Given the following data: Substrate $$\Delta H^o$$(kJ/mol) $$S^o$$(J/mol K) $$\Delta G^o$$ (kJ/mol) FeO(s) $$-266.3$$ $$57.49$$ $$245.12$$ C(Graphite) $$0$$ $$5.74$$ $$0$$ Fe(s) $$0$$ $$27.28$$ $$0$$ CO(g) $$-110.5$$ $$197.6$$ $$-137.15$$ Determine at what temperature the following reaction spontaneous? $$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$$

$$\Delta G^o$$ is $$+$$ve, hence the reaction will never be spontaneous

Given the following data: Substance $$\Delta H^{o}(kJ/mol)$$ $$\Delta S^{o}(kJ/mol)$$ $$\Delta G^{o}(kJ/mol)$$ $$FeO(s)$$ $$-266.3$$ $$57.49$$ $$-245.12$$ $$C$$(Graphite) $$0$$ $$5.74$$ $$0$$ $$Fe(s)$$ $$0$$ $$27.28$$ $$0$$ $$CO(g)$$ $$-110.5$$ $$197.6$$ $$-137.15$$ Determine at what temperature the following reaction is spontaneous? $$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$$

$$\Delta G^{o}$$ is +ve the reaction will never be spontaneous

Given the following data: Substance $$\triangle H^0 (kJ/mol)$$ $$S^0 (J/molK)$$ $$\triangle G^0(kJ/mol)$$ FeO(s) $$-266.3$$ $$57.49$$ $$-245.12$$ C(Graphite) 0 $$5.74$$ 0 Fe(s) 0 $$27.28$$ 0 CO(g) $$-110.5$$ $$197.6$$ $$-137.15$$ Determine at what temperature the following reaction is spontaneous? $$FeO(s) + C(Graphite) \rightarrow Fe(s) + CO(g) $$

$$ \triangle G^0$$ is +ve, hence the reaction will never be spontaneous.

MCQ Questions for Class 11 Medical Chemistry Thermodynamics Quiz 13

The value of

$\triangle G$ for the process

$H_{2}O(s) \rightarrow H_{2}O(l)$ at

$1\ atm$ and

$260\ K$ is:-

0%
$< 0$
0%
$= 0$
0%
$> 0$
0%
Unpredictable

Statement I: In adsorption process, the value of

$\Delta H$ is always negative.
Statement II: During adsorption surface area of adsorbent decreases.
Which of the above statement is/are true? Choose the correct option.

0%
Only I
0%
Only II
0%
I and II
0%
None of these

Following reaction occurs at

${25}^{o}C$ :

$2NO(g, 1\times { 10 }^{ -5 }atm)+{ Cl }_{ 2 }(g, 1\times { 10 }^{ -2 }atm)\rightleftharpoons 2NOCl\ \left( g, 1\times { 10 }^{ -2 }atm \right)$

$\Delta { G }^{ o }$ is_______________.

0%
$-45.65kJ$
0%
$-28.53kJ$
0%
$-22.82kJ$
0%
$-57.06kJ$

Explanation

$\underset { 1 \times 10^{-5} atm}{2NO}+\underset {1 \times 10^{-2}atm}{Cl_2} \rightleftharpoons \underset { 1 \times 10^{-2} atm}{2NOCl}$

$K_p= \cfrac {(P\quad NOCl)^2}{(P\quad NO)^2(P\quad Cl_2)}$

$=\cfrac {(1 \times 10^{-2})^2}{(1 \times 10^{-5})^2\times 1 \times 10^{-2}}$

$=\cfrac { 1 \times 10^{-2}}{1 \times 10^{-10}}=1 \times 10^8$

$\Delta u= -RTln K_p= -8.314 \times 298 \times ln 10^8$

$= -8.314 \times 298 \times 8 \times ln10$

$= -8.314 \times 298 \times 8 \times 2.3$

$= -45.587 J/Kg$

$=-45.6 kJ$ .

For the process

${ H }_{ 2 }O(1,1bar,373.15k)\rightarrow { H }_{ 2 }O(g,1bar,373.15k)$ , the correct set of thermodynamic parameters is:

0%
$\Delta G\ =\ 0,\ \Delta S\ =\ +\ ve$
0%
$\Delta G\ =\ 0,\ \Delta S\ =\ -ve$
0%
$\Delta G\ =\ +ve,\ \Delta S\ =\ 0$
0%
$\Delta G\ =\ -ve,\ \Delta S\ =\ +ve$

When two moles of Hydrogen atoms join together to form a mole of hydrogen molecules in closed rigid vessel with diathermic walls:

$H(g)+H(g)\longrightarrow { H }_{ 2 }(g)$

0%
$w< 0$
0%
$\Delta U=negative$
0%
${ q }_{ system }=positive\quad$
0%
${ q }_{ surrounding }=negative\quad$

Explanation

$H_{2}$ has two atoms. So, a mole of

$H_{2}$ should have 1 mole of

$H_{2}$ molecules, and when broken down into atoms, 2 moles of atoms.

$H(g)+H(g)\rightarrow H_{2}(g)$ .

So, from the above reaction

$w<0$ .

The difference between the heat of reaction at constant pressure and constant volume for the reaction given below at

${25}^{o}C$ in KJ is:

$\quad 2{ C }_{ 6 }{ H }_{ 6(l) }+15{ O }_{ 2(g) }\longrightarrow 12{ CO }_{ 2(g) }+6{ H }_{ 2 }{ O }_{ (l) }$

0%
$-7.43$
0%
$+3.72$
0%
$-3.72$
0%
$+7.43$

Explanation

The difference between heats of reaction at constant pressure and constant volume for the following reaction at

$25^oC$ in kJ

$2C_6H_{6_(l)}+15O_{2_(g)}\longrightarrow12CO_{2_(g)}+6H_2O_{(l)}$

$\Delta n=12-15=-3$

$\Delta H=\Delta E+RT\Delta n$

$\Delta H-\Delta E=RT\Delta n$

$=(-3)\times8.314\times10^{-3}\times298$

$=-7.443\ kJ/mole$

For reversible isothermal expansion of one mole of an ideal gas at

$300 K$ , from a volume of

$10 L$ to

$20 L$ ,

$\Delta H$ is :

0%
$1.73 kJ$
0%
$-1.73 kJ$
0%
$3.46 kJ$
0%
zero

The normal boiling point of a liquid A is 350 K.

$\Delta { H }_{ vap }$ at normal boiling point is 35 KJ/mole. Pick out the correct statement(s). (Assume

$\Delta { H }_{ vap }$ to be independent of pressure).

0%
$\Delta { S }_{ vaporisation }$ > 100 J/K mole at 350 K and 0.5 atm
0%
$\Delta { S }_{ vaporisation }$ < 100 J/K mole at 350 K and 0.5 atm
0%
$\Delta { S }_{ vaporisation }$ < 100 J/K mole at 350 K and 2 atm
0%
$\Delta { S }_{ vaporisation }$ = 100 J/K mole at 350 K and 2 atm

Explanation

The equilibrium at the boiling point is

$A(1)\leftrightharpoons A(g); \Delta_{vap}H=35\ kJ/mol$

$\Delta_{vap}S=\frac{\Delta_{vap}H}{350 K}=100\ J/Kmol$

At 350 K and 0.5 atm

The boiling point

$(T_{vap})$ will be lower at the decreased pressure.

Since

$T_{vap}<35K, \Delta_{vap}S>100\ J/Kmol$

Mayuri was performing thermometric titration and she took 100 ml of

$1 M$ sulphuric acid and started adding

$1 M$ calcium hydroxide. When she plotted a graph of temperature vs volume of the titrant added, she found that the temperature was initially increasing and then it started decreasing. The maximum of the graph is obtained at 100 ml of calcium hydroxide. What will be the enthalpy change of this reaction?

[Given:

$\Delta H\ = -13.7 kcal/eq$ ]

0%
-13.7 kcal
0%
-27.4 kcal
0%
-1.37 kcal
0%
-2.74 kcal

The lattice energy of

$CsI(s)$ and the enthalpy of solution is

$33\ kJ/mol$ . Calculate the enthalpy of hydration

$(kJ)$ of 0.65 moles of

$CsI$ .

0%
$738 kJ$
0%
$10 kJ$
0%
$-371 kJ$
0%
$-822 kJ$

The heat liberated on complete combustion of 1 mole of

${ CH }_{ 4 }$ gas to

${ CO }_{ 2 }\left( g \right)$ and

${ HO }_{ 2 }\left( l \right)$ is 890 KJ. Calculate the heat evolved by 2.4 L of

${ CH }_{ 4 }$ on complete combustion.

0%
95.3 KJ
0%
8900 KJ
0%
89 KJ
0%
8.9 KJ

Explanation

Given

The volume of

$!\ mol$ of methane at

$298\ K$ and

$1\ atm$ pressure is

$24L$ . Let's calculate the moles of methane for it's

$2.4L$ as

$=2.4\left(\frac{1\ mole}{24\ L}\right)$

$=0.10\ mol$

Now, it says that combustion of

$1\ mol$ of methane generates

$890\ J$ of heat. We could calculate the heat generated by the combustion of

$0.10\ moles$ of methane as:

$=0.10\ mol\left(\frac{890\ kJ}{1\ mole}\right)$

$=89kJ$

Therefore, 89kJ of heat will be evolved by the combustion of 2.4L of methane

For the reaction,

${ X }_{ 2 }{ O }_{ 4 }\left( l \right) \longrightarrow 2X{ O }_{ 2 }\left( g \right)$

$\Delta U=2.1\ k\ cal$ ,

$\Delta S=20\ cal\ { K }^{ -1 }$ at 300 K Hence ,

$\Delta G$ is_________.

0%
+2.7 kcal
0%
-2.7 kcal
0%
+9.3 kcal
0%
-9.3 kcal

Explanation

The change in Gibbs free energy is given by

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

where,

$\Delta H=$ enthalpy of the reaction

$\Delta S=$ entropy of reaction

Thus in order to determine

$\Delta G$ , the value of

$\Delta H$ must be known. The value of $$\Delta H$ can be calculated by the reaction,

$\Delta H=\Delta U+\Delta n_gRT$

where,

$\Delta U=$ change in internal energy

$\Delta n_g=$ (number of moles of gaseous product)-(number of moles of gaseous reactant)

$=2-0=2$

$R=$ gas constant

$=2\ cal$

$\Delta U=2.1\ kcal=2.1\times10^3\ cal$

therefore,

$\Delta H=(2.1\times10^3)+(2\times2\times300)=3300\ cal$

Hence,

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

$\Delta G=3300 -(300\times20)$

$\Delta G=-2700\ cal$

$\Delta G=-2.7\ kcal$

The value of

$\Delta G$ for the process

${ H }_{ 2 }O\left( s \right) \rightarrow { H }_{ 2 }O\left( l \right)$ at 1 atm and 260 K is:

0%
<0
0%
=0
0%
>0
0%
unpredictable

Explanation

$T<273K$ ice remains as ice and water converts into ice. So, ice to water conversion is not a spontaneous process.

$\Rightarrow$

$\Delta G>0$ for

${ H }_{ 2 }O\left( s \right) \longrightarrow { H }_{ 2 }O\left( l \right)$ at

$260K$ .

Ans :- C

For an ideal gas

$\displaystyle \frac {C{p,m}}{C{v,m}}\,=\,\gamma$ . The molecular mass of the gas is M , its specific heat capacity at constant volume is:

0%
$\displaystyle \frac{R}{M(\gamma\,-\,1)}$
0%
$\displaystyle \frac{M}{R(\gamma\,-\,1)}$
0%
$\displaystyle \frac{\gamma\,RM}{\gamma\,-\,1}$
0%
$\displaystyle \frac{\gamma R}{M(\gamma\,-\,1)}$

Explanation

Solution:- (A)

$\cfrac{R}{M \left( \gamma - 1 \right)}$

Let

As we know that,

${C}_{p, m} - {C}_{v, m} = R ..... \left( 1 \right)$

Dividing equation

$\left( 1 \right)$ by

${C}_{v, m}$ we have

$\cfrac{{C}_{p, m}}{{C}_{v, m}} - \cfrac{{C}_{v, m}}{{C}_{v, m}} = \cfrac{R}{{C}_{v, m}}$

$\Rightarrow \cfrac{{C}_{p, m}}{{C}_{v, m}} - 1 = \cfrac{R}{{C}_{v, m}}$

Here,

${C}_{p, m} =$ molar specific heat capacity at constant pressure

${C}_{v, m} =$ molar specific heat capacity at constant volume

Given:-

$\cfrac{{C}_{p, m}}{{C}_{v, m}} = \gamma$

$\therefore \gamma - 1 = \cfrac{R}{{C}_{v, m}}$

As we know that,

${C}_{v, m} = M {C}_{v}$

here,

${C}_{v}$ is specific heat capacity

$\therefore \gamma - 1 = \cfrac{R}{M {C}_{v}}$

$\Rightarrow {C}_{V} = \cfrac{R}{M \left( \gamma - 1 \right)}$

Hence the specific heat capacity of gas at constant volume is

$\cfrac{R}{M \left( \gamma - 1 \right)}$ .

For the reversible isothermal expansion of one mole of an ideal gas at

$300$

$K$ , from a volume of

$10 L$ to

$20 L$ ,

$\Delta H$ is :

0%
$1.73 \ kJ$
0%
$-1.73 \ kJ$
0%
$3.46 \ kJ$
0%
0

Consider the following process

$\Delta H(kJ/mol)$

$\frac{1}{2}A\rightarrow B +50$

$3B \rightarrow 3C +D -125$

$E + A \rightarrow 2D + 350$
For

$B + D \rightarrow E +2C, \Delta H$ will be:

0%
325 kJ/mol
0%
525 kJ/mol
0%
-375 kJ/mol
0%
-325 kJ/mol

Explanation

Solution:-

$\cfrac{1}{2} A \longrightarrow B + 50$

$2 \times \left[ \cfrac{1}{2} A \longrightarrow B + 50 \right]$

$A \longrightarrow 2B + 100 ..... \left( 1 \right)$

$3B \longrightarrow 2C + D - 125 ..... \left( 2 \right)$

$E + A \longrightarrow 2D + 350$

$2D \longrightarrow E + A - 350 ..... \left( 3 \right)$

Adding equation

$\left( 1 \right), \left( 2 \right) \& \left( 3 \right)$ , we have

$A + 3B + 2D \longrightarrow 2B + 100 + 2C + D - 125 + E + A - 350$

$B + D \longrightarrow E + 2C - 375$

Hence the value of

$\Delta{H}$ will be

$-375 \; {KJ}/{mol}$ .

It for reaction

$N_2(g)+3H_2(g) \rightarrow 2NH_2(g), \Delta H_{1}^{0}=-30$ KJ/mole at temprature

$300$ K and it specific heat capacities of different species are

$S_{P,N_{2}}=1J/g^{0}C$ and

$S_{P,NH_{2}}=j/g^{0}C$ , then

$\Delta H_{2}^{0}$ at

$400$ K for the same reaction will be (assume heat capacities to be constant in given temperature range)

0%
$-32 k j/$ mole
0%
$-28 k J/$ mole
0%
$-32.7 kJ/$ mole
0%
$-27.3 kJ/$ mole

${ NH }_{ 2 }{ CN }_{ \left( s \right) }+\dfrac { 3 }{ 2 } { O }_{ 2\left( g \right) }\rightarrow { N }_{ 2\left( g \right) }+{ CO }_{ 2\left( g \right) }+{ H }_{ 2 }{ O }_{ \left( l \right) }$
This reaction is carried out in a bomb calorie-meter. The heat released was

$743\ KJ\ { mol }^{ -1 }$ . The value of

${ \Delta H }_{ 300 }$ for this reaction would be:

0%
$-740\ KJ\ { mol }^{ -1 }$
0%
$-741.75\ KJ\ { mol }^{ -1 }$
0%
$-743.0\ KJ\ { mol }^{ -1 }$
0%
$-744.25\ KJ\ { mol }^{ -1 }$

Explanation

In a bomb calorimeter, heat released=

$- \Delta U$

$\therefore \Delta U= -743 KJmol^{-1}$

$\therefore \Delta H= \Delta U-\Delta ngRT$

where,

$\Delta ng$ = difference between gaseous moles of products and reactants

$\therefore \Delta ng=(1+1)-\cfrac {3}{2}=\cfrac {1}{2}$

$\therefore \Delta H= \Delta U+\cfrac {1}{2} RT$

$= -743+\cfrac {1}{2}\times 8.314 \times 300 \times 10^{-3}$

$\Delta H_{300}=-741.75 KJ mol^{-1}$

The three thermodynamic states

$P, Q$ and

$R$ of a system are connected by the paths shown in the figure given on the right. The entropy change in the processes

$P \rightarrow Q , Q \rightarrow R$ and

$P\rightarrow R$ along the paths indicated are

$\triangle S_{P Q}, \triangle S_{QR}$ and

$\triangle S_{PR}$ respectively. If the process

$P \rightarrow Q$ is adiabatic and irreversible, while

$P\rightarrow R$ is adiabatic and reversible, the correct statement is:

0%
$\triangle S_{QR} > 0$
0%
$\triangle S_{PR} > 0$
0%
$\triangle S_{QR} < 0$
0%
$\triangle S_{PQ} > 0$

Explanation

It is given that

$P\rightarrow R$ is an adiabatic and reversible process.

For reversible adiabatic process,

${q}_{reversible}=0$

$\triangle S=\int { \cfrac { { dq }_{ reversible } }{ T } } =0$

Hence

$\triangle {S}_{PR}=0$

The entropy of irreversible process is always going to increase. So, in an adiabatic irreversible process, change of entropy due to internal irreversibility is greater than zero.

Hence

$\triangle {S}_{PQ}=\cfrac{\triangle q}{T}\gt 0$

entropy change of a cyclic process is always zero.

Since entropy is a state function i.e. it doesn't depend on the path

and

$\triangle {S}_{PR}=0\\ \triangle {S}_{PQ}\rightarrow (+)ve$

Then

$\triangle {S}_{QR}$ must be

$-ve$ .

Hence

$\triangle {S}_{QR}\lt 0$

Given the following data:

Substrate	$\Delta H^o$ (kJ/mol)	$S^o$ (J/mol K)	$\Delta G^o$ (kJ/mol)
FeO(s)	$-266.3$	$57.49$	$245.12$
C(Graphite)	$0$	$5.74$	$0$
Fe(s)	$0$	$27.28$	$0$
CO(g)	$-110.5$	$197.6$	$-137.15$

Determine at what temperature the following reaction spontaneous?

$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$

0%
$298$ K
0%
$668$ K
0%
$966$ K
0%
$\Delta G^o$ is $+$ ve, hence the reaction will never be spontaneous

If the bond dissociation energies of

$XY$ ,

${ X }_{ 2 }$ and

${ Y }_{ 2 }$ (all diatomic molecules) are in the ratio of

$1:1:0.5$ and

${ \Delta }_{ 1 }H$ for the formation of XY is

$-200\ kJ\ { mol }^{ -1 }$ . The bond dissociates energy of

${ X }_{ 2 }$ will be:

0%
$100\ kJ\ mol^{-1}$ `
0%
$200\ kJ\ mol^{-1}$
0%
$800\ kJ\ mol^{-1}$
0%
$400\ kJ\ mol^{-1}$

Explanation

$X_2+Y_2\rightarrow2XY$

$\Delta H=(BE)X_2+(BE)Y_2 - 2(BE)X Y$

$(BE)$ of

$X - Y=a$

then

$(BE)$ of

$Y - Y=\dfrac{a}{2}$

and

$(BE)$ of

$X - X=a$

So,

$\Delta H_f(X - Y)=-200\ kJ$

So,

$-400\ (for\ 2\ moles\ XY)=a+\dfrac{a}{2}-2a$

$\Rightarrow-400=\dfrac{-a}{2}$

$a=800\ kJ$

The bond dissociation energy of

$X_2=800\ kJ/mol$

For the reaction at

$25, X_{ 2 }O_{ 4 } { O }_{ 4_{ (l) } } \longrightarrow 2X { O }_{ 2_{ (g) } }$ .

$\Delta H =$ 2.1 kcal and

$\Delta S =$ 20 cal

${ K }^{ -1 }$ . The reaction would be:

0%
spontaneous
0%
non-spontaneous
0%
at equilibrium
0%
unpredictable

Explanation

By considering 2 things i.e if reaction is exothermic or endothermic we can justify if reaction is spontaneous or not. If,

$\Delta$ G is negative then reaction is said to be spontaneous. Now,

$\Delta$ G =

$\Delta$ H-

$\Delta$ S. Therefore,

$\Delta$ G will be negative, hence it is a spontaneous reaction.

When a block of iron floats in mercury at

$0^o$ C, a fraction

$k_1$ of its volume is submerged, while at the temperature

$60^o$ C, a fraction

$k_2$ is seen to be submerged. If the coefficient of volume expansion of iron is

$\gamma _{Fe}$ and that of mercury is

$\gamma_{Hg}$ , then the ratio

$k_1/k_2$ can be expressed as.

0%
$\dfrac{1+60\gamma_{Fe}}{1+60\gamma_{Hg}}$
0%
$\dfrac{1-60\gamma_{Fe}}{1+60\gamma_{Hg}}$
0%
$\dfrac{1+60\gamma_{Fe}}{1-60\gamma_{Hg}}$
0%
$\dfrac{1+60\gamma_{Hg}}{1+60\gamma_{Fe}}$

For the reaction given below the values of standard Gibbs free energy of formation at 298 K are given.
What is the nature of the reaction?

$I_2 + H_2S \rightarrow 2HI + S$

$\Delta G_f^0 (HI) = 1.8 \ kJ\ mol^{-1}$ ,

$\Delta G_f^0 (H_2S) = 33.8 \ kJ\ mol^{-1}$

0%
Non-spontaneous in forward direction
0%
Spontaneous in forward direction
0%
Spontaneous in backward direction
0%
Non-spontaneous in both forward and backward directions

Explanation

$I_2 + H_2S \rightarrow 2HI + S$

$\Delta G^0 = \sum G_{f(products)}^0 - \sum G_{f(Reactants)}^0$ = -30.2 kJ
Hence, the reaction is spontaneous in forward direction.

What will be

$\Delta G$ for the reaction at

$25^0C$ when partial pressures of reactants

$H_2$ ,

$CO_2$ ,

$H_2O$ and

$CO$ are 10, 20, 0.02 and 0.01 atm respectively?

[Given :

$G^0_{H_2O}$

$= -228.58$ kJ,

$G^0_{CO}$

$= - 137.15$ kJ and

$G^0_{CO_2}$

$= -394.37$ kJ.]

0%
+5.61 kJ
0%
-5.61 kJ
0%
7.09 kJ
0%
-8.13 kJ

Explanation

$H_{2(g)} + CO_{2(g)} \rightleftharpoons H_2O_{(g)} + CO_{(g)}$

$\Delta G^0$ for reaction

$= [G^0_{H_2O} + G^0_{CO}] - [G^0_{H_2} + G^0_{CO_2}]$

$= [-228.58 - 137.15] - [0 - 394.37] = 28.64\ kJ$

Also,

$\Delta G = \Delta G^0 + 2.303\ RT\ log Q$

$= 28.64 + 2.303 \times 8.314 \times 10^{-3} \times 298\ log \cfrac{0.02 \times 0.01}{10 \times 20}$

$\Delta G = -5.61 kJ$

Thus, the reaction will occur in the forward direction.

Which of the following statements is correct for a reverse process in a state of equilibrium ?

0%
$\Delta G=2.30\quad RT\quad log\quad K$
0%
$\Delta { G }^{ o }=-2.30\quad RT\quad log\quad K$
0%
$\Delta { G }^{ o }=2.30\quad RT\quad log\quad K$
0%
$\Delta { G }=-2.30\quad RT\quad log\quad K$

Explanation

The correct option is C

$\Delta G^0=-2.303 RT log K$

$\Delta=\Delta G^0+2.30 RT log K log Q$

At equilibrium when

$\Delta=O and Q$

$\Delta G =\Delta G^0+2.303 RT log K=0$

$\Delta G^0=-2.303 RT log K$

$\Delta G=\Delta G^0+2.303 RT log KQ$

At equilibrium when

$\Delta G= \Delta G^0+2.303 RT log K=0$

$\Delta^0=-2.303 RTK$

Which thermochemical law is represented by the following figure?

0%
Standard enthalpy of a reaction
0%
Born - Haber cycle of lattice enthalpy
0%
Hess's law of constant heat summation
0%
Standard enthalpy of a solution

Explanation

According to Hess' Law of Constant Heat Summation,

$X \xrightarrow {\Delta H} Y$ ,
And

$X \xrightarrow {\Delta H_1} P \xrightarrow {\Delta H_2} Q \xrightarrow {\Delta H_3} Y$
then ,

$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3$

$\Delta S$ will be highest for the reaction:

0%

$2Ca(s)+O_2(g) \rightarrow 2CaO(s)$
0%

$CaCO_3(s) \rightarrow CaO(s) +CO_2(g)$
0%

$C(s)+O_2(g)\rightarrow CO_2(g)$
0%

$N_2(g)+O_2(g)\rightarrow 2NO(g)$

Explanation

The entropy of a system increases whenever its particles have more freedom of motion. Thus, the entropy increases whenever you have more moles of gaseous products than of reactants and whenever you have more product particles in solution than you have of reactant particles.

Hence among the given reactions, the entropy increases in-

${CaC{O}_{3}}_{\left( s \right)} \longrightarrow {CaO}_{\left( s \right)} + {C{O}_{2}}_{\left( g \right)}$

Since the reactant side has no gaseous particle while the product side has

$1$ gaseous particle.

Hence the entropy increases and

$\Delta{S}$ will be highest.

In which of the following pair of reactions first reaction is spontaneous while second reaction is non spontaneous?

0%
(i) $^{-}SH+H_2O\rightarrow H_2S+OH^-$

(ii) $NH^-_2+H_2O\rightarrow NH_3+OH^-$
0%
(i) $^{-}OR+H_2SO_4\rightarrow ROH+HSO^-_4$

(ii) $R^-+NH_3\rightarrow RH+NH^-_2$
0%
(i) $Cl^{-}+HF\rightarrow HCl+F^-$

(ii) $^-OH+HCl\rightarrow H_2O+Cl^-$
0%
(i) $^-OH+HBr\rightarrow H_2O+Br^-$

(ii) $RO^-+NH_3\rightarrow ROH+NH^-_2$

Explanation

(A) (i)

$\underset {base} {^{\circleddash}SH}+\underset {acid}{H_2O} \longrightarrow H_2S+OH^{\circleddash}$ (ii)

$\underset {base}{NH_2^{\circleddash}}+\underset {acid}{H_2O}\longrightarrow NH_3+OH^{\circleddash}$

Here both reactions are spontaneous. As

$SH^{\circleddash}$ and

$NH_2^{\circleddash}$ are highly basic.

(B) (i)

$\underset {base}{^{\circleddash} OR} +\underset {acid}{H_2SO_4} \longrightarrow ROH+HSO_4^{\circleddash}$ (ii)

$\underset {base}{R^{\circleddash}}+\underset {acid}{NH_3}\longrightarrow RH+NH_2^{\circleddash}$

Both reactions are spontaneous

$R^{\circleddash}$ and

$OR^{\circleddash}$ are highly basic.

$\underset {base}{Cl^-}+\underset {acid}{HF}\longrightarrow HCl+F^{\circleddash}$ (ii)

$\underset {base}{^{\circleddash}OH}+\underset {acid}{HCl} \longrightarrow H_2O+Cl^{\circleddash}$

Here also both reactions are spontaneous.

(D) (i)

$\underset {base}{OH^{\circleddash}}+\underset {acid}{HBr}\longrightarrow H_2O+Br^{\circleddash}$ (ii)

$\underset {base}{RO^{\circleddash}}+\underset {acid}{NH_3} \nrightarrow ROH+NH_2^{\circleddash}$

(i) reaction is spontaneous but (ii) is not spontaneous. Because basicity of

$RP^{\circleddash}$ is not high enough to abstract proton from weak acid

$NH_3$ .

$\Delta H> 0$ and

$\Delta S> 0$ , the reaction proceeds spontaneously when:

0%
$\Delta H > T \Delta S$
0%
$\Delta H < T \Delta S$
0%
$\Delta H=T\Delta S$
0%
$None\ of\ the\ above$

Which of the following conditions regarding a chemical process ensures its spontanlity at all temperature?

0%
$\triangle H>0,\quad \triangle G<0$
0%
$\triangle H<0,\quad \triangle S>0$
0%
$\triangle H<0,\quad \triangle S<0$
0%
$\triangle H>0,\quad \triangle S<0$

Explanation

$\triangle H<0$ ,

$\triangle S>0$

$\Rightarrow$ Reaction spontaneous at all temperature.

$\rightarrow$ For spontaneity,

$\triangle G<0$

$\Rightarrow \triangle H- T \triangle S <0$

where

$\triangle H$ is negative and

$\triangle S$ is positive.

In a constant volume calorimeter, 5 g of gas with molecular weight 40 was burnt in excess of oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.75 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

$\triangle U$ of combustion of the gas in kJ

${ mol }^{ -1 }$ is:

0%
15
0%
12
0%
90
0%
8

Explanation

Solution:- (A)

$15 \; {kJ}/{mol}$

As we know that, for ideal gas under any process,

$\Delta{U} = \cfrac{\text{heat capacity} \times \text{change in temprature}}{\text{no.of moles}}$

Given:-

Mol. wt. of gas

$= 40 \; g$

Wt. of gas

$= 5 \; g$

$\therefore$ No. of moles

$= \cfrac{5}{40} = 0.125 \text{ mole}$

Heat capacity

$= 2.5 \; {kJ}/{K}$

Change in temperature

$= 298 - 298.75 = 0.75 \; K$

$\therefore \Delta{U} = \cfrac{2.5 \times 0.75}{0.125} = 15 \; {kJ}/{mol}$

Hence the numerical value for the

$\Delta{U}$ of combustion is

$15 \; {kJ}/{mol}$ .

For which reaction will

$\Delta H=\Delta U$ ?

0%
$H_2(g)+Br_2(g)\rightarrow 2HBr(g)$
0%
$C(s)+2H_2O(g)\rightarrow 2H_2(g)+CO_2(g)$
0%
$PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)$
0%
$2CO(g)+O_2(g)\rightarrow 2CO_2(g)$

Explanation

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

$\Delta H = \Delta U + \Delta n_{g}RT$

In only reaction (A) the

$\Delta n_g$ = 0 and hence

$\Delta H = \Delta U$

A system is taken along paths A and B as shown. If amounts of heat given in these processes are respectively QA and QB, then:

0%
QA=QB
0%
QA>QB
0%
$QB<QA$
0%
none of these

Explanation

Solution:- (A)

${Q}_{A} = {Q}_{B}$

As we know that heat is a state function, i.e., it does not depends on the path.

Hence

${Q}_{A} = {Q}_{B}$

For a particular reaction

$\triangle H^0 = -76.6 KJ$ and

$\Delta S^0 = 226 JK^{-1}$ . This reaction is:

0%
spontaneous at all temperatures
0%
non-spontaneous at all temperatures
0%
spontaneous at temperature below $66^0$ C
0%
spontaneous at temperature above $66^0$ C

Explanation

Given: Change in enthalpy

$\triangle H^o=-76.6 \ KJ$

Change in Entropy

$\triangle S^o= 226 \ JK^{-1}$

Using the relation,

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

$\triangle G = -76.6 \ KJ - T \times 226 \ JK^{-1}$

Since, both enthalpy

$(\triangle H^o)$ and Entropy term

$(T \triangle S^o)$ is negative. Hence

$\triangle G$ is negative whatever be the temperature.

Given the following data:

Substance	$\Delta H^{o}(kJ/mol)$	$\Delta S^{o}(kJ/mol)$	$\Delta G^{o}(kJ/mol)$
$FeO(s)$	$-266.3$	$57.49$	$-245.12$
$C$ (Graphite)	$0$	$5.74$	$0$
$Fe(s)$	$0$	$27.28$	$0$
$CO(g)$	$-110.5$	$197.6$	$-137.15$

Determine at what temperature the following reaction is spontaneous?

$FeO(s)+C(Graphite)\rightarrow Fe(s)+CO(g)$

0%
$298\ K$
0%
$668\ K$
0%
$966\ K$
0%
$\Delta G^{o}$ is +ve the reaction will never be spontaneous

For the following concentration cell,to be spontaneous

$Pt({H}_{2}){P}_{1} atm |HCl\ || \ Pt({H}_{2}), {P}_{2} atm$ .

Which of the following is correct?

0%
${P}_{1}={P}_{2}$
0%
${P}_{1}<{P}_{2}$
0%
${P}_{1}>{P}_{2}$
0%
Can't be predicted

Explanation

Solution:- (B)

${P}_{1} < {P}_{2}$

${E}_{cell} = {{E}^{0}}_{cell} + \cfrac{0.059}{2} \log{\cfrac{{P}_{2}}{{P}_{1}}}$

For cell reaction to be spontaneous,

${E}_{cell} > 0$

$\therefore {E}_{cell}$ will be

$+ve$ if

${P}_{2} > {P}_{1}$ .

Hence for the cell reaction to be spontaneous,

${P}_{2} > {P}_{1}$

Given the following data:

Substance	$\triangle H^0 (kJ/mol)$	$S^0 (J/molK)$	$\triangle G^0(kJ/mol)$
FeO(s)	$-266.3$	$57.49$	$-245.12$
C(Graphite)	0	$5.74$	0
Fe(s)	0	$27.28$	0
CO(g)	$-110.5$	$197.6$	$-137.15$

Determine at what temperature the following reaction is spontaneous?

$FeO(s) + C(Graphite) \rightarrow Fe(s) + CO(g)$

0%
$298$ K
0%
$668$ K
0%
$966$ K
0%
$\triangle G^0$ is +ve, hence the reaction will never be spontaneous.

Consider the following reactions. In which case the formation of product is favoured by decrease in pressure?
(1)

$CO_2(g)+C(s) \rightarrow 2CO(g)$ ;

$\Delta H=+172.5$ kJ
(2)

$N_2(g)+3H_2(g) \rightarrow 2NH_3(g)$ ;

$\Delta H=-91.8$ kJ
(3)

$N_2(g)+O_2(g) \rightarrow 2NO(g)$ ;

$\Delta H=+181$ kJ
(4)

$2H_2O(g) \rightarrow 2H_2(g)+O_2(g); \Delta H=484.6$ kJ

0%
2,3
0%
3,4
0%
2,4
0%
1,4

Explanation

If pressure is decreased, then equilibrium shift to more gas molecules side.

Here product is favoured if product gas mole are more than reactant gas mole.

(1)

${CO_2}_{(g)}+C_{(s)}\rightleftharpoons 2CO$

Here reactant gas mole=

$1$

Product gas moles=

$2$

If we reduce pressure, equilibrium shift to product side.

(4)

${2H_2O}_{(g)}\rightleftharpoons {2H_2}_{(g)}+{O_2}_{(g)}$

Reactant gas moles=

$2$

Product gas moles=

$3$

If we reduce pressure, equilibrium shift to product side

Standard entropy of

${X}_{2},{Y}_{2}$ and

$X{Y}_{3}$ are

$60,40$ and

$50$

$J{K}^{-1}$

${mol}^{-1}$ , respectively. For the reaction,

$\cfrac{1}{2}{X}_{2}+\cfrac{3}{2}{Y}_{2}\rightarrow {XY}_{3}.\Delta H=-30kJ$ to be at equilirbium, the temperature will be:

0%
$1250K$
0%
$500K$
0%
$750K$
0%
$1000K$

Explanation

Solution:- (C)

$750 \; K$

$\cfrac{1}{2} {X}_{2} + \cfrac{3}{2} {Y}_{2} \longrightarrow X{Y}_{3} \quad \Delta{H} = -30 \; kJ$

For the above reaction-

$\Delta{S} = {\sum{\Delta{S}}}_{\left( Product \right)} - {\sum{\Delta{S}}}_{\left( reactant \right)}$

$\Delta{S} = 50 - \left( \cfrac{1}{2} \times 60 + \cfrac{3}{2} \times 40 \right)$

$\Rightarrow \Delta{H} = 50 - 90 = -40$

As we know that,

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

But at equilibrium,

$\Delta{G} = 0$

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

$\Rightarrow T = \cfrac{\Delta{H}}{\Delta{S}}$

Given:-

$\Delta{H} = -30 \; kJ = -30000 \; J$

$\therefore T = \cfrac{-30000}{-40} = 750 \; K$

Hence the temperature of the given reaction is

$750 K$ .

Pressure of

$10$ moles of an ideal gas is changed from

$2\ atm$ to

$1\ atm$ against constant external pressure without change in temperature. If surrounding temperature (

$300\ K$ ) and pressure (

$1\ atm$ ) always remains constant then calculate total entropy change (

$\Delta {S}_{system}+\Delta {S}_{surrounding}$ ) for given process.
[Given:

$\ln{2}=0;70$ and

$R=8.0J/mol/K$ ]

0%
$56J/K$
0%
$14J/K$
0%
$16J/K$
0%
None of these

The change in entropy of

$2$ moles of an ideal gas upon isothermal expansion at

$243.6K$ from

$20$ litre until the pressure becomes

$1atm$ is:

0%
$1.385\ cal/K$
0%
$-1.2\ cal/K$
0%
$1.2\ cal/K$
0%
$2.77\ cal/K$

Explanation

The change in entropy of 2 moles of an ideal gas upon isothermal expansion at

$243.6\ K$ from 20 litre until the pressure becomes 1atm is:

Given:

$P_1 = 1 \,atm$

No. of moles = 2 mol

T = 243.6 K

V = 20 litre

We know,

$PV = nRT$

$P=\dfrac{nRT}{V}$

$P=\dfrac{2\times 0.0821\times 243.6}{20}$

$P = 2\, atm$

$\Delta S=-4R\,ln\dfrac{P_2}{P_1}$

$\Delta S=-4\times \dfrac{8.314}{4.19}\,ln\dfrac{2}{1}$

$\Delta S=2\times 0.6931\times 1.9=2.77\,cal/K$

Change in entropy is

$2.77\, cal/K$

Two samples of DNA, A and B have melting points

$340K$ and

$350K$ respectively. This is because

0%
B has more GC content than A
0%
A has more GC content than B
0%
B has more AT cotent than A
0%
both have same AT content

Explanation

The B sample of DNA having higher melting point must be having more GC content as comppared to sample A. Since GC base pair having 3 hydrogen bonds as compare to AT base pair having only 2 hydrogen bonds, results in stronger bonding.

So, option

$A$ is correct

$\Delta {H}_{vaporisation}$ of substance

$X(l)$ (molar mass :

$30g/mol$ ) is

$300J/g$ at its boiling point

$300K$ , then molar entropy change for reversible condensation process is:

0%
$30J/mol.K$
0%
$-300J/mol.K$
0%
$-30J/mol.K$
0%
None of these

Explanation

Solution:- (A)

$30 \; {J}/{mol-K}$

Given:-

${\Delta{H}}_{vap.} = 300 \; {J}/{g}$

Molar mass of

$X = 30 \; {g}/{mol}$

$\therefore {\Delta{H}}_{molar} = 300 \times 30 = 9000 \; {J}/{mol}$

Boiling point of substance

$X, \; \left( T \right) = 300 K$

As we know that,

Molar entropy change,

${\Delta{S}}_{molar} = \cfrac{{\Delta{H}}_{molar}}{T} = \cfrac{9000}{300} = 30 \; {J}/{mol-K}$

Hence the molar entropy change for the given process will be

$30 \; {J}/{mol-K}$ .

In conversion of line-stone to lime,

$Ca{CO}_{3}(s)\rightarrow CaO(s)+{CO}_{2}(g)$
the values of

$\Delta {H}^{o}$ and

$\Delta {S}^{o}$ are

$+179.1kJ{mol}^{-1}$ and

$160.2J/K$ respectively at

$298K$ and

$1$ bar. Assuming that

$\Delta {H}^{o}$ and

$\Delta {S}^{o}$ do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is:

0%
$1008K$
0%
$1200K$
0%
$845K$
0%
$1118K$

Which of the following statements/relationships is not correct in thermodynamic changes?

0%
$w=-nRT\ln{\cfrac{{V}_{2}}{{V}_{1}}}$ (isothermal reversible expansion of an ideal gas)
0%
For a system at constant volume, heat involved is equal to change in internal energy.
0%
$w=nRT\ln{\cfrac{{V}_{2}}{{V}_{1}}}$ (isothemal reversible expansion of an ideal gas)
0%
$\Delta U=0$ (isothermal reversible expansion of an ideal gas)

An intimate of ferric oxide

$({Fe}_{2}{O}_{3})$ and aluminum (Al) is used as solid rocket fuel. Calculate fuel value per gram of the mixture Heats of formation are as follows:

$\triangle { H }_{ f }\left( { Al }_{ 2 }{ O }_{ 3 } \right) =399\quad kcal/mole$

$\triangle { H }_{ f }\left( { Fe }_{ 2 }{ O }_{ 3 } \right) =199\quad kcal/mole$

0%
$0.9345 Kcal/g$
0%
$200 Kcal/g$
0%
$3.94Kcal/g$
0%
None

Ethyl chloride (

${C}_{2}{H}_{5}Cl$ ), is prepared by reaction of ethylene with hydrogen chloride:

${C}_{2}{H}_{4}(g)+HCl(g)\rightarrow {C}_{2}{H}_{5}Cl(g)$

$\Delta {H}=-72.3kJ/mol$
What is the value of

$\Delta {E}$ (in kJ), if

$70g$ of ethylene and

$73g$ of

$HCl$ are allowed to react to

$300K$

0%
$-69.8$
0%
$-180.75$
0%
$-174.5$
0%
$-139.6$

Assuming that water vapour is an ideal gas, the internal energy change (

$\Delta U$ ) when

$1mol$ of water is vaporised at

$1$ bar pressure and

${100}^{o}C$ will be:

(Given: Molar enthalpy of vapourisation of water at

$1$ bar and

$373K=41\ kJ.{mol}^{-1}$ and

$R=8.3J{mol}^{-1}$

${K}^{-1}$ )

0%
$4.100\ kJ$ ${mol}^{-1}$
0%
$3.7904\ kJ$ ${mol}^{-1}$
0%
$37.904\ kJ$ ${mol}^{-1}$
0%
$41.00\ kJ$ ${mol}^{-1}$

Explanation

The change :

$H_2O(l) → H_2O(g)$

$\Delta H = \Delta U + \Delta ngRT$

"or"

$\Delta U = \Delta H - \Delta ngRT$

Substituting the values, we get

$\Delta U = 41.00\, kJ/mol -1\times 8.3\, J/mol/K \times 373\, K\times \frac{1}{1000}$

$= 41.00\, kJ mol^{-1} − 3.096\, kJ mol^{-1}$

$= 37.904\, kJ/mol$

Heat of reaction for,

$CO(g)+\dfrac{1}{2}{O}_{2}(g)\rightarrow {CO}_{2}(g)$ at constant

$V$ is

$-67.71Kcal$ at

${17}^{o}C$ . The heat of reaction at constant

$P$ at

${7}^{o}C$ is :

0%
$-68.0KCal$
0%
$+68.0KCal$
0%
$-67.42Kcal$
0%
None

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 13 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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