}-P\Delta{V}$

$\Delta{U}=-92\times{10^3}-(40\times{1.013}\times{-1.25})$ [Since,

$1atm = 1.013bar, \ P=40\times{1.013}atm$ ]

$\Delta{U}=-92,000+50.65=-91,949J=-91.949\times{10^3} J$

$\Delta{U}=-91.949 \ kJ$

$1g$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298\ K$ and $1$ atmospheric pressure according to the equation.
$C (graphite) + O_{2}(g) \rightarrow CO_{2}(g)$
During the reaction, temperature rises from $298 k$ to $299 k$ . If the heat capacity of the bomb calorimeter is $20.7 kJ/K$ . What is the enthalpy change for the above reaction at $298 K$ and $1 atm$ ?

Hint: Enthalpy of the system defines the heat content of the system and enthalpy change defines the gain or loss of energy in a system. It is expressed as $\Delta H$ .

Step 1: Calculation

The given reaction in closed vessel is-

$C(graphite)+O_2(g)\rightarrow CO_2(g)$

Change in temperature is-

$\Delta T=299-298=1\ K$

The heat capacity of the bomb calorimeter is $20.7\ KJ/K$ .

So, $C_v=20.7\ KJ/K$

Now, heat absorbed $=C_v\times{\Delta T}$

$=20.7\times1$ $=20.7\ KJ$

So, $20.7\ KJ$ of heat is involved in the combustion of $1\ gram$ of graphite.

$\therefore$ The heat involved in the combustion of $12\ grams$ of graphite is $=20.7\times12$

$=248.4\ KJ$

In this process heat is involved, so $\Delta H$ must be negative.

$\Delta H=-248.4\ KJ$ .

Final Step: The enthalpy change of the reaction is $-248.4\ KJ$ .

Calculate the $\Delta_{r}H^{o}$ at $298\ K$ for the reaction:
$C(graphite)+2H_{2}(g)\rightarrow CH_{4}(g)$
from the following date:
$C(graphite)+O_{2}(g)\rightarrow CO_{2}(g) \\ \Delta_{r}H^{o}=-393.5\ kJ\ mol^{-1} \\ H_{2}(g)+\dfrac {1}{2}O_{2}(g)\rightarrow H_{2}O(l)\\ \Delta_{r}H^{o}=-285.8\ kJ\ mol^{-1}$
$CO_{2}(g)+2H_{2}O(l)\rightarrow CH_{4}(g)+2O{2}(g)\\ \Delta_{r}H^{o}=+890.3\ kJ\ mol^{-1}$

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Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?

A spontaneous process should continue taking place by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working

A chemical reaction is spontaneous at 298K and non-spontaneous at 350K. Which of the following is correct for the reaction?
s.no
$\Delta G$
$\Delta H$
$\Delta S$
A)
-
-
+
B)
+
+
+
C)
-
+
-
D)
+
-
+
E)
-
-
-

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When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by $10^0$ C. What is the specific heat of the substance?

We know that

$Q=msdt$
Q is 600 caloriee
M is 3 kg
dt is 10 degree
600cal=3×10×s×kgcelsius
s=20. Cal/kg celsius

Calculate $\triangle G$ for the reaction at 298 K
$2SO_2(g)+O_2(g)\longrightarrow 2SO_3(g)$ , $\triangle { G }^{ o }=-142 kJ{ mol }^{ -1 }$
If partial pressures of $SO_2$ , $O_2$ and $SO_3$ are 50 atm 75 atm and 2 atm respectively at $298 K$

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Explain why:
Ice at zero degrees centigrade has a greater cooling effect than water at $0 ^\circ C$ .

Ice at zero degree centigrade has a greater cooling effect than water at

$0 ^\circ C$ because Ice it has low specific heat which means it stores less heat inside it. but water has a higher specific heat capacity so it stores more heat inside it and unable the cool very easily at

$0 ^\circ C$ so that is why ice is more cooling effect than water at

$0 ^\circ C$

What is the need to heat solution ?

During the process of crystallization the solution is heated after filtering it as heating evaporates the water that is used to dissolve the impure substance. It does converts the unsaturated solution into a saturated one.

At $298 \mathrm{K}$ . $\mathrm{K}_{\mathrm{p}}$
for the reaction $N_{2} O_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)$ is
0.98 . Predict whether the reaction is spontaneous or not.

$\Delta _r G^{\ominus}=\Delta _r H^{0}-T \Delta _r S^{\ominus}=-R T ln K$

For the
process to be spontaneous AG should be negative.

Since $\Delta, G^{\ominus}=-R T I n K_{p}$

In this
case, $\mathrm{K}_{\mathrm{p}}=0.98$

$\therefore \Delta _r G$ is - ve

Hence, the
reaction is spontaneous.

Given that $\Delta H=0$ for mixing of two gases. Explain whether the diffusion of these gases
into each other in a closed container is a spontaneous process or not?

The diffusion will be a
spontaneous process. The greater the disorder in an isolated system, the higher
is the entropy. Hence the process will be spontaneous.

Explain the meaning of thermochemical equation for the reaction:
$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)+136kcal$ .

When two moles i.e,

$4$ grams of gaseous

$H_{2}$ reacts with

$1$ mole

$(32g)$ of gaseous

$O_{2}$ to give

$2$ moles of liquid water, i.e,

$36$ grams of liquid water is formed and

$136\ kcal$ of heat is liberated.

What is spontaneous process? Give example.

Process which takes place itself, without any external aid under the given condition is called spontaneous process. Example: Flow of heat from higher to lower temperature.

Justify the following statements:
(a) An exothermic reaction is always thermodynamically spontaneous.
(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.

(a) It is false. Exothermic process are not always spontaneous. If

$\Delta S=-ve$ and

$T\Delta S>\Delta H$ , the process will be non-spontaneous even if it is exothermic.
(b) The entropy of vapour is more than that of liquid therefore entropy increases during vaporization.

Evaporation of water is an endothermic process but spontaneous. Explain.

It is because entropy increases during this process because water vapour have more entropy than liquid water.

$\Delta G$ becomes

$-ve$ because

$T\Delta S>\Delta H$ . Hence process becomes spontaneous.

What is meant by entropy driven reaction? How can a reaction with positive changes of enthalpy and entropy be made entropy driven?

Those reaction which are endothermic and entropy is increasing such that

$T\Delta S>\Delta H$ are entropy driven reactions. If

$\Delta S_{total}$ is

$+ve$ , the reaction will be spontaneous. If

$\Delta S$ is

$+ve$ and is small, but

$T$ is large, the reaction will be spontaneous.

A reaction which is endothermic, i.e,

$\Delta H$ is

$+ve$ and

$\Delta S$ is

$-ve$ ,

$T$ must be large,

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ will be

$-ve$ if

$T\Delta S>\Delta H$ , i.e, the reaction will be spontaneous at higher temperature because entropy will increase with increase in temperature which will make this reaction spontaneous.

Define non-spontaneous process.

Process which does not takes place itself or on its own but with the help of external aid under the given condition is called non-spontaneous process.

Example: Flow of heat energy from lower to higher temperatures.

What are the applications of Hess's law of constant heat summation?

1. It helps in calculating the enthalpies of formation of those compounds which cannot be determined experimentally.
2. It helps in determining the enthalpy of allotropic transformation like

$C(graphite) \rightarrow C(diamond)$
3. It helps in calculating the enthalpy of hydration.

What is meant by the free energy of a system? What will be the direction of chemical reaction when
(i) $\Delta G=0$
(ii) $\Delta G>0$
(iii) $\Delta G<0$

Free energy is defined as energy which can be converted into useful work.
(i)

$\Delta G=0$ , the reaction will be in equilibrium.
(ii)

$\Delta G>0$ , the reaction will not take place.
(iii)

$\Delta G<0$ , the reaction will be spontaneous.

Give Hess's Law of constant Heat summetion and write its uses.

This law is given by a russian chemist G.H. Hess , in

$1840$ the law is known after his name as Hess's law . it states thatl:
"If a reaction takes place in several steps then its standard reaction entahlpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divide at the same temperature".
In general if the enthalpy of an overall reaction

$A \rightarrow B$ along one route is

$\Delta H$ and

$H_1, H_2, H_3 .......$
Representing enthalpies of reactions leading to same product B along another route , if the reaction is taking place in three steps, then we have

$\Delta H = H_1 +H_2 + H_3$
It can be represented as :
example : Combustion of carbon dioxide can be done in two ways :

$Path I : C_{(s)} + O_{ 2(g)} CO_{ 2(g) } \Delta H = - 393.5 kJ$

$Path II : C_{(s)} + \frac {1}{2} O_{2(g)} \Delta H_1 = -110.5 kJ$

$CO_{(g)} + \frac {1}{2}O_{2(g)} \rightarrow CO_{2(g)} \Delta H_2 = -283.0 kJ$
In both the cases,

$\Delta H_1 +\Delta H_2 = ....393.5kJ$
Thus the heat of reaction in single step and that in two steps is same. this proves Hess's law .
uses of Hess's law of constant heat :
1. Determination of transition : it helps in the determination of entahlpy of transition during allotropic modification.
2. Determination of enthalpy of formation : it helps in the determination of enthalpy of formation which cannot be determined experimentally e.g it is not possible to calculate enthalpy of formation of CO experimentally but can be calculated by hess's law.
Bond energy : It may be defined as "the quantity of heat evolved when a bond is formed between two free atoms in a gaseous state to form a molecular product in a gaseous state".It is also known as enthalpy of formation of the bond . it mat also be defined as "the average quantity of heat required to break (dissociate) bonds of that type present in one mole of the compound."

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1738821_1829706_ans_0307cda066ae48a9a67ab40be4c6c76d.png

What will be the sign of $G$ for melting of ice at $267 K$ and $276 K$ ? (Melting point of ice= $272 K$ ).

$267 K$ , the process is non -spontaneous

$\Delta G = +ve$ while at

$276 K$ it become spontaneous

$\Delta G = -ve$

Explain the steps involved in Born Haber cycle for the formation of NaCl.

Step 1: Sublimation of metallic sodium of gaseous sodium atom with the enthalpy of sublimation =

$\triangle H_{s}$

$Na(s) \rightarrow Na(g); \triangle H = \triangle H_{s}$

Step 2 : Dissociation of molecule of chlorine to chlorine atoms with enthalpy of dissociation =

$\triangle H_{d}$

$\dfrac{1}{2} Cl_{2} (g) \rightarrow Cl(g) ; \triangle H = \triangle H_{d}$

Step 3: Ionization of gaseous sodium with the enthalpy of ionization =

$\triangle H_{i}$

$Na(g) \rightarrow Na^{+} (g)$

Step 4: Addition of electron to gaseous chlorine atom with the enthalpy of electron affinity =

$\triangle HEa$

$Cl_{(g)} + e^{-} \rightarrow Cl^{-} (g) , \triangle H = \triangle H_{Ea}$

Step 5: Close packing of gaseous sodium ion and chloride ion to form the lattice structure of NaCl, with lattice chloride ion to form a lattice structure of NaCl with Lattice energy = U.

$NaCl^{+}(g) + Cl^{-}(g) \rightarrow NaCl \triangle; \triangle H = U$

Step 6: But the sum of all the energies will be equal to the heat of formation of one mole of sodium chloride from its reluctant i.e,

$\triangle Hf$

$Na(s) + 1/2 Cl_{2}(g) \rightarrow NaCl(s) ; \triangle H = \triangle H_{f}$

Why most of the exothermic processes (reactions) are spontaneous?

$\Delta G=\Delta H-T\Delta S$ ;
For exothermic reactions,

$\Delta H$ is

$-ve$ . For a spontaneous process,

$\Delta G$ ishould be

$-ve$ . Thus, decrease in enthalpy

$(-\Delta H)$ contributes significantly to make

$\Delta G$ negative.

In a constant volume calorimeter, $3.5$ g of a gas with molecular weight $28$ was burnt in excess oxygen at $298.0$ K. The temperature of the calorimeter was found to increase from $298.0$ K to $298.45$ K due to the combustion process. Given that the heat capacity of the calorimeter is $2.5$ kJ K $^{-1}$ , the numerical value for the enthalpy of combustion of the gas in kJmol $^{-1}$ is:

$C_v = 2.5 kJ/K = 2500 J/K$

$\Delta {T} = T_2 - T_1 = 298.45 - 298 = 0.45 K$

Energy release at constant volume due to combustion of 3.5 gm of a gas

$= C\times \Delta T=2.5 \times 0.45\ kJ$

Hence, energy released due to the combustion of 28 gm (i.e., 1 mole) of a gas

$=2.5\displaystyle \times 0.45\times\frac{28}{3.5}=9\ \mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}1^{-1}$

$200$ ml of $1$ M $HCl$ is mixed with $400$ ml of $0.5$ M $NaOH$ . The temperature rise in the calorimeter was found to be $4.0^{o}C.$ Water equivalent of calorimeter is $25$ g and the specific heat of the solution is $1$ cal/mL/degree. If the theoritical heat of neutralization of a strong acid and strong base is $-13.5$ kcal, then the percentage error (as nearest integer) incurred in this experiment while calculating the heat of neutralization is

Total volume

$200+400=600 ml$
200 ml of 1 M acid

$=$ 400 ml of 0.5 M

$NaOH$
200 m. eq. of acid reacts with 200 m.eq. of base

$=\Delta H$

$\therefore$ Heat of neutralization

$=5\times \Delta H$
Heat produced during neutralization

$=$ heat taken up by calorimeter + solution

$=m_{1}s_{1}\Delta T+m_{2}s_{2}\Delta T$

$=(25\times 1 \times 4)+(600\times 1 \times 4)=2500 cal$

$\therefore$ Heat of neutralization

$=-5\times 2500$

$=-12500 cal=-12.5 kcal$
% error

$=(\dfrac{13.5-12.5}{13.5})\times 100=7.4$

It is observed that the functioning of neuronal circuits in our brain is driven by the energy available from the combustion of glucose. Calculate the amount of glucose (in gm) which should be burnt per hour to produce sufficient energy for the brain, which operates at $\dfrac{128}{9}$ watts.

Given : $|\Delta H^{0}|$ of combustion of glucose at 400 K = $3000$ kJ/mol
$|\Delta S^{0}|$ of combustion of glucose at 400 K= $180$ J/mol K

$C_{6}H_{12}O_{6}(s)+6O_{2}(g)\rightarrow 6CO_{2}(g)+6H_{2}O(g)$

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$

Useful energy

$=-3000-400 \times 0.18$

$=-3072 \ kJ/mol$

Energy required per hour

$= \dfrac{128}{9}\times 60\times 60 J$

$\therefore$ Glucose needed per hour

$=$

$\dfrac{\dfrac{128}{9}\times 60\times 60}{3072\times 1000}mol=\dfrac{\dfrac{128}{9}\times 60\times 60}{3072\times 1000}\times 180 g = 3\ g$

From the following data at $\displaystyle 25^{\circ}C$ , The standard enthalpy of formation of FeO(s) is X, then 0.3-X is:

Reaction $\displaystyle \Delta _{r}H^{\circ}$ ( kJ / mole )
$\displaystyle Fe_{2}O_{3}(s)+3C (graphite)\rightarrow 2Fe(s)+3CO(g)$ 492.6
$\displaystyle FeO(s)+C (graphite)\rightarrow Fe(s)+CO(g)$ 155.8
$\displaystyle C (graphite)+O_{2}(g)\rightarrow CO_{2}(g)$ $\displaystyle -$ 393.51
$\displaystyle CO(g)+1/2\: O_{2}(g)\rightarrow CO_{2}(g)$ $\displaystyle -$ 282.98

$FeO\left( s \right) +C\left( graphite \right) \longrightarrow Fe\left( s \right) +CO\left( g \right) ,\quad 155.8$

$CO\left( g \right) +\cfrac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) ,\quad -282.98$
+

$FeO\left( s \right) +C\left( graphite \right) +\cfrac { 1 }{ 2 } { O }_{ 2 }\left( g \right) \longrightarrow Fe\left( s \right) +C{ O }_{ 2 }\left( g \right) ,\quad \triangle H=-127.18$

$C+{ O }_{ 2 }\longrightarrow C{ O }_{ 2 },\quad -393.51$
-

$FeO-\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow Fe,\quad 266.33$

$\Rightarrow Fe+\cfrac { 1 }{ 2 } { O }_{ 2 }\longrightarrow FeO,\quad \triangle H=-266.33$

$0.3-\left( 266.33 \right) =266.63\approx 267$

For the ionic solid $MX_{2}$ ,where X is monovalent ,the enthalpsy of formation of the solid from M(s) and $X_{2}$ (g) is 1.5 times the electron gain enthalpsy of X(g).the first and second ionisation enthalpsy of the metal(M) are 1.2 and 2.8 times of the enthalpsy of sublimation of M(s).the bond dissociation enthalpy of $X_{2}$ (g) is 0.8 times the first ionisation enthaply of metal and it is also equal to one-fifth of the magnitude of lattice enthalpy of $MX_{2}$ . If the electron gain enthalpy of X(g) is $96$ Kcal/mol, find the enthalpy of sublimation of metal(M) in Kcal/mol.
(Round off answer to integer).

The Born Haber cycle is shown below.

$\Delta H_f =S +D+IE_1 + IE_2 + 2EA +U$

Substitute values in the above expression.

$1.5 \times (-96 ) = S+0.8 IE_1+IE_1+IE_2+2(-96) +5IE_1$

$1.5 \times (-96) = S + 6.8 (1.2S) + 2.85 +2(-96)$

Hence,

$S=41.38 \ kcal/mol$ .

If X is the lattice entahlpy of $Ca{Cl}_{2}$ ,find $X$ . Given that the enthalpy of
(i) sublimation of $Ca$ is $121kJ{ mol }^{ -1 }$
(ii) dissociation of ${Cl}_{2}$ to $Cl$ is $242.8kJ{ mol }^{ -1 }$ ;
(iii) ionization of $Ca$ to ${Ca}^{2+}$ is $2422kJ{ mol }^{ -1 }$ ;
(iv) electron gain enthalpy for $Cl$ to ${Cl}^{-}$ is $-355kJ{ mol }^{ -1 }$ ;
(iv) ${ \Delta }_{ f }H$ overall is $-795kJ{ mol }^{ -1 }$ .

Given:

${ \Delta }_{ sub }{ H }_{ Ca }=121kJ{ mol }^{ -1 }$

$\Delta { H }_{ Cl-Cl }=242.8kJ{ mol }^{ -1 }$

${ \Delta }_{ f }{ H }_{ { Ca }^{ 2+ } }=2422kJ{ mol }^{ -1 }$

$\quad { \Delta }_{ eg }{ H }_{ Cl }=-355kJ{ mol }^{ -1 }\quad$

${ \Delta }_{ f }{ H }_{ Ca{ Cl }_{ 2 } }=-795kJ{ mol }^{ -1 }$

${ \Delta }_{ f }{ H }_{ Ca{ Cl }_{ 2 } }={ \Delta }_{ sub }{ H }_{ Ca }+\Delta { H }_{ Cl-Cl }+{ \Delta }_{ f }{ H }_{ { Ca }^{ 2+ } }+2\times { \Delta }_{ eg }{ H }_{ Cl }+{ \Delta }_{ l }{ H }_{ Ca{ Cl }_{ 2 } }$
or

$\quad -795=121+242.8+2422-(355\times 2)+{ \Delta }_{ l }{ H }_{ Ca{ Cl }_{ 2 } }$

$\therefore$

${ \Delta }_{ l }{ H }_{ Ca{ Cl }_{ 2 } }=-2870.8kJ$

${ mol }^{ -1 }$

Calculate the lattice enthalpy(nearest integer value) in $kJ{ mol }^{ -1 }$ of $LiF$ , given that enthalpy of,

(i) sublimation of lithium is $155.3kJ$ ${ mol }^{ -1 }$ ;

(ii) dissociation of $1/2$ mole of ${F}_{2}$ is $75.3kJ$ ;

(iii) ionization enthalpy of lithium is $520kJ$ ${ mol }^{ -1 }$ ;

(iv) electron gain enthalpy of 1 mole of $F(g)$ is $-333kJ$ ;

(v) ${ \Delta }_{ f }H$ overall is $-594kJ$ ${ mol }^{ -1 }$ .

Given:

${ \Delta }_{ sub }{ H }_{ Li }=155.2kJ{ mol }^{ -1 }$

$\quad \cfrac { 1 }{ 2 } \Delta { H }_{ F-F }=75.3kJ{ mol }^{ -1 }$

${ \Delta }_{ f }{ H }_{ Li }=520kJ{ mol }^{ -1 }$

${ \Delta }_{ eg }{ H }_{ F }=-333kJ{ mol }^{ -1 }$

${ \Delta }_{ f }{ H }_{ LiF }=-594.1kJ{ mol }^{ -1 }$

For

$LiF$ :

${ \Delta }_{ f }{ H }_{ LiF }={ \Delta }_{ sub }{ H }_{ Li }+\Delta \cfrac { 1 }{ 2 } { H }_{ F-F }+{ \Delta }_{ i }{ H }_{ Li }+{ \Delta }_{ eg }{ H }_{ F }+{ \Delta }_{ l }{ H }_{ LiF }$
or

$-594.1=155.2+75.3+520-333+{ \Delta }_{ l }{ H }_{ LiF }$

$\therefore$

${ \Delta }_{ l }{ H }_{ LiF }=-1011.6kJ$

${ mol }^{ -1 }$

The standard enthalpy of formation of $FeO$ and $Fe_2O_3$ is $-65 \: kcal \: mol^{-1}$ and $-197 \: kcal \: mol^{-1}$ respectively. If a mixture containing $FeO$ and $FeO_3$ in $2 : 1$ mole ratio on oxidation is changed into $1 : 2$ mole ratio, thermal energy in kcal released per $6$ mole of mixture will be (nearest integer)____________.

$\displaystyle Fe+\frac{1}{2}O_2\longrightarrow FeO; \: \Delta H=-65 \: kcal \: mol^{-1}$ . ..(i)

$\displaystyle 2Fe+\frac{3}{2}O_2\longrightarrow Fe_2O_3; \: \Delta H=-197 \: kcal \: mol^{-1}$ ...(ii)
By eq,

$[(ii) - 2 \times (i)]$
Also,

$\displaystyle 2FeO+\frac{1}{2}O_2\longrightarrow Fe_2O_3; \: \Delta H=-67 \: kcal \: mol^{-1}$ ...(iii)
Let

$a$ mole of FeO and

$b$ mole of

$Fe_2O_3$ , are present such that

$\displaystyle a+b=1 \: and \:\frac{a}{b}=2$

$\displaystyle a=\frac{2}{3} \:and \:b=\frac{1}{3}$

$2FeO+\frac{1}{2}O_2\longrightarrow Fe_2O_3$
Initial

$a$

$b$
After oxidation

$(a - 2a')$

$b+ a'$
Given,

$\displaystyle \frac{a}{b}=2 \:and \: \frac{a - 2a'}{b+ a'}=\frac{1}{2}$

$\displaystyle \therefore \frac{a - 2a'}{\frac{a}{2}+ a'}=\frac{1}{2}$

$\displaystyle \therefore 2a-4a'=\frac{a}{2}+ a'$

$\displaystyle \therefore a'=\frac{3a}{10}=\frac{3\times 2}{10\times 3}=\frac{1}{5}$

$\therefore FeO \: used = 2a' = 2/5 \: mol$

$\because 2 \: mole \: FeO \: gives \: 67 \: kcal$ .

$\displaystyle \therefore \frac{2}{5} mole \: FeO \: gives \: \frac{67\times 2}{2\times 5}= 13.4 \: kcal \: heat$ .

Answer

$= \dfrac{13.4}{6} \approx 2$

The endothermic reaction of carbon (graphite) in super heated steam at 500 K
$[i.e., C_{graphite} +H_2O(g)\longrightarrow CO(g)+ H_2 (g)]$

is made by providing heat by burning a part of carbon (graphite) in $O_2(g)$ followed by an exothermic reaction :
$[C_{graphite} +O_2(g)\longrightarrow CO_2(g);\:\Delta H= -393.7 \: kJ \: mol^{-1} \: at \: 500K]$ .

If a total of 1.34 mole of graphite is required for the production of 1 mole of $H_2$ at 500 K, the heat of reaction for carbon (graphite) with $H_2O$ in kJ is ________.

One mole of

$H_2$ will be formed by 1 mole graphite following the reaction:

$C\:(graphite) +H_2O(g)\longrightarrow CO(g)+ H_2 (g);\: \Delta H^{\circ}=?$ ... (i)

Thus, mole of graphite used by the reaction (ii)

$=(1.34- 1)=0.34$

$C\:(graphite) +O_2(g)\longrightarrow CO_2(g);\: \Delta H^{\circ}= -393.7 \:kJ$ ... (ii)

$Therefore, \: heat \: produced \: in \: (ii) \: by \: 0.34 \: mole \: C \: (graphite)=heat \: absorbed \: in \: (i) \: by \: 1 \: mole \: C \: (graphite)$

$0.34\times \Delta H_{(ii)}=1\times \Delta H_{(i)}$
or

$0.34\times393.7=1\times \Delta H_{(i)}$

$\therefore \Delta H_{(i)}=133.86 \:kJ$

6.80 g of $NH_3$ ; is passed over heated CuO. The standard heat enthalpIes of $NH_{3}$ , $CuO(s)$ and $H_20(l)$ are -46.0, -155.0 and $-285 0 \: kJ \: mol^{-1}$ respectively and the change is.
$NH_3 +(3/2)CuO\longrightarrow \frac{1}{2}N_2(g)+ (3/2)H_2O(l) +(3/2)Cu(s)$ .
If the enthalpy change is $x$ kJ then $-\dfrac{x}{15}$ (nearest integer) is___________.

$NH_3 + (3/2)CuO\longrightarrow \frac{1}{2}N_2(g)+ (3/2)H2O(l)+ (3/2)Cu(s) ; \: \Delta H=?$

$\Delta H^{\circ}=\Sigma H^{\circ}_{Products}-\Sigma H^{\circ}_{Reactants}$

$\displaystyle =[\frac{1}{2}\times H^{\circ}_{N_2}+( 3 / 2)\times H^{\circ}_{H_2O(l)}+~(3 /2) H^{\circ}_{Cu}]-[(3/2)H^{\circ}_{CuO}+H^{\circ}_{NH_3}]$

$=[0+ (3/2)\times (-285)+ (3/2)\times 0]-[(3/2) \times(-155)+ (-46)]$

$=-149\,kJ$

$[\because H^{\circ} for \: pure \: element = 0 \: and \: H^{\circ}_{compound} =H^{\circ}_{formation}]$

$\because 17\ g\ NH_3$ brings in change in

$\Delta H = -149 \: kJ$

$\because 6.8 \ g \ NH_3$ brings in change in

$\Delta H = (-149 \times 6.8)/17 = -59.6 \,kJ$

$\therefore \Delta H for \: 6.89 \: g \: NH_3 = - 59.6 \: kJ=x$

$\therefore \dfrac{-x}{15}=-59.6 \approx 4$

The answer is 4.

The dissolution of 1 mole of NaOH(s) in $100$ mole of $H_2O(l)$ give rise to evolution of heat as -42.34 kJ. However, if 1 mole of NaOH(s) is dissolved in $1000$ mole of $H_2O(l)$ the heat given out is 42.76 kJ. If the enthalpy change when $900$ mole of $H_2O(l)$ are added to a solution containing 1 mole of NaOH(s) in $100$ mole of $H_2O$ is $x$ $kJ$ then the value of $-100x$ is _____.

$NaOH(s) + 100H_2O\longrightarrow NaOH\ (100H_2O); \: \Delta H= -42.34 \: kJ$ ... (i)

$NaOH(s) + 1000H_2O\longrightarrow NaOH\ (1000H_2O); \: \Delta H= -42.76 \: kJ$ ... (ii)

By subtracting equations

$[(ii) - (i)]$ , we get-

$NaOH(s) + 900H_2O\longrightarrow NaOH\ (900H_2O); \: \Delta H= -0.42 \: kJ$

$\therefore -100x =42$

Calculate the resonance energy of $C_6H_6$ , using Kekule formula for $C_6H_6$ from the following data: (i) $\Delta H^{\circ}$ for $C_6H_6 = -358.5 \: kJ \: mol^{-1}$ (ii) $Heat \: of \: atomisation \: of \: C = 716.8\: kJ \: mol^{-1}$ (iii) Bond energy of $C-\! \! \! -H, \;C-\! \! \! -C, \:C=\! \! =C$ and $H-\! \! \! -H$ are 490, 340, 620 and $436.9 \: kJ \: mol^{-1}$ . Report your answer in kj (only value).

For

$C_6H_6$ ;

$6C(s)+ 3H_2(g) \longrightarrow C_6H_6;\ ; \Delta H_{exp} = -358.5 \:kJ$
Also, using Bond energy

$\Delta H_{cal}$ can be given as

$\Delta H_{cal} -Bond \: energy \: data \: used \: for \: formation \: of \: bond + B.E. \: data \: used \: for \: dissociation \: of \: bonds$

$= - [3\times e _{(C-\!\! \! -C)} + 3\times e_{(C=\! \! =C)}+6\times e_{(C-\!\! \! -H)}]+[6C_{s\rightarrow g}+3\times e_{(H-\!\! \! -H)} ]$

$= -[3\times 340+3\times 620+6\times 490]+[6\times 716.8+3\times 436.9] \: \Delta H_{cal}= -208.5kJ$

$\therefore Resonance \: energy =\Delta H_{exp} -\Delta H_{cal}$

$= -358.5 - (-208.5) = -150.0 \:kJ$

$2C_4H_{10}\longrightarrow C_8H_{18}+H_2$ . Given that bond energy of $C -C,\:C -H$ are $347.3$ and $414.2$
kJ mol $^{-1}$ and the heat of formation of hydrogen atom is $217.55$ kJ mol $^{-1}$ . Find $\Delta H$ for the above reaction in kJ mol $^{-1}$ .
(If $\Delta H=x$ , input the answer to the nearest integer as $\dfrac{x}{9}$ ).

For

$2C_4H_{10}\longrightarrow C_8H_{18}+H_2; \: \Delta H=?$

$\therefore \Delta H=$ bond energy data used for the formation of bond + bond energy data used for dissociation of bond

$\Delta H= - [7\times E _{(C-\!\! \! -C)} + 18\times E_{(C-\!\! \! -H)}+E_{(H-\!\! \! -H)}]+2[3\times E_{(C-\!\! \! -C)} +10\times E_{(C-\!\! \! -H)} ]$

$\Delta H = -e _{(C-\!\! \! -C)} +2\times e_{(C-\!\! \! -H)}-e_{(H-\!\! \! -H)}$

$\because$ Given,

$\dfrac{1}{2}H_2\longrightarrow H; \: \Delta H =217.55$ kJ

$\therefore e_{(H-\!\! \! -H)}=2\times 217.55=435.10$ kJ

$\therefore \Delta H = -347.3+ 2\times (414.2)- 1\times (435.1)$

$\Delta H = 46.0$ kJ mol

$^{-1}$

Now,

$\dfrac {46}{9} = 5.11 \equiv 5$ kJ mol

$^{-1}$

The heat of combustion of acetylene is 312 kcal. If heat of formation of $CO_2$ and $H_2O$ are 94.38 and 68.38 kcal respectively, $C \equiv\!\! \equiv C \:$ bond energy is $x \:kcal$ . Given that heat of atomisation of C and H are 150.0 and 51.5 kcal respectively and $C-\! \! \! -H$ bond energy is 93.64 kcal.

Value of $x$ (nearest integer value) is _______.

Given,

$C_2H_2 + \dfrac{5}{2}O_2\longrightarrow 2CO_2 + H_2O; \: \Delta H = -312 \:kcal$ ... (i)
(

$\because$ Heat of combustion are exothermic)

$C+ O_2\longrightarrow CO_2; \: \Delta H = -94.38 \:kcal$ ... (ii)

$H_2+\frac{1}{2}O_2\longrightarrow H_2O;\: \Delta H = -68.38 \:kcal$ ... (iii)

Multiply eq. (ii) by 2 and add in eq. (iii)

$2C+ H_2+ (5/2)O_2\longrightarrow 2CO_2+ H_2O \: \Delta H = -257.14 \:kcal$ ...( iv)
Subtracting eq. (i) from eq. (iv)

$C_2H_2 + (5/2)O_2\longrightarrow 2CO_2 + H_2O    \:            \Delta H = -312 \:kcal$ ...( i)
-            -            -           -                                               +
........................................................................................................................

$2C+ H_2\longrightarrow C_2H_2; \: \Delta H = +54.86 \:kcal$ ... (v)

Also,

$\Delta H$ for

$C+ H_2\longrightarrow C_2H_2$

$\Delta H=$ Bond energy data of formation of bond + Bond energy data of dissociation of bond

$\Delta H= - [e _{(C \equiv\!\! \equiv C)} + 2\times e_{(C-\!\! \! -H)}]+[2C_{s\rightarrow g}+e_{(H-\!\! \! -H)}]$

Also given,

$C_{s\rightarrow g}=150 \:kcal$

$\displaystyle \frac{1}{2}H_2(g)\longrightarrow H (g)- 51.5 \:kcal$

$\therefore e_{H-\!\! \! -H}=51.5\times 2 \:kcal= 103.0 \:kcal$

$e_{C-\!\! \! -H}= 93.64 \:kcal$

$\therefore 54.86 = -[e _{(C \equiv\!\! \equiv C)}+2\times 93.64] +[2\times 150+1\times 103.0]$

$\therefore e _{C \equiv\!\! \equiv C}=160.86 \:kcal \simeq 161$

Exactly 3.0 g of carbon was burned to $\displaystyle { CO }_{ 2 }$ in copper calorimeter. The mass of calorimeter is 1.5kg and the mass of water in calorimeter is 2kg. The initial temperature was $\displaystyle { 20 }^{ o }C$ and the final temperature was $\displaystyle { 31.3 }^{ o }C$ . If specific heat of copper is $\displaystyle 0.0826\,cal/g-K$ , the heat value to closest integer of carbon in kcal/g is

The quantity of heat transferred

$\displaystyle q = m_{water} \times S_{water} \times \Delta T_{water} + m_{copper} \times S_{copper} \times \Delta T_{copper}$

$\displaystyle q = 2000 \times 1 \times 11.3 + 1500 \times 0.0826 \times 11.3$

$\displaystyle q = 22600 + 1400 = 24000 \: cal = 24 \: kcal$
This heat is for 3 g of carbon.
For 1 g of carbon, the heat transferred will be

$\displaystyle \frac { \: 24kcal}{3 \: g} =8 \: kcal/g$ .

A new element $E$ forms a compound with chlorine which contains 1.455 g $\displaystyle Cl$ per $\displaystyle g\ E$ . The specific heat of $E$ was found $0.05\ cal/g$ . If eq. weight of element is 24, then the value of $\displaystyle x$ in $\displaystyle { ECl }_{ x }$ is:

Atomic mass of

$E$

$\displaystyle = \frac {6.4}{\text {specific heat in cal}} = \frac {6.4}{0.05} = 128$
1 g of

$E$ combines with 1.455 g of chlorine
Equivalent mass of

$E$ is the mass of

$E$ that combines with 35.5 g of chlorine. The equivalent mass of

$E$ will be

$\displaystyle \frac {1}{1.455} \times 35.5 = 24.4$
Valence

$\displaystyle = \frac {\text {atomic mass}}{\text {equivalent mass}} = \frac {128}{24.4} = 5.2 \simeq 5$
Hence, the value of

$x$ in

$\displaystyle ECl_x$ is 5.

The enthalpy changes of some process are given below :

$\alpha\! -\! D\! -\! glucose_{(s)}+ H_2O\longrightarrow \alpha\! -\! D\! -\! glucose_{(aq)}; \quad \Delta H_{dissolution}= 10.84 \:kJ$ .

$\beta \! -\! D\! -\! glucose_{(s)}+ H_2O\longrightarrow \beta \! -\! D\! -\! glucose_{(aq)}; \quad \Delta H_{ dissolution}= 4.68 \:kJ$ .

$\alpha\! -\! D\! -\! glucose_{(aq)}\longrightarrow \beta \! -\! D\! -\! glucose_{(aq)}; \: \quad \Delta H_{mutarotation} = -1.16 \:kJ$

The $\Delta H^{\circ}$ for $\alpha\! -\! D\! -\! glucose\longrightarrow \beta \! -\! D\! -\! glucose$ is :

$\alpha\! -\! D\! -\! glucose_{(s)}+ H_2O\longrightarrow \alpha\! -\! D\! -\! glucose_{(aq)}; \:Heat \: of \: dissolution= 10.84 \:kJ$ ......(1)

$\beta \! -\! D\! -\! glucose_{(s)}+ H_2O\longrightarrow \beta \! -\! D\! -\! glucose_{(aq)}; \:Heat \: of \: dissolution= 4.68 \:kJ$ ......(2)

$\alpha\! -\! D\! -\! glucose_{(aq)}\longrightarrow \beta \! -\! D\! -\! glucose_{(aq)}; \: Heat \: of \: mutarotation = -1.16 \:kJ$ .....(3)

$\alpha\! -\! D\! -\! glucose (s) \longrightarrow \beta \! -\! D\! -\! glucose(s)$ ......(4)

Equation (1) - equation (2) + equation (3) gives equation (4).

Hence,

$\Delta H^o = 10.84 \: kJ-4.68 \: kJ-1.15 \: kJ=5 \: kJ$

If $\displaystyle \Delta G$ and $\displaystyle \Delta S$ for the reaction: $\displaystyle A(g)+B(g)\rightarrow P(g)$ , at 300 K are $\displaystyle -600$ cal and $\displaystyle -10\;cal/K$ . Then $\displaystyle \Delta H$ for the reaction in kcal is ______.

$\displaystyle \Delta G$ and

$\displaystyle \Delta S$ for reaction:

$\displaystyle A(g)+B(g)\rightarrow P(g)$ at 300 K are

$\displaystyle -600$ cal and

$\displaystyle -10\;cal/K$ ,

$\displaystyle \Delta G = \Delta H -T \Delta S$

$\displaystyle -600 = \Delta H -300(-10)$

$\displaystyle \Delta H = -3600 \: cal = -3.6 \: kcal$

$\Delta H^{\circ}_f$ of $Fe_2O_3$ and $Al_2O_3$ are -189 kJ and -405 kJ respectively. How much heat (in kJ) is given out during reaction of 1 g Al according to
$2Al +Fe_2O_3\longrightarrow Al_2O_3 +2Fe$ ?

The enthalpy change for the reaction is,

$\Delta H = \Delta H_{f, Al2O3}+2\Delta H_{f, Fe}-[2\Delta H_{f, Al}+\Delta H_{f, Fe2O3}]$

$= -405 \: kJ/mol +2(0 \: kJ/mol)-[2(0 \: kJ/mol)-189 \: kJ/mol]=-216 \: kJ/mol$

The heat given out during reaction of 1 g Al is

$\frac {-216 \: kJ/mol}{2 \times 27 g/mol} = -4 \: kJ/g$

$KCl_{(s)} \longrightarrow K^ + + Cl^-$

The heat evolved in kcal for the process at $250^{\circ}C$ of dissolving 1.0 mole of KCl in excess of water is __________.

[Given: $\Delta H^{\circ}_f$ of $K^+_{(aq)}, \:Cl^-_{(aq)}$ and $KCl_{(s)}$ are -60, -40 and -104 kcal/mole]

The heat evolved for the process is

$\Delta H = \Delta H _f (K^+) + \Delta H _f (Cl^-) - \Delta H _f (KCl) = -60 \: kcal/mol-40 \: kcal/mol-(-104 \: kcal/mol)=4 \: kcal/mol$

If heat of formation of $CaCl_2$ and $NaCl$ are 191 and 97.5 kcal, the heat of reaction for $CaCl_2 + 2Na \longrightarrow 2NaCl + Ca$ is:

The heats of formation of elements in standard states are zero.
Thus, the heats of formation of Na and Ca are zero.

The heat of the reaction is:

$\Delta H_{rxn} = 2\Delta H_{NaCl}+\Delta H_{Ca} - [ \Delta H_{CaCl2}+2\Delta H_{Na}]$

$=2 (97.5 \: kcal)+0 \: kcal-[191 \: kcal+2(0 \: kcal)] = 4 \: kcal$

Match the Following:

The matched pairs are shown below.

The enthalpies of neutralisation of $NaOH$ and ${NH}_{4}OH$ by $HCl$ are $-13680$ calories and $-12270cal$ respectively. What would be the enthalpy change of one gram equivalent of $NaOH$ is added to one gram equivalent of ${NH}_{4}Cl$ is solution? Assume that ${NH}_{4}OH$ and $NaCl$ are quantitatively obtained (only magnitude in cal).

Enthalpy of neutralization is change in enthaply when 1 g equivalent of an acid reacts completely with 1 gram equivalent of base in dilute solution.

$R_1: NaOH(aq.) +HCl(aq) \rightarrow NaCl(aq) + H_2O;\; \Delta H_1=-13680 cal$

$R_2: NH_4OH(aq.) +HCl(aq) \rightarrow NH_4Cl(aq) + H_2O;\; \Delta H_2=-12270 cal$

$R_1-R_2 \Rightarrow NH_4Cl(aq.) +NaOH(aq) \rightarrow NH_4OHl(aq) + NaC(aq.)$
Given that 1 gram equivalent of

$NaOH$ is added to 1 gram equivalent of

$NH_4Cl$

$\Rightarrow \Delta H=\Delta H_1-\Delta H_2=-13680+12270 = -1410 cal$
Answer = 1410

The standard heats of formation of ${CH}_{4}(g), {CO}_{2}(g)$ and ${H}_{2}O(l)$ are $-76.2, -398.8, -241.6kJ{mol}^{-1}$ . Calculate amount of heat evolved by burning $1\ {m}^{3}$ of methane measured under normal (STP) conditions (nearest number in MJ).

$R_1 : C(s) + 2H_2(l) \rightarrow CH_4 (g)$ ;

$\Delta H_1 = -76.2 kJmol^{-1}$

$R_2 : C(s) + O_2(g) \rightarrow CO_2 (g)$ ;

$\Delta H_2 = -398.8 kJmol^{-1}$

$R_3 : H_2(g)+ \frac{1}{2}O_2(g) \rightarrow H_2O (l)$ ;

$\Delta H_3 = -241.6 kJmol^{-1}$

Now, burning methane

$R_4 : CH_4(g) + 2O_2(g) \rightarrow CO_2 (g)+2H_2O(l)$ ;

$\Delta H_4$

And

$R_4 = -R_1+R_2+2R_3$

According to Hess's Law of constant heat summation,

$\Delta H_4 = -\Delta H_1+\Delta H_2+2\Delta H_3$

$\Delta H_4= 76.2-398.8+2\times(-241.6) = -805.8\ kJmol^{-1}$

1 mole of methane gas at STP occupies 22.4

$l$ volume (assuming ideal gas behavior)

$22.4 l \rightarrow 1 mole$

$1000 l (1 m^{3}) \rightarrow ?\; mole$

$\Rightarrow 1 m^{3}$ of methane

$=\frac{1000}{22.4}$ mole of methane

Therefore, the heat evolved by burning

$1\ m^3$ of methane

$=\frac{1000}{22.4}\times{805.8}$

$=35973\; kJ = 35.973\; MJ$

Answer = 36

Calculate the enthalpy change when infinitely dilute solution of $Ca{Cl}_{2}$ and ${Na}_{2}{CO}_{3}$ mixed $\Delta {H}^{o}_{r}$ for ${Ca}^{2+}(aq)$ , ${CO}_{3}^{2-}(aq)$ and $Ca{CO}_{3}(s)$ are $-129.80, -161.65, -288.5 kcal$ ${mol}^{-1}$ respectively (nearest number in kcal).

Dilute solution of

$CaCl_2$ contains

$Ca^{+2}$ and

$Cl^{-}$ ions
Dilute solution of

$Na_2CO_3$ contains

$Na^{+}$ and

$CO_3^{-2}$ ions
When both are mixed

$Ca^{+2}(aq.)+Cl^{-}(aq.)+Na^{+}(aq.)+CO_3^{-2}(aq.) \rightarrow CaCO_3(s)+Na^{+}(aq.)+Cl^{-}(aq.); \quad \Delta H$

$\Delta H = \Sigma \Delta H^o_f(products)- \Sigma \Delta H^o_f(reactants)$

$\Rightarrow \Delta H = \Delta H^o_{f, CaCO_3(s)} +\Delta H^o_{f, Na^+(aq.)}+\Delta H^o_{f, Cl^-(aq.)}-[\Delta H^o_{f, Ca^{+2}(aq.)}+H^o_{f, Cl^{-}(aq.)}+H^o_{f, Na^{+}}+H^o_{f, Ca^{+2}(aq.)}]$
on simplification

$\Delta H = \Delta H^o_{f, CaCO_3(s)}-\Delta H^o_{f, Ca^{+2}(aq.)}-\Delta H^o_{f, CO_3^{-2}(aq.)}$

$\Rightarrow \Delta H = -288.5+129.8+161.65=2.95\; kcal/mol$
Answer = 3

Calculate the electron affinity of fluorine atom using the following data (only magnitude in nearest integer in kj/mol).
All the values are in $kJ$ ${mol}^{-1}$ at ${25}^{o}C$ , $\Delta{H}_{diss}({F}_{2})=160$ ; $\Delta {H}^{o}_{r}$ ( $NaF(s))=-571;\ I.E$ $[Na(g)]=494, \ \Delta {H}_{vap}[Na(s)]=101$ . Lattice energy of $NaF(s)=-894$ .

From Born-Hyber cycle in the diagram,

$\Delta H^o_f = \Delta H^o_{vap,\; Na(s)} +\frac{1}{2}\Delta H^o_{dis,\; F}+I.E_{Na}+E.A_{F}+L.E_{NaF}$

On simplifying

$E.A_{F} = \Delta H^o_f -\Delta H^o_{vap,\; Na(s)} -\frac{1}{2}\Delta H^o_{dis,\; F}-I.E_{Na}-L.E_{NaF}$

$\Rightarrow E.A_{F}=-571-101-\frac{1}{2}\times 160-494+894= -352 \; kJ/mol$

List equation/law (in Column I) with statement (in Column II).
Match the following.

(A) Arrhenius equation gives the variation of rate constant with temperature.
(B) Kirchhoff equation gives the variation of enthalpy of a reaction with temperature.
(C) Second law of thermodynamics gives entropy of an isolated system tends to increase and reach a maximum value.
(D) Hess's law of constant heat summation gives enthalpy change in a reaction is always constant and independent of the manner in which the reaction occurs.

Becker and Ruth measured the heat evolved in the following processes at ${20}^{o}C$ : (1) $1$ mole of solid ${({COONH}_{4})}_{2}{H}_{2}O$ is burned in oxygen, (2) $1$ mole of solid ${(COOH)}_{2}{({H}_{2}O)}_{2}$ is burned in oxygen, (3) $1$ mole of solid ${({COONH}_{4})}_{2}{H}_{2}O$ is dissolved in a large excess of water, (4) $1$ mole of solid ${(COOH)}_{2}{({H}_{2}O)}_{2}$ is dissolved in a large excess of water, (5) $1$ mole of oxalic acid in dilute solution is neutralized with gaseous ammonia. They found (1) $189.86kcal$ , (2) $53.10kcal$ , (3) $-11.47kcal$ , (4) $-8.62kcal$ , (5) $43.13kcal$ ; (1) and (2) were measured at constant volume, the others at constant pressure. The end products of (1) and (2) were nitrogen, carbon-di-oxide and water. The heat of formation for $1$ mole of water from the elements had previously been determined as $68.35kcal$ at constant pressure and ${20}^{o}C$ . Find the change in enthalpy when $1$ mole of ${NH}_{3}$ is formed from the elements at ${20}^{o}C$ (only magnitude in nearest integer in kcal)

$R_1 : (COONH_4)_2H_2O(s)+2O_2(g) \rightarrow N_2(g)+2CO_2(g)+5H_2O(l); \Delta Q_1=189.86\; kcal$

$R_2 : (COOH)_2(H_2O)_2(s)+\frac{1}{2}O_2(g) \rightarrow 2CO_2(g)+3H_2O(l); \Delta Q_2=53.10\; kcal$

$R_3 : (COONH_4)_2H_2O(s) \rightarrow (COONH_4)_2(aq) +H_2O; \Delta Q_3=-11.47\; kcal$

$R_4 : (COOH)_2(H_2O)_2(s) \rightarrow (COOH)_2(aq)+2H_2O(l); \Delta Q_4=-8.62\; kcal$

$R_5: (COOH)_2(aq) +2NH_3(g) \rightarrow (COONH_4)_2(aq); \Delta Q_5= 43.13\; kcal$

$R_6: H_2(g)+\frac{1}{2}O_2(g) \rightarrow H_2O(l); \Delta Q_6= 68.35\; kcal$
Formation of ammonia from the elements

$R_7: \frac{1}{2}N_2(g)+\frac{3}{2}H_2(g) \rightarrow NH_3 (g); \Delta Q_7$
rearranging there reactions give

$2R_7=-R_5-R_4+R_3+R_2-R_1+3R_5$
Therefore, according to Hess's Law of constant heat summation

$2\Delta Q_7 =-\Delta Q_5-\Delta Q_4+\Delta Q_3+\Delta Q_2-\Delta Q_1+3\Delta Q_6$

$2\Delta Q_7 =-43.13+8.62-11.47+53.10-189.86+3\times 68.35= 22.31 kcal$

$\Rightarrow \Delta Q_7 =\frac{22.31}{2} = 11.15 kcal$
Answer = 11

Note:

$R_5$ reaction happens in two steps

$NH_3(g)+H_2O(l) \rightarrow NH_4OH(aq)$

$(COOH)_2(aq)+NH_4OH(aq) \rightarrow (COONH_4)_2(aq) +H_2O(l)$

For the hypothetical reaction,
$\displaystyle A_{2}g+B_{2}\left ( g \right )\rightleftharpoons 2AB\left ( g \right )$
If $\displaystyle \Delta _{r}G^{\circ}\: and\: \Delta _{r}S^{\circ}$ are 20 kj/mol and -20 $\displaystyle JK^{-1}mol^{-1}$ respectively at 200 K
$\displaystyle \Delta _{r}C_{P}\: is\: 20JK^{-1}mol^{-1}$ then value $\displaystyle \frac{\Delta _{r}H^{\circ}}{10}$ of (KJ/mol) at 400 K is:

$\Delta H^o_r (200)=\Delta G^o_r (200)+T\Delta S^o_r (200)=20000 \: J/mol + 200 \: K \times (-20 \: J/K/mol)=16000 \: J/mol = 16 \: kJ/mol$

$\Delta H^o_r (400)=\Delta H^o_r (200)+\Delta _r C_P (T_2-T_1)=16 \:kJ/mol+\frac {20 \: J/mol(400 \: K-200 \:K)}{1000}=16 \ :kJ/mol + 4 \ :kJ/mol=20 \ :kJ/mol$

$\dfrac {\Delta H^o_r (400)}{10}=\dfrac {20 \ :kJ/mol}{10}=2 \ :kJ/mol$

For the hypothetical reaction, $\displaystyle { A }_{ 2 }\left( g \right) +{ B }_{ 2 }\left( g \right) \rightleftharpoons 2AB\left( g \right)$ , if $\displaystyle { \Delta }_{ r }{ G }^{ o }$ and $\displaystyle { \Delta }_{ r }{ S }^{ o }$ are $\displaystyle 20\ { kJ }/{ mol }$ and $\displaystyle -20 \ { JK }^{ -1 }{ mol }^{ -1 }$ respectively at 200 K. $\Delta _rC_p$ is $20\ JK^{-1}mol^{-1}$ .

Then value $\displaystyle \frac { { \Delta }_{ r }{ H }^{ o } }{ 10 }$ of ( $kJ/mol$ ) at 400 K is:

The relationship between the standard Gibbs free energy change, the standard enthalpy change and the standard entropy change is as shown.

$\displaystyle { \Delta G }_{ 200 }^{ o }={ \Delta H }_{ 200 }^{ 0 }-T\Delta { S }_{ 200 }^{ 0 }$

$\displaystyle 20 \times 10^3 = \Delta H^0_{200} - 200 \times (-20)$

$\displaystyle { \Delta H }_{ 200 }^{ 0 }=16000$ J mol

$^{ -1 } =16$ kJ mol

$^{ -1 }$ .

The relationship between the enthalpy change at two different temperatures and the heat capacity at constant pressure is as shown.

$\displaystyle { \Delta H }_{ { T }_{ 2 } }^{ 0 }={ \Delta H }_{ { T }_{ 2 } }+{ \Delta C }_{ p }\left[ { T }_{ 2 }-{ T }_{ 1 } \right]$

$\displaystyle { \Delta H }_{ 400 }^{ 0 }={ \Delta H }_{ 200 }^{ 0 }+\frac { 20\times 200 }{ 1000 }$ kJ mol

$^{ -1 }$ .

$\displaystyle =16+4=20$ kJ mol

$^{ -1 }$ .

Hence, the standard enthalpy change for the reaction at 400 K is 20 kJ/mol.

$\displaystyle \frac {{ \Delta H }_{ 400 }^{ 0 }}{10} = \frac {20}{10} = 2$ kJ mol

$^{ -1 }$ .

For the reaction at 298 K, 2A+B $\displaystyle \rightarrow C$

$\displaystyle \bigtriangleup$ H = 400 KJ $\displaystyle mol^{-1}$ and $\displaystyle \bigtriangleup$ S = 0.02 KJ $\displaystyle K^{-1}$ $\displaystyle mol^{-1}$

At what temperature will the reaction become spontaneous considering $\displaystyle \bigtriangleup$ H and $\displaystyle \bigtriangleup$ S to be constant over the temperature range?

$\displaystyle \Delta G = \Delta H - T \Delta S$

Assuming that equation at equilibrium then

$\displaystyle \Delta G$ =

$0$

$\displaystyle \Delta G = \Delta H - T \Delta S = 0$

$\displaystyle T = \frac {\Delta H}{\Delta S} = \frac {400}{0.2}=2000 K$ .

For the reaction to be spontaneous

$\Delta G$ must be negative.

Therefore temprature should be greater than 2000 K.

The equilibrium constant for a reaction iswhat will be value of $\displaystyle \bigtriangleup G^{\ominus }$ ? R =8.314 $\displaystyle JK^{-1} mol^{-1}$ , T=300 K

$\displaystyle \Delta G^0 = -2.303 RT log K = -2.303 \times 8.314 \times 300log10 = -5.744 kJ/mol$

Calculate the value of ${ \Delta H }/{ kJ }$ for the following reaction using the listed thermochemical equations:
$2C\left( s \right) +{ H }_{ 2 }\left( g \right) \longrightarrow { C }_{ 2 }{ H }_{ 2 }\left( g \right)$
$2{ C }_{ 2 }{ H }_{ 2 }\left( g \right) +5{ O }_{ 2 }\left( g \right) \longrightarrow 4C{ O }_{ 2 }\left( g \right) +2{ H }_{ 2 }O\left( I \right) { \Delta H }/{ kJ }=-2600kJ$
$C\left( s \right) +{ O }_{ 2 }\left( g \right) \longrightarrow C{ O }_{ 2 }\left( g \right) { \Delta H }/{ kJ }=-390kJ$
$2{ H }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \longrightarrow 2{ H }_{ 2 }O\left( I \right) { \Delta H }/{ kJ }=-572kJ$

$\Delta { H }_{ r }=\left[ -\displaystyle\frac { 1 }{ 2 } { \left( \Delta { H }_{ f } \right) }_{ { C }_{ 2 }{ H }_{ 2 } }+2{ \left( \Delta { H }_{ f } \right) }_{ C{ O }_{ 2 } }+\displaystyle\frac { 1 }{ 2 } { \left( \Delta { H }_{ f } \right) }_{ { H }_{ 2 }O } \right]$

$\Delta { H }_{ r }=-\displaystyle\frac { 1 }{ 2 } \left( -1300 \right) +2\left( -390 \right) -\displaystyle\frac { 1 }{ 2 } \times 572$

$\Delta { H }_{ r }=234$

Calculate $E_{cell}$ and $\Delta$ G for the following at $28^o$ C

$Mg(s)+Sn^{2+}(0.025M)\rightarrow Mg^{2+}(0.06M)+Sn_{(s)}$

$E^0_{cell}=2.23$ V

Is the reaction spontaneous?

Given: Cell reaction

$Mg_{(s)}+Sn^{2+}_{(aq)}\rightarrow Mg^{2+}_{(aq)}+Sn_{(s)}$

$E_{Cell}=2.23; [Sn^{2+}]=0.025$ M

$[Mg^{2+}]=0.06M; n=2$

$E_{cell}=E_{cell}-\displaystyle\frac{0.0592}{2}log_{10}\frac{[Mg^{2+}]}{[Sn^{2+}]}$

$=2.23-\displaystyle\frac{0.0592}{2}log_{10}\frac{0.06}{0.025}$

$=2.23-0.296log 24$

$=2.23-0.296\times 0.380=2.118$ V

$\Delta G=-nFE_{cell}$

As,

$[1F=96500C]$

$\Delta G=-2\times 96500\times 2.118$

$=408774$ J

$=-408.8$ kJ

Since

$\Delta$ G for a given cell reaction is negative, the given reaction is spontaneous.

Give the characteristics of free energy.

Characteristics of Free energy:
i) G is defined as (H-Ts) where H and S are the enthalpy and entropy of the system respectively. T

$=$ temperature. Since H and S are state functions, G is the state function.
ii) G is an extensive property while

$\Delta G=(G_2-G_1)$ which is the free energy change between the initial

$(1)$ and final

$(2)$ states of the system becomes the intensive property when mass remains constant between initial and final states (or) when the system is a closed system.
iii) G has a single value for the thermodynamic state of the system.
iv) G and

$\Delta$ G values correspond to the system only. There are three cases of

$\Delta$ G in predicting the nature of the process. When,

$\Delta G < 0$ (negative), the process is spontaneous and feasible;

$\Delta G=0$ . The process is in equilibrium and

$\Delta G > 0$ (positive), the process is non-spontaneous and not feasible.
v)

$\Delta G=\Delta H-T\Delta S$ . But according to

$Ist$ law of thermodynamics,

$\Delta H=\Delta E+P\Delta V$ and

$\Delta E=q-w$ .
But

$\Delta S=\displaystyle\frac{q}{T}$ and

$T\Delta S=q=$ heat involved in the process.

$\therefore \Delta G=q-w+P\Delta V-q=-w+P\Delta V$
(or)

$-\Delta G=w-P\Delta V=$ network.
The decrease in free energy

$-\Delta G$ , accompanying a process taking place at constant temperature and pressure is equal to the maximum obtainable work from the system other than work of expansion. This quantity is called as the "net work" of the system and it is equal to

$(w-P\Delta V)$

$\therefore$ Network

$=-\Delta G=w-P\Delta V$ .

$-\Delta G$ represents all other forms of work obtainable from the system such as electrical, chemical or surface work etc other than P-V work.

$\therefore \Delta G=q-w+P\Delta V-T\Delta S$ .

For a chemical reaction the values of $\Delta H$ and $\Delta S$ at 400 K are -10 K cal mol $^{-1}$ and 20 cal. deg $^{-1}$ mol $^{-1}$ respectively. Calculate the value of $\Delta G$ of the reaction.

$\Delta H$ and

$\Delta S$ values of a reaction at 400 K are -10 K cal mol

$^{-1}$ and 20 cal deg

$^{-1}$ mole

$^{-1}$ respectively. Calculate

$\Delta G$ value.

$\Delta S$ = enthalpy change = - 10 K cal mol

$^{-1}$

$\Delta S$ = entropy change = 20 cal deg

$^{-1}$ mol

$^{-1}$
T = 400 K

$\Delta G = \Delta H - T \Delta S$

$= -10000 \ cal \ mol^{-1} - (400 \times 20) cal \ deg^{-1} mol^{-1}$

$= - 18,000 \ cal \ deg^{-1} mol^{-1}$

$= 18 K cal \ deg^{-1} mol^{-1}$

Calculate $q.w.\Delta U$ and $\Delta H$ for the reversible isothermal expansion of one mole of an ideal gas at ${127}^{o}C$ from a volume of $10\ {dm}^{3}$ to $20\ {dm}^{3}$ .

Since, the process is isothermal,

$\Delta U=\Delta H=0$
From the first law of thermodynamics

$\Delta U=q+w=0$

$q=-w$

$w=-2.303nRT\log { \cfrac { { V }_{ 2 } }{ { V }_{ 1 } } } =-2.303\times 1\times 8.314\times 400\log { \cfrac { 20 }{ 10 } }$

$=-2.303\times 1\times 8.314\times 0.3010=-2305.3J$ (Work done by system)

$q=-w=2305.3J$ (Heat absorbed by the system)

Pressure over $1000\ ml$ of a liquid is gradually increases from $1$ bar to $1001$ bar under adiabatic conditions. If the final volume of the liquid is $990\ ml$ , calculate $\triangle U$ and $\triangle H$ of the process, assuming linear variation of volume with pressure.

1240587_689039_ans_1214ba92d78b40719fcfe79d28fb8b43.jpg

Explain spontaneous and non-spontaneous processes, giving two examples of each.

A spontaneous processes takes place on its own without the external influence. Once started, a spontaneous processes proceeds on its own, without the continuous external help. For spontaneous processes, the change in Gibbs free energy is negative (

$\displaystyle \Delta G < 0$ ).
Examples of spontaneous processes:
1) Heat flows from hotter body to a colder body.
2) A solid KCl spontaneously dissolves in water.

A non-spontaneous processes does not takes place on its own. It needs continuous external influence.
Once started, a non-spontaneous processes will stop, when the continuous external force is withdrawn.
For non-spontaneous processes, the change in Gibbs free energy is positive (

$\displaystyle \Delta G > 0$ ).
Examples of non-spontaneous processes:
1) Flow of heat from inside of refrigerator to the room. Room is at higher temperature than refrigerator.
2) Boiling of water.

One mole of ethane was burnt in oxygen at constant volume to give $C{O}_{2}$ at 298 K. The heat evolved was found to be 1554 kJ. Calculate the heat of reaction at constant pressure.

$H=1554$

$kj$

$T=298K$

$C_2H_6+\cfrac{7}{2}O_2\longrightarrow 2CO_2+3H_2O$

$\triangle ng=(\cfrac{9}{2}-2)=\cfrac{5}{2}$

$\triangle H=\triangle U+\triangle n_gRT\\\triangle H=1334-\cfrac{5}{2}(8.3)(298)\\\triangle H=1554-6183=-4629.5 \text{ kj}$

Five moles of an ideal gas at $300K$ , expanded isothermally from an initial pressure of $4\ atm$ to a final pressure of $1\ atm$ against a cont. ext. pressure of $1\ atm$ . Calculate $q, w, \triangle U$ and $\triangle H$ . Calculate the corresponding value of all if the above process is carried out reversibly.

$w=2.303nRTlog\dfrac { { P }_{ 1 } }{ { P }_{ 2 } }$

$=2.303\times 5mole\times 8.314J{ mol }^{ -1 }{ K }^{ -1 }\times 300Klog\dfrac { 4atm }{ 1atm }$

$=17232.4J$

Since for isothermal reversible expansion of a ideal gas,

$\Delta H=0,\quad \Delta U=0$

now,

$\Delta U=Q+w$

$\therefore \quad Q=-w=-17232J$

The volume of a gas at STP is 2 L, 300 J heat is given to it, as a result of which volume changes to 2.5 L at 1 atm. Calculate the change in internal energy of the system.

$\Delta U = q +w$

$W = - P \Delta V$

$W = -1( 2.5 -2) = 0.5 L atm$

$= -0.5 \times 101.3 J$

$= -50.65 J$

$Q = 330 J$

$\Delta U = 300 J + ( -50.65 J)$

$= 249.35 J$

What will be the final temperature attained if all the heat released in neutralization of $1$ L of $0.2$ M $NH_4OH$ with $2$ L of $0.1$ M HCl increases the temperature of the final solution having density $0.95$ gm/ml and specific heat capacity = $13J/g^0C$ , if original temperature was $27^0C$ ? Assume weak base to be completely unionized.

$NH_4 OH + HCl \rightarrow NH_4 Cl + H_2O$

Number of mole of

$NH_4 OH = 1L \times 0.2 M$

$n_{NH_4OH} = 0.2 \, mole$

Number of mole of

$HCl = 2L \times 0.1 \, M$

$n_{HCl} = 0.2 \, mole$

Total volume of final solution =

$2L + L = 3L$

Mass =

$0.95 \times 3 L \times 1000 = 2850 gm$

total heat released =

$2850 \times 13 J \times \Delta T = 37.05 kJ\Delta T$

$37.05 \, \Delta T = 51.46 kJ$

$\Delta T = \dfrac{51.56}{37.05} = 1.4$

$T_2 - T_1 = 1.4$

$T_2 - 300 = 1.4$

$T_2 = 301.4 k$

The cell in which the following reaction occurs:

$2{\text{F}}{{\text{e}}^{{\text{3 + }}}}\left( {{\text{aq}}} \right) + 2{{\text{I}}^{\text{ - }}}\left( {{\text{aq}}} \right) \to 2{\text{F}}{{\text{e}}^{{\text{2 + }}}}\left( {{\text{aq}}} \right) + {{\text{I}}_{\text{2}}}\left( {\text{s}} \right)\,{\text{has}}\,{\text{E}}_{\left( {{\text{cell}}} \right)}^ \odot = 0.236\,{\text{V}}$ at 298 K.

Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

$\Delta G = nFE^\circ$

$= - 2 \times 96500 \times 0.236 = - 45.548\,kJ$

$log K = \dfrac{{nFE^\circ }}{{2.303RT}} = \dfrac{{2 \times 0.236}}{{0.0591}} = 7.986$

$K = {\left( {10} \right)^{7.986}} = 96827785.6$

$= 9.68 \times {10^7}$

$\Delta { H }^{ 0 }\quad and \quad \Delta { S }^{ 0 }\quad for\quad CC{ l }_{ 4 }$ are $2.5{ KJ }^{ -1 }{ mol }^{ -1 }\quad $\quad 9.99{ JK }^{ -1 }{ mol }^{ -1 }$ respectably at $29{ 8 }^{ 0 }K$ find the temperature at which solid ${ CCl }_{ 4 }$ and its liquid are in equilibrium at one atmosphere.

As we know

$\Delta G^{ 0 }=\Delta { H }^{ 0 }-\Delta { S }^{ 0 }$

for solid

$\&$ liquid to be in equilibrium,

$\Delta G^{ 0 }=0\\ \therefore Q=\Delta H^{ 0 }-T\Delta S^{ 0 }\\ \Delta H^{ 0 }=T\Delta S\\ T=\cfrac { \Delta H^{ 0 } }{ \Delta S } =\cfrac { 2.5KJ/mol }{ 9.99J{ K }^{ -1 }{ mol }^{ -1 } } \\ \cfrac { 2.5\times { 10 }^{ 3 } }{ 9.99 } K=250.2K$

$100kJ$ heat is transferred from a larger heat reservoir at $400K$ to another large heat reservoir at $300K$ . Suppose there is no change in temperature due to exchange of heat:
Find (a) $\Delta {S}_{source}$ , (b) $\Delta{S}_{sink}$ and (c) $\Delta {S}_{total}$ . Comment on spontaneity of process:

$\Delta S_{seure}=\cfrac{q_{reversible}}{T}=\cfrac{-100\times 1000}{400}=-250$

$JT^{-1}\quad$ (Heat is Transferred)

$\Delta S_{sink}=\cfrac{q_{reversible}}{T}=\cfrac{100\times 1000}{300}=333.3$

$JT^{-1}$

$\Delta S_{Total}=\Delta S_{sink}+\Delta S_{source}$

$\implies \Delta S_{Total}=333.3-250$

$\implies \Delta S_{Total}=83.3$

$JT^{-1}$

Since,

$\Delta S>0$ process is spontaneous

Determine whether the following reaction will be spontaneous or non-spontaneous under standard conditions.
$Zn\left( s \right) +{ Cu }^{ 2+ }\left( a \right) \longrightarrow { Zn }^{ 2+ }\left( aq \right) +Cu\left( s \right) \triangle { H }^{ 0 }=-219kJ,\triangle { S }^{ 0 }=-21{ JK }^{ -1 }\left( Spontaneous \right)$

$\Delta G=\Delta H-T\Delta S$

$T=29$

$K$

$\Delta G<0$ for spontaneous process.

$\Delta G=-219\times 1000-[(-21)\times 298$

$K]$

$\implies \Delta G=-212.742$

$kJ$

Therefore, process is spontaneous.

A heat engine absorbs $760kJ$ heat from a source at $380K$ . It rejects (i) $650kJ$ , (ii) $560kJ$ , (iii) $504kJ$ of heat to sink at $280K$ . State which of these represent a reversible, an irreversible or an impossible cycle.

From the second law of thermodynamics, we know that

$\\ \Delta (T/{ T }^{ 1 }) = A/Q$

$\Delta *Q/{ T }^{ 1 }=100*760/380\quad =200K$

The rejected:

$760-200=560 k$

Hence

$560KJ$ is reversible.

In the following reaction $2{H_2}{O_2}\left( {aq} \right) \to 2{H_2}O\left( \ell \right) + {O_2}\left( g \right)$
(a) What is the rate of formation of ${H_2}O$
(b) What is the rate of disappearance of ${H_2}O$ .

$2{H}_{2}{O}_{2}$ (aq)

$\rightarrow$

$2{H}_{2}{O}(l)+{O}_{2}$

$(g)$

Now, Rate=

$-\cfrac{1}{2}$

$\cfrac{d}{dt}$

${H}_{2}$

${O}_{2}$

$-\cfrac{1}{2}$

$\cfrac{d}{dt}$

${H}_{2}$

${O}$

$\cfrac{d}{dt}$

$[{O}_{2}]$

Rate of formation of

${O}_{2}$ =

$\cfrac{d}{dt}$

${O}_{2}$ =36 g

${min}^{-1}$

Rate of formation of

${H}_{2}$ O=

$\cfrac{d}{dt}$ [

${H}_{2}$ O]

$\times$

$\cfrac{d}{dt}$

$[{O}_{2}]$

$\times 36 g{min}^{-1}$

Rate of formation of

${H}_{2}$ O=

$72 g{min}^{-1}$

Rate of disappearance of

${H}_{2}$ O=

$-\cfrac{d}{dt}$ [

${H}_{2}$

${O}_{2}$ ]

$\times$

$\cfrac{d}{dt}$

$[{O}_{2}]$

$\times 36 g{min}^{-1}$

Rate of formation of

${H}_{2}$ O=

$72 g{min}^{-1}$

a) The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are $-890.3 kJ\quad { }^{ -1 }$ , $-393.5 kJ\quad { }^{ -1 }$ and $-285.8 kJ\quad { }^{ -1 }$ respectively. Calculate the enthalpy of formation of ${ CH }_{ 4 }\left( g \right)$
b) Write the relationship between Gibb's energy, equilibrium constant and change in enthalpy.

(a) (a)

$CH_4+2O_2CO_2+2H_2O; \Delta H=-890.3 kJ/mol \longrightarrow (1)$

(b)

$C+O_2\longrightarrow CO_2, \Delta H= -393.5 \longrightarrow (2)$

(c)

$H_2+\cfrac {1}{2}O_2 \longrightarrow H_2O, \Delta H= -285.8 \longrightarrow (3)$

$C+2H_2\longrightarrow CH_4, \Delta H=?$

$= \text{Equation 2+2} \times \text{ Equation 3}-\text{ Equation 1 }$

$=-393.5+2 \times (-285.8)-(-890.3)=-74.8 kJ/mol$ .

(b)

$\Delta G= \Delta G^o+RT\log Q$

At equilibrium=

$\Delta G^o=0$

$\Delta G= RT \log K \Rightarrow K=$ equilibrium constant

$\Delta G=\Delta H-T \Delta S$ .

For the reaction $NH_4Cl(S) \rightarrow NH_3(g) + HCl(g)$ at $25^0$ C, enthalpy change $\triangle = +177 kJ mol^{-1}$ and entropy change $\triangle S = +285 JK^{-1}mol^{-1}$ . Calculate free energy change $\triangle G$ at $25^0$ C and product whether the reaction is spontaneous or not.

Given,

$\triangle H= 177 \ kJ/mol, \ \triangle S=285 \ JK^{-1}mol^{-1}, \ T=298 \ K$

Using the relation,

$\triangle G=\triangle H- T\triangle S$

$\triangle G= (177 \times 10^{3} \ J/mol)- (298 \ K) (285 \ JK^{-1}mol^{-1})$

$\triangle G= 92070 \ J/mole$

$\boxed {\triangle G= 92.07 \ kJ/mole}$

$\boxed {\triangle G >0}$

Since

$\triangle G$ is positive, Hence reaction is non spontaneous.

State and explain Hesss law of constant heat summation. What is the value of $\triangle { S }^{ o }_{ surr. }$ ] for the following reaction at $298\ K$
${ 6CO }_{ 2\left( g \right) }+{ 6H }_{ 2 }{ O }_{ \left( 1 \right) }\rightarrow { C }_{ 6 }{ H }_{ 12 }{ O }_{ 6\left( g \right) }+{ 6CO }_{ 2\left( g \right) }$ Given
$\triangle { G }^{ o }=2879kJ{ mol }^{ -1 }$
$\triangle { S }^{ o }=-205Jk^{-1}{ mol }^{ -1 }$

1346438_1135177_ans_525fa30918ef4067967d4dc198e14f0d.jpg

Calculate the electron gain enthalpy of fluorine from the data given below. $\Delta H_f$ of KF is - 560.8 KJ/ mol, dissociation energy of $F_2$ is 158.9 KJ / mol. Lattice energy of KF is 807.5 KJ/mol, ionization energy of potassium is 414.2 KJ/mol and enthalpy of sublimation of $K = 87.8 KJ/ mol.$

Accordin to hess law

$\Delta H_F = \Delta H_{sub} + \dfrac{1}{2} \Delta H_{diss} + IE + E.A + \Delta H_{lattice}$

$-560.8 = 87.8 + \dfrac{158.9}{2} + 414.2 + EA - 807.5$

$-560.8 = 581.45 + EA - 807.5$

$-560.8 = -226.05 + EA$

$EA = -560.8 + 226.05$

$EA = -334.75 KJ$

If the molar heat capacity of $O_2$ gas is $640Jk^{-1}mol^{-1}$ . Then calculate its specific heat capacity.

Molar heat capacity of

${O_2}$ gas

$640\,J\,{K^{ - 1}}mo{l^{ - 1}}$

${C_N} = 640\,J\,{K^{ - 1}}mo{l^{ - 1}}$

${C_P} = ?$

${C_V} = M{C_P}$

$\Rightarrow 640 = 16\,{C_P}$

${C_P} = \frac{{640}}{{16}} = 40\,J/K/mol$

We get

$,$

$40\,J/K/mol$

Calculate $\Delta H_{lattice}$ of $SnO_2$ , If $\Delta H_f$ of $SnO_2$ is -588 KJ / mol, Enthalpy of Sublimation (Sn) = 292 KJ / mol, Enthalpy of Dissociation $(O_2) = 454 \, KJ / mol$ . Total Electron gain enthalpy for O = 636 KJ / mol, Ionization enthalpy $(Sn \rightarrow Sn^{4+})$ = 8990 kJ / mol.

Total gain = Total loss

$\Delta H_{I.E} + \Delta H_{ss} + \Delta H_{sub} = |\Delta H_F| + |\Delta H_{eg} | + L. E$

$\Rightarrow 8990 + 454 + 292 = 588 + 636 + L.E.$

$L.E = 11596 KJ$
1037591_1172334_ans_f8db9830b7d1469bb3950b107dfae47e.PNG

A gas expands from 14 liters to 19 liters against a constant pressure of 2 atm. During the process 600 J of heat flows in the gas from the surroundings. Calculate the internal energy change of its system.

1351598_1200908_ans_617e2a8138614c8aa20100cd5d196b8e.jpg

Calculate w and $\Delta U$ for the conversion of 0.5 mol of water at 100-degree celsius to steam at 1atm pressure. The heat of vaporization of water at 100degree celsius is 40670 J/mol.

1350727_1221149_ans_931d31d7a07c471bac1f40136c2742f5.jpg

During an adiabatic process, the pressure of a gas is found to be proportional to cube of its absolute temperature. Calculate the Poisson's ratio of gas.

1162597_1312117_ans_7703ebddff21433782d909e192339e18.jpg

Work done by a sample of an ideal gas in a process A double the work done in another process B. The temperature rises through the same amount in the two processes. If ${C_A}$ and ${C_B}$ are the molar heat capacities for the two processes, then the relation between $C_A$ and $C_B$ will be:

2041316_1367206_ans_81d9861982284154b4916554578fb8d6.jpeg

Calculate $\Delta U$ and $\Delta H$ in calories if 1 mole of a monoatomic ideal gas is heated at constant pressure of 1 atm from $25^oC$ to $50^oC$ .

For monoatomic ideal gas,

$C_v=\dfrac 32R$

$C_p=\dfrac{\Delta H}{\Delta T}$

$\therefore \Delta H=C_p\Delta T\ mol^{-1}$ ( or

$nC_p\Delta T$ for

$n$ moles )

$=\dfrac 52 \times 1.987\times 25=124.2\ cal$
Work done,

$w=-P\Delta V=-P(V_2-V_1)$

$=-(PV_2 -PV_1)=-(nRT_2-nRT_1)$

$=-mR(T_2-T_1)$

$=-1\times 1.987(323-298)\ cal$

$=-49.7\ cal$

$\Delta E=q+w$

$=124.2-49.7\ cal=74.5\ cal$ .

At $O^oC$ ice and water are in equilibrium and enthalpy change for the process:
${ H }_{ 2 }{ O }_{ (S) }\rightleftharpoons { H }_{ 2 }{ O\quad (l) }$ is 6.0 kJ $mol^{-1}$
Calculate the entropy change for this conversion of ice into liquid water.

2106353_1587929_ans_46c16aecc1794b9eb9e67523bb0c911e.jpg

An athlete in the weight room lifts a $50kg$ mass through a vertical distance of $2.0m$ . The mass is allowed to fall through the $2.0m$ distance while coupled to an electrical generator. The electrical generator produces an equal amount of electrical work which is used to produce aluminium by Hall electrolytic process.

${ Al }_{ 2 }{ O }_{ 3 }(\textrm{solution})+3C(\textrm{graphite})\rightarrow 2Al(l)+3CO(g);\Delta { G }^{ o }=593kJ$

How many times must the athlete lift the $50kg$ mass to provide sufficient Gibbs energy to produce $27g$ $Al$ ? $\left( g=10m/{ s }^{ 2 } \right)$

1958985_1679487_ans_e58fc82b97bd4562b752b415652cf395.jpg

The increase in Gibbs free energy (in kJ) of $13g$ of ethanol (density $=0.78g{cm}^{-3}$ ), when the pressure is increased isothermally from $1$ bar to $3001$ bar, is _____.

1961753_1679467_ans_dcea7eb940a74cc1a7f230fbd851e723.jpg

State and explain Hess's law of constant heat summation.

Hess's law of constant heat of summation states that regardless of the multiple stages or steps of a reaction the total enthalpy change of the reaction is the sum of all changes.

For

$R\rightarrow P$ ;

$\Delta H_r=\Delta H_1+\Delta H_2+\Delta H_3$ .

1495283_1699999_ans_b98484a22b3e4042968a1c30891f23c0.png

The first ionisation enthalpy of Li is 5.4 eV and the electron gain enthalpy of Cl is 3.6 eV. Calculate $\triangle H$ in kcal mol $^{-1}$ for the reaction,

$Li(g) +Cl(g) \rightarrow Li^++ Cl^-$

Carried out at such low pressure that resulting ions do not combine with each other.

1713384_1725691_ans_09d0508aef1c46ea9c09b91c5e278b46.jpg

Explain the term lattice energy as applied to an ionic solid. Calculate the lattic energy of caesium chloride using the following data:

Ammonium nitrate can decompose with explosion by the following reaction :
$NH_{4}NO_{3(s)}\rightarrow N_{2}O_{(g)} + 2H_2O_{(g)}; \, \Delta H = -37.0\ kJ$
Calculate the heat produced when $2.50\ g$ of $NH_{4}NO_{3}$ decomposes.

$\because 80 g$

$NH_{4}NO_{3}$ gives heat

$= 37.0\ kJ$

$\therefore 2.50g$

$NH_{4}NO_{3}$ gives heat

$=\dfrac{(37.0 \times 2.50)}{80} = 1.16\ kJ$

A cooking gas cylinder is assumed to contain $11.2 \ kg$ isobutane. The combustion of isobutane is given by:

$C_{4}H_{10(g)} + (13/ 2)O_{2(g)} \rightarrow 4CO_{2(g)} + 5H_{2}O_{(l)} ; \Delta H = -2658\ kJ$

(a) If a family needs $15000\ kJ$ of energy per day for cooking, how long would the cylinder last?
(b) Assuming that $30\%$ of the gas is wasted due to incomplete combustion, how long would the cylinder last?

(a)

$\because 58\ g$ isobutane provides energy

$= 2658 \ kJ$

$11.2\times 10^{3} g$ isobutane provides energy

$= \dfrac {26558\times 11.2\times 10^{3}}{58} kJ=513268.9\ kJ$

The daily requirement of energy

$= 15000\ kJ$

$\therefore$ Cylinder will last

$= \dfrac{513268.9}{15000} = 34\ days$

(b) Total loss of energy due to wastage

$= \dfrac{513268.9\times 70}{100}kJ$ is

$30\%$ and thus energy used for work

$\therefore$ Cylinder will last

$= \dfrac{513268.9\times 70}{100\times 15000} = 24\ days$

For the reaction, $C(s) +O_2(g) \rightarrow CO_2(g),$ $\Delta H ^{\circ}=-393.51\ kJ/mol$ and $\Delta S ^{\circ}=2.86\ J/mol. K$ at $25^{\circ}C$ . Does the reaction become more or less favourable as the temperature increases?

$C(s)+O_2(g)\rightarrow CO_2(g)$

$\Delta H^o=−393.51\ kJ/mol$

$\Delta S^o=2.86\ J/mol.K$ at

$T=25^oC$

$\Delta G^o=\Delta H^o−T\Delta S^o$

$\Delta G^o=−393.51\times 1000−298\times2.86=−394.36\ kJ<0$ favourable at

$25^oC$ as temp increases,

$\Delta G$ become more -ve so it becomes more favourable at the higher temperature.

Show how the Born-Haber cycle may be used to estimate the enthalpy of the hypothetical reaction:
$Ca(s)+ Cl_2(g)\to CaCl(s)$
explain why $CaCl(s)$ has never been made even though the enthalpy for this reaction is negative.

Calculate $\Delta G^0$ for the following reaction.
$Ag^+(aq)+Cl^-(aq)\rightarrow AgCl(s)$
Given: $\Delta G^0_f(AgCl)=-109$ kJ/mol
$\Delta G^0_f(Cl^-)=-129$ kJ/mol
$\Delta G^0_f(Ag^+)=77$ kJ/mol
Represent the given reaction in the form of a cell. Also calculate $E^0$ of the cell and find $log K_{sp}$ of $AgCl$ .

1624585_1741066_ans_75f8541d94df499589e9f1728647dcc9.jpg

What is the relationship between $\Delta G, \Delta H \ \& \ \Delta S$ ?

The relationship between

$\Delta G, \Delta H \ \& \ \Delta S$ is

$\Delta G = \Delta H \ -T \Delta S$

From the following data of AH, of the following reactions,
$C_{( s)}+\dfrac{1}{2}O_2(g) \longrightarrow CO_{(g)}$ ; $\Delta H = -110 \ kJA$
$C_{(s)} + H_2O_{(g)} \longrightarrow CO_{(g)} + H_{2(g)}; \Delta H = 132kJ$

Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at $1273\ K$ , keeping the temperature constant.

1763351_1746804_ans_ed6c1a0ce0074b2ea64a1a79faaf1822.PNG

The Born-Haber's cycle for rubidium chloride $(RbCl)$ is given above , ( the energies are in $kcal \,mol^{-1}$ .
Find out the electron affinity of chloride in $kJ\,mol^{-1}$ .

Heat of formotion = Heat of sublimation + Dessolution energy +Ionzation + energy + (electron affnity)

Substituting values from the question , we get

$\Delta H_{f} = \Delta H_{S} + \Delta H_{D} + \Delta IE + \Delta E_{a} + \Delta U$

$−105 = +20.5 + 28.75 + 96.0 + X − 159.5$

$\therefore X = -90.75 kcal/mol$

What is Heat capacity?

Heat capacity of a system is defined as quantity of heat required to raise temperature by

$1^{0}$ .

How non-spotaneous process be converted to spontaneous process?

$AG + \Delta H +TAS$ when the signs for

$\Delta H$ and

$\Delta S$ Are both same , then temperature determines that the process is spontaneous . when

$\Delta H$ is positive (unfavourable ) and

$\Delta S$ is positive (favourable) , high temperatures are needed so the favourable

$|Delta S$ term dominates making the process spontaneous

$( \Delta F < 0)$ . When

$\Delta H$ is negative (favourble) and

$\Delta S$ is negative (unfavourable) Low temperature are needed so the favourable

$\Delta H$ term dominates making the process spontaneous

$( \Delta G < 0 )$

At $0^o$ and $1 atm,$ heat absorbed by the system by melting of one mle of ice is $6.05 KJ/mol.$ The molar volume of ice and water are $0.0196 L$ and $0.0180 L$ respectively. Calculate $\Delta H$ and $\Delta U.$

Heat is absorbed here at constant pressure so

$\Delta H = 1440 cal$

$\Delta U$ can be calculated by

$\Delta U - \Delta H - P \Delta V$

$P = 1 atm \Delta V = 0.196 L - 0.180 L$

$= 0.0016 L$

$P \Delta V = 1 \times 0.0016 = 0.0016 L atm$

$= 0.0016 \times 24.217$ calories

$1 I atm = 24.217 calories$
So

$\Delta I = 1440 cal - 0.0016 \times 24.217 cal$

$= 1439.96$ calories
So

$\Delta H = \Delta U$

Taking a specific example show that $\Delta S_{total}$ is a criterion for spontaneity of a change.

$\Delta G=-T\Delta S_{total}$

$\Delta S_{total}$ is

$+ve$ ,

$\Delta G$ will be

$-ve$ , reaction will be spontaneous.

$\Delta S_{total}$ is

$-ve$ ,

$\Delta G$ will be

$+ve$ , reaction will be non-spontaneous.

$\Delta S_{total}=\Delta S_{syst}+\Delta S_{surroundings}$

For example in case of freezing of water at

$273K$ ,

$T\Delta S_{total}$ is

$+0.08J/K/mol$ , whereas at

$274K$ it is

$-0.080J/K/mol$ . Since the

$\Delta S_{total}$ is

$+ve$ at

$273K$ , freezing of water will take place. On the other hand at

$274K$ , the total entropy change is negative the freezing will not occur.

State and illustrate Hess's law.

Hess's law: Whether a chemical reaction takes place in a single step or in a several steps, the total change in enthalpy remains the same.

Consider the formation of

$CO_{2}$ from

$C$ .

I. Method (Single step):

$C(s)+O_{2}(g)\rightarrow CO_{2}(g)$

$\Delta H=-xkJ$ , Heat liberated =

$xkJ$ .

II. Method (Two steps):

$C(s)+1/2O_{2}(g)\rightarrow CO(g)$ ;

$\Delta H=-x_{1}kJ$

$CO(g)+1/2O_{2}(g)\rightarrow CO_{2}(g)$

$\Delta H=x_{2}kJ$

Total heat liberated =

$(x_{1}+x_{2})kJ$

According to Hess's law,

$x=x_{1}+x_{2}$

What you mean by Born-Haber cycle ?

Born-Haber cycle is a cycle of operation needed for the formation of an ionic crystalline solid from its gaseous ions as an element.

What is Thermodynamics? Explain the Zeroth law of Thermodynamics and give its importance.

It is based on the principle of temperature.
To understand this law, consider two systems

$A$ and

$B$ , separated by an adiabatic wall, each of these are in contact with a third system

$C$ by a conducting wall. The states of the system will change until

$A$ and

$B$ are in equilibrium with

$C$ . Now, let the adiabatic wall between

$A$ and

$B$ be replaced by a conducting wall and

$C$ is insulated from

$A$ and

$B$ by an adiabatic wall. It is found that

$A$ and

$B$ are in thermal equilibrium with each other. This is the basis of Zeroth law of thermodynamics. This law states that the two systems are in thermal equilibrium with a third system separately, then both are in thermal equilibrium with each other.
The zeroth law apparently suggests that when two systems

$A$ and

$B$ are in thermal equilibrium, then , there must be a physical quantity that has the same value for both. The thermodynamic variable whose value is equal for the two systems (in thermal equilibrium) is called temperature

$(T)$ . If

$A$ and

$B$ are separately in equilibrium with

$C$ , then,

$T_A=T_C$ and

$T_B=T_C-$ This means that

$T_A=T_B$ . So, the systems

$A$ and

$B$ are in thermal equilibrium.
1755898_1833383_ans_de9571862794413bb56256ae8e2d9152.png

Enthalpies of formation of $CO(g),CO_2(g), NaO(g)$ and $N_2O_4(g)$ are $-110,-393.81$ and $9.7 kJ \quad mol^{-1}$ respectively. Find the value of $\Delta H$ for the reaction
$N_2O_{3(g)} \rightarrow N_2O_{(g)}+3CO_{(g)}$

$\Delta rH$ for a reaction is calculated as the difference between

$\Delta_fH$ (heat of formation) va;ue of products and

$\Delta_fH$ (heat of formation) value of reactants . for the given reaction,

$N_2O_{4(g)} + 3CO_{(g)} \rightarrow N_2O_{2(g)} + 3CO_{ 2(g) }$

Substituting the values of

$\Delta_f H$ for

$N_2O, CO_2, N_2O_4$ and

$CO$ from the question , we get :

$\Delta rH = [ \left\{ \Delta_f H(N_2O) + 3 \Delta_f H (CO_) \right\} ] - [ \left\{ \Delta_f H (N_2O_4) + 3 \Delta_f H (CO) \right\} ]$

$= [ \left\{ (81) + 3( -393) \right\} ] - [ \left\{ ( 9.7) + 3( -110) \right\} ]$

$= -777.7 kJ mol^{-1}$

For the reaction
$2A_{(g)}+B_{(g)} \rightarrow 2D_{(g)}$
$\Delta U^o =-10.5 kJ$ and $\Delta S^o =-44.1 JK_{-1}$
Calculate $\Delta G^o$ for the reaction and predict whether the reaction may occur spontaneously.

For the given reaction

$2A_{(g)} + B_{(g)} \longrightarrow 2D_{(g)}$

$\Delta n_g = 2-3 = -1 mole$

Substituting the value of

$\Delta U^{ \theta}$ in the expansion of

$\Delta H$ :

$\Delta H^{ \theta} = \Delta U^{ \theta} + \Delta n_g RT$

$= ( -10.5 kJ) +( +1)( 8.314 \times 10^{-3} kJ K^{-1}) ( 298 K)$

$= -10.5 kJ -2.48 kJ$

$= -12.98 kJ$

When we substitute the values of

$\Delta H^{ \theta }$ and

$\Delta S^{ \theta}$ in the expression of

$|delta G^{\theta}$ then according to Gibbs' Helmholtz equation

$\Delta G^{ \theta} = \Delta H^{ \theta} - T \Delta S^{ \theta}$

$= -12.98 kJ -(298 K) ( -44.1 J K^{-1})$

$= -12.98 kJ -( 13.14kJ )$

$\Delta G^{ \theta} = 0.16 kJ$

Since

$\Delta G^{ \theta}$ for the reaction is positive , the reaction will not occur spontaneously .

The reaction of cyanamide, $NH_2CN(s),$ with dioxygen was carried out in a bomb clorimeter, and $\Delta U$ was found to be $-742.7 kJ mol^{-1}$ at $298 K.$ Calculate enthalpy change for the reaction at

$298 K. NH_2CN(g) + 3/2O_2(g) \rightarrow N_2(g)+CO_2(g)+H_2O(l)$

Entahlpy change for a reaction

$( \Delta H )$ is given by the expressuon,

$\Delta H = \Delta U + \Delta n_gRT$
Where,

$\Delta U =$ change in internal energy

$\Delta n_g$ = change in number of moles
For the given reaction

$\Delta n_g = \Sigma n_g$ (products) -

$\Sigma n_g$ ( reactants)

$= 9 2- 2.5 )$ moles

$\Delta U = -0.5$ moles
And

$\Delta U = -742.7 kJ mol^{-1}$

$T = 298 K$

$R = 8.314 \times 10^{-3} kJ mol^{-1}$
Substituting the values in the expression of

$\Delta H$

$\Delta H = ( -742.7 kJ mol^{-1}) + ( -0.5 mol) ( 298 K) ( 8.314 \times 10^{-3} mol^{-1}K^{-1}) = -742.7 -1.2$

$\Delta H = -743.9 kJ mol^{-1}$

Explain heat, work and internal energy of a system.

Temperature determines the direction of flow of heat when two bodies are in thermal contact. Heat flows from the body at a higher temperature to a lower temperature. The flow stops when the two bodies are in thermal equilibrium. This type of heat transfer is a non-mechanical energy transfer.
When thermal energy is transferred from an object into the atmosphere then the thermal energy of the object is taken to be negative while on giving heat on an object from the atmosphere the thermal energy of the object is taken to be positive. If the temperature of the object remains constant due to the exchange of heat in the object the state of the object changes.

A chemist while studying the properties of gaseous $CCl_2F_2$ , a chlorofluorocarbon refrigerant cooled a $1.25\ g$ sample at constant atmospheric pressure of $1.0\ atm$ from $320\ K$ to $293\ K$ . During cooling, the sample volume decreased from $274$ to $248\ mL$ . Calculate $\Delta H$ and $\Delta U$ for the chorofluorocarbon for this process. For $CCl_2F_2, C_p \simeq 80.7\ J/(mol\ K)$ .

$\Delta H=q_p$ and

$C_p$ is heat evolved or absorbed per mole for

$1^o$ fall or rise in temperature. Here, fall in temperature

$=320-293=27\ K$ .
Molar mass of

$CCl_2F_2=12+2\times 35.5+2\times 19$

$=121\ g\ mol^{-1}$

$\therefore$ Heat evolved from

$1.25\ g$ of the sample on being cooled from

$320\ K$ to

$293\ K$ at constant pressure

$=\dfrac {80.7}{121}\times 1.25\times 27\ J=22.51\ J$
Further

$\Delta H=\Delta U+P\Delta V=-22.51\ J$

$P \Delta V=1\ atm \times \dfrac {(248-274)}{1000}L=-0.026\ L\ atm$

$=-0.26\times 101.325\ J=-2.63\ J$

$\therefore -22.51=\Delta U-2.63\ J$
or

$\Delta U=-22.51+2.63\ J=-19.88\ J$

In a gobar gas plant, gobar gas is obtained by bacterial fermentation of animal refuse. The main combination gas present in the gobar gas is found to be methane ( $80\%$ by weight )whose heat of combustion is $809\ mol^{-1}$ . How much gobar gas would have to produced per day for a village of $100$ families if average consumption of a family is $20.000\ kJ$ per day to meet all its energy requirements.

Total energy requirement for

$100$ families per day

$=20,000\times 100\ kJ=2\times 10^6 kJ$
Methane

$(CH_4)$ undergoes combination as fllows:

$CH_4 (g)+2O_2(g)\to CO_2(g)+2H_2O(l), \Delta H=-809\ kJ\ mol^{-1}$
For

$809\ kJ$ of energy,

$CH_4$ required

$=16\ g$
(Molar mass of

$CH_4=16\ g\ mol^{-1}$ )
For

$2\times 10^6\ kJ$ of energy

$CH_4$ required

$=\dfrac {16}{809}\times 2\times 10^6g =39.56\ kg$
As gobar contains

$80\%$ by weight of methane, therefore gobar gas required

$=\dfrac {100}{80}\times 39.56\ kg$

$=49.45\ kg$

Reaction	$\displaystyle \Delta _{r}H^{\circ}$ ( kJ / mole )
$\displaystyle Fe_{2}O_{3}(s)+3C (graphite)\rightarrow 2Fe(s)+3CO(g)$	492.6
$\displaystyle FeO(s)+C (graphite)\rightarrow Fe(s)+CO(g)$	155.8
$\displaystyle C (graphite)+O_{2}(g)\rightarrow CO_{2}(g)$	$\displaystyle -$ 393.51
$\displaystyle CO(g)+1/2\: O_{2}(g)\rightarrow CO_{2}(g)$	$\displaystyle -$ 282.98

Thermodynamics - Class 11 Medical Chemistry - Extra Questions

The standard heat of formation (ΔfH0298)(\Delta_f H^0_{298}) of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite are −1560,−393.5-1560, -393.5 and −286 kJ/mol-286\ kJ/mol, respectively is ______.

The degree of freedom of a three atomic molecule can not be 5. true=1, false=0

What is meant by an equilibrium?

The specific heat of a gas is 0.125 and 0.075 cal/g. The value of its molar mass is ______.

Thermochemical equations can be reversed by changing the sign of \Delta H\Delta H.If true enter 1, else enter 0.

\displaystyle \Delta G=\Delta H-T\Delta S\displaystyle \Delta G=\Delta H-T\Delta S.If true enter 1, else enter 0.

Which has more heat capacity: a bucket full of water or a mug full of water?

Given equal masses of aluminium and copper, which would require more heat if both are heated to the same temperature. Justify your answer.[Specific heat capacity of Al= 899 Jkg^{-1} C^{-1}Al= 899 Jkg^{-1} C^{-1} and Cu= 387 Jkg^{-1} C^{-1}Cu= 387 Jkg^{-1} C^{-1}]

Differentiate between heat capacity and specific heat capacity.

Why is water used as a coolant in car radiations and factories to keep the engine and other parts cool?

What is the quantity of heat released per kg of water per 1^o^oC fall in temperature?

Calculate the heat energy that will be released when 5.0 kg of steam at 100^o^oC condenses to form water at 100^o^oC. Express your answer in S.I. unit (specific heat of vapourisation of steam is 2268 kJ/kg)

Discuss the change in energy and arrangement of molecules on increasing the temperature of ice from -5^o^oC to 10^o^oC at 1 atm pressure.

Water falls from a height of 50 cm. Calculate the rise in temperature of water when it strikes the bottom. (g = 10 ms^{-2}g = 10 ms^{-2}, sp. heat capacity of water =4200 J/kg^oC=4200 J/kg^oC)

State Hess's law of constant heat summation and explain it with an example.

State Hess's law.

A gas contained in a cylinder is filled with a frictionless piston expands against a constant pressure 11 atmosphere from a volume of 44 litre to a volume of 1414 litre. In doing so, it absorbs 800J800J thermal energy from surroundings. Determine \Delta U\Delta U for the process.

In the reaction{ A }^{ + }+B\longrightarrow A+{ B }^{ + }{ A }^{ + }+B\longrightarrow A+{ B }^{ + }, there is no entropy change. If enthalpy change is 22kJ22kJ of {A}^{+}{A}^{+}, calculate \Delta G\Delta G for the reaction.

Calculate the temperature at which liquid water will be in equilibrium with water vapour.\Delta { H }_{ vap }=40.73\ kJ{ mol }^{ -1 }\Delta { H }_{ vap }=40.73\ kJ{ mol }^{ -1 } and \Delta { S }_{ vap }=0.109\ kJ{ mol }^{ -1 }{ K }^{ -1 }\Delta { S }_{ vap }=0.109\ kJ{ mol }^{ -1 }{ K }^{ -1 }

Commercial sample of H_2O_2H_2O_2 is labeled as 10\ Volume10\ Volume. Its % strength is nearly:

The following data is known about the melting of KClKCl:\Delta H=7.25kJ\quad { mol }^{ -1 };\Delta S=+0.007KJ{ K }^{ -1 }{ mol }^{ -1 }\Delta H=7.25kJ\quad { mol }^{ -1 };\Delta S=+0.007KJ{ K }^{ -1 }{ mol }^{ -1 }Calculate its melting point.

The Born-Haber cycle for formation of rubidium chloride (RbClRbCl)is given below (the enthalpies are in kCalkCal {mol}^{-1}{mol}^{-1}).Find the value of XX.

Calculate the heat change during the reaction 2424g C and 128128g S following :the change C+S_2\rightarrow CS_2C+S_2\rightarrow CS_2; \Delta\DeltaH=22=22K cal.

For a chemical reaction, \Delta H\Delta H and \Delta S\Delta S are negative. State, giving reason, under what conditions this reaction is expected to occur spontaneously.

For a reaction both enthalpy change and entropy change are positive. Under what conditions the reaction will be spontaneous?

For the reaction:2A(g)+B(g)\rightarrow 2D(g)2A(g)+B(g)\rightarrow 2D(g), \Delta U^{\ominus}=-10.5\Delta U^{\ominus}=-10.5 kJ and \Delta S^{\ominus}=-44.1\Delta S^{\ominus}=-44.1 JK^{-1}JK^{-1}.Does the reaction occur spontaneously?[Give your answer in terms of Yes or No]

How do you define specific heat ? What is specific heat of water ?

Explain Third law of thermodynamics.

Many of the photochemical changes have positive sign of \delta G\delta G, yet they are spontaneous. Why?

What is specific heat?

If 1.0 kcal1.0 kcal of heat is added to 1.2 L1.2 L of { O }_{ 2 }{ O }_{ 2 } in a cylinder at constant pressure of 11 atm, the volume increase to 1.5 L1.5 L. Calculate \Delta U\Delta U and \Delta H\Delta H of the process (1 L-atm=100 J)(1 L-atm=100 J)

State and explain Hess's law of constant heat summation? Give its application.

Derive relationship between H and U.

Write the importance of Hess's law of constant heat summation.

A gas cylinder contains 14.2 Kg butane gas. it is consumption of energy in a family is 10000 KJ for cooking purposes how long this cylinder will be sufficient to supply. If the enthalpy of combustion of butane is 2658 KJ mol

How it is possible to find the value of \Delta U\Delta U and not the value of UU

Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?

A chemical reaction is spontaneous at 298K and non-spontaneous at 350K. Which of the following is correct for the reaction?s.no\Delta G\Delta G\Delta H\Delta H\Delta S\Delta SA)--+B)+++C)-+-D)+-+E)---

When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by 10^010^0C. What is the specific heat of the substance?

Explain why:Ice at zero degrees centigrade has a greater cooling effect than water at 0 ^\circ C0 ^\circ C.

What is the need to heat solution ?

At 298 \mathrm{K}298 \mathrm{K}. \mathrm{K}_{\mathrm{p}}\mathrm{K}_{\mathrm{p}}for the reaction N_{2} O_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)N_{2} O_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) is0.98 . Predict whether the reaction is spontaneous or not.

Given that \Delta H=0\Delta H=0 for mixing of two gases. Explain whether the diffusion of these gasesinto each other in a closed container is a spontaneous process or not?

Explain the meaning of thermochemical equation for the reaction:2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)+136kcal2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)+136kcal.

What is spontaneous process? Give example.

Justify the following statements:(a) An exothermic reaction is always thermodynamically spontaneous.(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.

Evaporation of water is an endothermic process but spontaneous. Explain.

What is meant by entropy driven reaction? How can a reaction with positive changes of enthalpy and entropy be made entropy driven?

Define non-spontaneous process.

What are the applications of Hess's law of constant heat summation?

What is meant by the free energy of a system? What will be the direction of chemical reaction when(i) \Delta G=0\Delta G=0(ii) \Delta G>0\Delta G>0(iii) \Delta G<0\Delta G<0

Give Hess's Law of constant Heat summetion and write its uses.

What will be the sign of G G for melting of ice at 267 K 267 K and 276 K 276 K ? (Melting point of ice= 272 K 272 K ).

Explain the steps involved in Born Haber cycle for the formation of NaCl.

Why most of the exothermic processes (reactions) are spontaneous?

\Delta H^{\circ}_f\Delta H^{\circ}_f of Fe_2O_3Fe_2O_3 and Al_2O_3Al_2O_3 are -189 kJ and -405 kJ respectively. How much heat (in kJ) is given out during reaction of 1 g Al according to 2Al +Fe_2O_3\longrightarrow Al_2O_3 +2Fe2Al +Fe_2O_3\longrightarrow Al_2O_3 +2Fe ?

If heat of formation of CaCl_2CaCl_2 and NaClNaCl are 191 and 97.5 kcal, the heat of reaction for CaCl_2 + 2Na \longrightarrow 2NaCl + CaCaCl_2 + 2Na \longrightarrow 2NaCl + Ca is:

Match the Following:

List equation/law (in Column I) with statement (in Column II).Match the following.

The equilibrium constant for a reaction iswhat will be value of \displaystyle \bigtriangleup G^{\ominus } \displaystyle \bigtriangleup G^{\ominus } ? R =8.314 \displaystyle JK^{-1} mol^{-1} \displaystyle JK^{-1} mol^{-1} , T=300 K

Calculate E_{cell}E_{cell} and \Delta\DeltaG for the following at 28^o28^oCMg(s)+Sn^{2+}(0.025M)\rightarrow Mg^{2+}(0.06M)+Sn_{(s)}Mg(s)+Sn^{2+}(0.025M)\rightarrow Mg^{2+}(0.06M)+Sn_{(s)}E^0_{cell}=2.23E^0_{cell}=2.23VIs the reaction spontaneous?

Give the characteristics of free energy.

For a chemical reaction the values of \Delta H\Delta H and \Delta S\Delta S at 400 K are -10 K cal mol^{-1}^{-1} and 20 cal. deg^{-1}^{-1} mol^{-1}^{-1} respectively. Calculate the value of \Delta G\Delta G of the reaction.

Calculate q.w.\Delta Uq.w.\Delta U and \Delta H\Delta H for the reversible isothermal expansion of one mole of an ideal gas at {127}^{o}C{127}^{o}C from a volume of 10\ {dm}^{3}10\ {dm}^{3} to 20\ {dm}^{3}20\ {dm}^{3}.

Explain spontaneous and non-spontaneous processes, giving two examples of each.

One mole of ethane was burnt in oxygen at constant volume to give C{O}_{2}C{O}_{2} at 298 K. The heat evolved was found to be 1554 kJ. Calculate the heat of reaction at constant pressure.

The volume of a gas at STP is 2 L, 300 J heat is given to it, as a result of which volume changes to 2.5 L at 1 atm. Calculate the change in internal energy of the system.

A heat engine absorbs 760kJ760kJ heat from a source at 380K380K. It rejects (i) 650kJ650kJ, (ii) 560kJ560kJ, (iii) 504kJ504kJ of heat to sink at 280K280K. State which of these represent a reversible, an irreversible or an impossible cycle.

If the molar heat capacity of O_2O_2 gas is 640Jk^{-1}mol^{-1}640Jk^{-1}mol^{-1}. Then calculate its specific heat capacity.

A gas expands from 14 liters to 19 liters against a constant pressure of 2 atm. During the process 600 J of heat flows in the gas from the surroundings. Calculate the internal energy change of its system.

Calculate w and \Delta U\Delta U for the conversion of 0.5 mol of water at 100-degree celsius to steam at 1atm pressure. The heat of vaporization of water at 100degree celsius is 40670 J/mol.

During an adiabatic process, the pressure of a gas is found to be proportional to cube of its absolute temperature. Calculate the Poisson's ratio of gas.

Work done by a sample of an ideal gas in a process A double the work done in another process B. The temperature rises through the same amount in the two processes. If {C_A}{C_A} and {C_B}{C_B} are the molar heat capacities for the two processes, then the relation between C_AC_A and C_BC_B will be:

Calculate \Delta U\Delta U and \Delta H\Delta H in calories if 1 mole of a monoatomic ideal gas is heated at constant pressure of 1 atm from 25^oC25^oC to 50^oC50^oC.

The increase in Gibbs free energy (in kJ) of 13g13g of ethanol (density =0.78g{cm}^{-3}=0.78g{cm}^{-3}), when the pressure is increased isothermally from 11 bar to 30013001 bar, is _____.

The standard heat of formation $(\Delta_f H^0_{298})$ of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite are $-1560, -393.5$ and $-286\ kJ/mol$ , respectively is ______.

Thermochemical equations can be reversed by changing the sign of $\Delta H$ .

If true enter 1, else enter 0.

$\displaystyle \Delta G=\Delta H-T\Delta S$ .

If true enter 1, else enter 0.

Given equal masses of aluminium and copper, which would require more heat if both are heated to the same temperature. Justify your answer.
[Specific heat capacity of $Al= 899 Jkg^{-1} C^{-1}$ and $Cu= 387 Jkg^{-1} C^{-1}$ ]

What is the quantity of heat released per kg of water per 1 $^o$ C fall in temperature?

Calculate the heat energy that will be released when 5.0 kg of steam at 100 $^o$ C condenses to form water at 100 $^o$ C. Express your answer in S.I. unit (specific heat of vapourisation of steam is 2268 kJ/kg)

Discuss the change in energy and arrangement of molecules on increasing the temperature of ice from -5 $^o$ C to 10 $^o$ C at 1 atm pressure.

Water falls from a height of 50 cm. Calculate the rise in temperature of water when it strikes the bottom. ( $g = 10 ms^{-2}$ , sp. heat capacity of water $=4200 J/kg^oC$ )

A gas contained in a cylinder is filled with a frictionless piston expands against a constant pressure $1$ atmosphere from a volume of $4$ litre to a volume of $14$ litre. In doing so, it absorbs $800J$ thermal energy from surroundings. Determine $\Delta U$ for the process.

In the reaction
${ A }^{ + }+B\longrightarrow A+{ B }^{ + }$ ,
there is no entropy change. If enthalpy change is $22kJ$ of ${A}^{+}$ , calculate $\Delta G$ for the reaction.

Calculate the temperature at which liquid water will be in equilibrium with water vapour.
$\Delta { H }_{ vap }=40.73\ kJ{ mol }^{ -1 }$ and $\Delta { S }_{ vap }=0.109\ kJ{ mol }^{ -1 }{ K }^{ -1 }$

Commercial sample of $H_2O_2$ is labeled as $10\ Volume$ . Its % strength is nearly:

The following data is known about the melting of $KCl$ :
$\Delta H=7.25kJ\quad { mol }^{ -1 };\Delta S=+0.007KJ{ K }^{ -1 }{ mol }^{ -1 }$
Calculate its melting point.

The Born-Haber cycle for formation of rubidium chloride ( $RbCl$ )is given below (the enthalpies are in $kCal$ ${mol}^{-1}$ ).
Find the value of $X$ .

Calculate the heat change during the reaction $24$ g C and $128$ g S following :the change $C+S_2\rightarrow CS_2$ ; $\Delta$ H $=22$ K cal.

For a chemical reaction, $\Delta H$ and $\Delta S$ are negative. State, giving reason, under what conditions this reaction is expected to occur spontaneously.

For the reaction:

$2A(g)+B(g)\rightarrow 2D(g)$ ,

$\Delta U^{\ominus}=-10.5$ kJ and $\Delta S^{\ominus}=-44.1$ $JK^{-1}$ .Does the reaction occur spontaneously?
[Give your answer in terms of Yes or No]

Many of the photochemical changes have positive sign of $\delta G$ , yet they are spontaneous. Why?

If $1.0 kcal$ of heat is added to $1.2 L$ of ${ O }_{ 2 }$ in a cylinder at constant pressure of $1$ atm, the volume increase to $1.5 L$ . Calculate $\Delta U$ and $\Delta H$ of the process $(1 L-atm=100 J)$

How it is possible to find the value of $\Delta U$ and not the value of $U$

A chemical reaction is spontaneous at 298K and non-spontaneous at 350K. Which of the following is correct for the reaction?
s.no
$\Delta G$
$\Delta H$
$\Delta S$
A)
-
-
+
B)
+
+
+
C)
-
+
-
D)
+
-
+
E)
-
-
-

When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by $10^0$ C. What is the specific heat of the substance?

Explain why:
Ice at zero degrees centigrade has a greater cooling effect than water at $0 ^\circ C$ .

At $298 \mathrm{K}$ . $\mathrm{K}_{\mathrm{p}}$
for the reaction $N_{2} O_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)$ is
0.98 . Predict whether the reaction is spontaneous or not.

Given that $\Delta H=0$ for mixing of two gases. Explain whether the diffusion of these gases
into each other in a closed container is a spontaneous process or not?

Explain the meaning of thermochemical equation for the reaction:
$2H_{2}(g)+O_{2}(g)\rightarrow 2H_{2}O(l)+136kcal$ .

Justify the following statements:
(a) An exothermic reaction is always thermodynamically spontaneous.
(b) The entropy of a substance increases on going from liquid to vapour state at any temperature.

What is meant by the free energy of a system? What will be the direction of chemical reaction when
(i) $\Delta G=0$
(ii) $\Delta G>0$
(iii) $\Delta G<0$

What will be the sign of $G$ for melting of ice at $267 K$ and $276 K$ ? (Melting point of ice= $272 K$ ).

$\Delta H^{\circ}_f$ of $Fe_2O_3$ and $Al_2O_3$ are -189 kJ and -405 kJ respectively. How much heat (in kJ) is given out during reaction of 1 g Al according to
$2Al +Fe_2O_3\longrightarrow Al_2O_3 +2Fe$ ?

If heat of formation of $CaCl_2$ and $NaCl$ are 191 and 97.5 kcal, the heat of reaction for $CaCl_2 + 2Na \longrightarrow 2NaCl + Ca$ is:

List equation/law (in Column I) with statement (in Column II).
Match the following.

The equilibrium constant for a reaction iswhat will be value of $\displaystyle \bigtriangleup G^{\ominus }$ ? R =8.314 $\displaystyle JK^{-1} mol^{-1}$ , T=300 K

Calculate $E_{cell}$ and $\Delta$ G for the following at $28^o$ C

$Mg(s)+Sn^{2+}(0.025M)\rightarrow Mg^{2+}(0.06M)+Sn_{(s)}$

$E^0_{cell}=2.23$ V

Is the reaction spontaneous?

For a chemical reaction the values of $\Delta H$ and $\Delta S$ at 400 K are -10 K cal mol $^{-1}$ and 20 cal. deg $^{-1}$ mol $^{-1}$ respectively. Calculate the value of $\Delta G$ of the reaction.

Calculate $q.w.\Delta U$ and $\Delta H$ for the reversible isothermal expansion of one mole of an ideal gas at ${127}^{o}C$ from a volume of $10\ {dm}^{3}$ to $20\ {dm}^{3}$ .

One mole of ethane was burnt in oxygen at constant volume to give $C{O}_{2}$ at 298 K. The heat evolved was found to be 1554 kJ. Calculate the heat of reaction at constant pressure.

A heat engine absorbs $760kJ$ heat from a source at $380K$ . It rejects (i) $650kJ$ , (ii) $560kJ$ , (iii) $504kJ$ of heat to sink at $280K$ . State which of these represent a reversible, an irreversible or an impossible cycle.

If the molar heat capacity of $O_2$ gas is $640Jk^{-1}mol^{-1}$ . Then calculate its specific heat capacity.

Calculate w and $\Delta U$ for the conversion of 0.5 mol of water at 100-degree celsius to steam at 1atm pressure. The heat of vaporization of water at 100degree celsius is 40670 J/mol.

Work done by a sample of an ideal gas in a process A double the work done in another process B. The temperature rises through the same amount in the two processes. If ${C_A}$ and ${C_B}$ are the molar heat capacities for the two processes, then the relation between $C_A$ and $C_B$ will be:

Calculate $\Delta U$ and $\Delta H$ in calories if 1 mole of a monoatomic ideal gas is heated at constant pressure of 1 atm from $25^oC$ to $50^oC$ .

The increase in Gibbs free energy (in kJ) of $13g$ of ethanol (density $=0.78g{cm}^{-3}$ ), when the pressure is increased isothermally from $1$ bar to $3001$ bar, is _____.