All of these options(A,B,C,D) are correct

When a directly measured enthalpy change of reaction is not available

To calculate an enthalpy change value through multiple steps

Because enthalpy is a State Function

As an easy way to calculate the enthalpy change of a reaction

For the change,

$C_{diamond}\longrightarrow +C_{graphite}; \:\Delta H=-1.89 \:kJ$ , if 6 g of diamond and 6 g of graphite are separately burnt to yield

$CO_2$ the heat liberated in first case is:

0%
less than in the second case by 1.89 kJ
0%
less than in the second case by 11.34 kJ
0%
less than in the second case by 14.34 kJ
0%
more than in the second case by 0.945 kJ

Explanation

$C_D\longrightarrow C_G; \: \Delta H= - 1.89$ ...( i)

$C_D+ O_2\longrightarrow CO_2; \: \Delta H= - q_1$ ...( ii)

$C_G+ O_2\longrightarrow CO_2; \: \Delta H= - q_2$ ...( iii)
Subtract equation (iii) from equation (ii).

$C_D\longrightarrow C_G; \: \Delta H= - q_1+ q_2$ ...(iv)
From equations (i) and (iv)

$q_1+ q_2=-1.89$
or

$q_2-q_1=-1.89 \: for \: 12 \: g \: C_{D\rightarrow G}$
Thus,

$for \: 6 \: g \: C_{D\rightarrow G}$

$\displaystyle q_2-q_1=\frac{-1.89}{2}= -0 .945 \:kJ$

The relationship between the free energy change

$(\Delta G)$ and entropy change

$(\Delta S)$ at constant temperature

$(T)$ is

0%
$\,\Delta G = \Delta H + T\Delta S$
0%
$\,\Delta H = \Delta G + T\Delta S$
0%
$\,T\Delta S = \Delta G + \Delta H$
0%
$\,\Delta G = -\Delta H - T\Delta S$

Explanation

As we know,

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

$\,\Delta H = \Delta G + T\Delta S$

Hess law is applicable for determination of enthalpy of

0%
Reaction
0%
Formation
0%
Transition
0%
All of above

A change in the free energy of a system at constant temperature and pressure will be:

$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$

At constant temperature and pressure

$\Delta_{ sys }G < 0 (spontaneous)$
$\Delta_{ sys }G = 0 (equilibrium)$
$\Delta_{ sys }G > 0 (non-spontaneous)$

The free energy for a reaction having

$\Delta H = 31400 cal$ ,

$\Delta S = 32 cal K^{ -1 } mol^{ -1 }$ at

$1000^{ \circ }C$ is?

0%
$-9336 cal$
0%
$-7006 cal$
0%
$-2936 cal$
0%
$+9006 cal$

Explanation

Change in the free energy of a system at constant temperature and pressure will be:

$\Delta_{ sys }G = \Delta_{ sys }H - T\Delta_{ sys }S$

As we know,

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

$\Delta G = 31400- 1273\times32 =- 9336cal$

The heat of hydrogenation of ethene is

$x_1$ and that of benzene is

$x_2$
Hence resonance energy of benzene is

0%
$x_1 - x_2$
0%
$x_1 + x_2$
0%
$3x_1 - x_2$
0%
$x_1 - 3x_2$

Explanation

$CH_2==CH_2 + H_2\longrightarrow CH_3--CH_3\quad\quad \Delta H = x_1$

$RE = 3x_1 - x_2$

$C(diamond) + 2H_2(g)\longrightarrow CH_4(g);\,\,\Delta H_1$

$C(g) + 4H(g)\longrightarrow CH_4(g);\,\,\Delta H_2$

Select the correct relation.

0%
$\Delta H_1 = \Delta H_2$
0%
$\Delta H_1 > \Delta H_2$
0%
$\Delta H_1 < \Delta H_2$
0%
$\Delta H_1 = \Delta H_2 + \Delta_{ vap }H(C) + \displaystyle \Delta_{ diss }H(H_2)$

Explanation

As we know,

$C(diamond) \longrightarrow C(g)\,\,$

$C(diamond) + 2H_2(g)\longrightarrow CH_4(g);\,\,\Delta H_1$

$\Delta H_1 = \Delta H_2 + x$

$C(g) + 4H(g)\longrightarrow CH_4(g);\,\,\Delta H_2$
So,

$\Delta H_1 > \Delta H_2$

Which one of the following correctly represents the physical significance of Gibb's energy change?

0%
$\displaystyle -\Delta G={ W }_{ \text{compression} }$
0%
$\displaystyle \Delta G={ W }_{ \text{expansion} }$
0%
$\displaystyle \Delta G={ -W }_{ \text{expansion} }={ W }_{ \text{non-expansion} }$
0%
$\displaystyle -\Delta G={ W }_{ \text{expansion} }$

Explanation

The decrease in Gibbs energy results in useful work done by the system.

Therefore,

$-\Delta G={ W }_{ \text{expansion} }$ is the correct expression.

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v :

${ \Delta H }_{ atomisation }$ K = 90 kJ/mol
w :

${ \Delta H }_{ ionisation }$ K = 418 kJ/mol
x :

${ \Delta H }_{ dissociation }$ H = 436 kJ/mol
y :

${ \Delta H }_{ electron affinity }$ H = 78 kJ/mol
z :

${ \Delta H }_{ lattice }$ KH = 710 kJ/mol

In terms of the letters v to z the expression for

${ \Delta H }_{ i }$ of K is

${ \Delta H }_{ i }$ =

$w/2$ .
If true enter 1, else enter 0.

0%
0
0%
1
0%
2
0%
3

$H_2(g) + Br_2(g)\longrightarrow 2HBr(g); \Delta H^{ \ominus } = -72.40 kJ$

$\Delta G^{ \ominus } = -106.49 kJ, T = 298 K$
The value

$\Delta S$ is .

0%
$114.3 J k^{ -1 }$
0%
$-114.3 J k^{ -1 }$
0%
$1363.6 J k^{ -1 }$
0%
Noe of these

Explanation

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

$-106.49 kJ = -72.40 - 298\times$

$\Delta S$

$-34.09 kJ = -298\times$

$\Delta S$

$\Delta S = \frac{ 34090 J }{ 298 } = 114.3 J k^{ -1 }$

Based on the first law of thermodynamics, which one of the following is correct ?

0%
$\text{For an isochoric process, E = -Q}$
0%
$\text{For an adiabatic process, E = -Q}$
0%
$\text{For an isothermal process, E = -Q}$
0%
$\text{For an cyclic process, W = -Q}$

A bomb calorimeter is used to measure the value of heat of reaction at a constant

0%
Volume
0%
Pressure
0%
Temperature
0%
None of these

All spontaneous process proceed in one direction only.

0%
True
0%
False

$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286 kJ$

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l)......... kJ(\pm?)$

0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 572 kJ$ .
0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) - 572 kJ$ .
0%
$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) +0 kJ$ .
0%
None of these

Explanation

Reaction:

$H_2(g) + \frac{ 1 }{ 2 }O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286 kJ$ .................1
multiply it by 2 we will get,

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l); \Delta H = -286*2 = -572 kJ$
which can be written as

$2H_2(g) + O_2(g)\longrightarrow 2H_2O(l) + 572 kJ$ .

The Born Haber cycle below represents the energy changes occurring at 298K when KH is formed from its elements
v :

${ \Delta H }_{ atomisation }$ K = 90 kJ/mol
w :

${ \Delta H }_{ ionisation }$ K = 418 kJ/mol
x :

${ \Delta H }_{ dissociation }$ H = 436 kJ/mol
y :

${ \Delta H }_{ electron affinity }$ H = 78 kJ/mol
z :

${ \Delta H }_{ lattice }$ KH = 710 kJ/mol

In terms of the letters v to z the expression for

${ \Delta H }_{ electron affinity }$ of H is

${ \Delta H }_{ electron affinity }$ is __.
I

0%
$y$
0%
$y/2$
0%
$2y$
0%
$y/3$

Explanation

The electron affinity

$\Delta { H }_{ electron\: affinity }$ of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion. It is exothermic as heat energy is released from the system.

$\Delta { H }_{ electron\: affinity }$ should be y not y/2.

The heat released when the requisite amount of ions in the gaseous state combine to give 1 mol of crystal lattice is known as:

0%
lattice energy
0%
hydration energy
0%
formation energy
0%
none of the above

The enthalpy change for reaction (i) is:

0%
+ 100.23 kJ/mol
0%
+ 127.5 kJ/mol
0%
- 127.5 kJ/mol
0%
- 100.23 kJ/mol

Explanation

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

$[({ \Delta H }_{ f })_{ C_{ 2 }{ F }_{ 4 } }]+2({ \Delta H }_{ f })_{ HCl }-2({ \Delta H }_{ f })_{ ^{ { CHClF }_{ 2 } } }]$

$= [658.3 + 2(92.3) + 2(485.2)]$

$= 127.5\ kJ/mole$

Gibbs- Helmholtz equation is

0%
$\Delta G = \Delta H - T\Delta S$ .
0%
$\Delta G = \Delta H +T\Delta S$ .
0%
$\Delta G = T\Delta H - \Delta S$ .
0%
none of these

Explanation

Gibbs-Helmholtz Equation

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

Measure of heat in the reaction

$\bullet$ In chemical reactions, reactions absorb energy to break bonds

$\bullet$ Energy is then released when bonds form between rearranged atoms of the product

What can be concluded about the values of

$\Delta H$ and

$\Delta S$ from this graph ?

0%
$\Delta H >0, \:\Delta S >0$
0%
$\Delta H >0, \:\Delta S <0$
0%
$\Delta H <0, \:\Delta S >0$
0%
$\Delta H <0, \:\Delta S <0$

Explanation

As we know,

$\Delta G = \Delta H - T \Delta S$
here,
as slope is negative so

$- T \Delta S < 0$
&

$T \Delta S > 0$

$\Delta S >0$
Also as intercept is positive so

$\Delta H > 0$

State True or False.

Gibbs- Helmholtz equation is

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

0%
True
0%
False

Explanation

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

The measure of heat in the reaction

1. In chemical reactions, reactions absorb energy to break bonds

2. Energy is then released when bonds form between rearrangement atom of the product

The Born Haber cycle below represents the energy changes occurring at 298K when

$KH$ is formed from its elements
v :

${ \Delta H }_{ atomisation }$

$K = 90 kJ/mol$
w :

${ \Delta H }_{ ionisation }$

$K = 418 kJ/mol$
x :

${ \Delta H }_{ dissociation }$

$H = 436 kJ/mol$
y :

${ \Delta H }_{ electron affinity }$

$H = 78 kJ/mol$
z :

${ \Delta H }_{ lattice }$

$KH = 710 kJ/mol$

Calculate the value of

$\Delta$

$H$ showing all your working.

0%
124 kJ/mol
0%
-124 kJ/mol
0%
124 J/mol
0%
None of these

Explanation

Born-Haber Cycle

It is a series of steps (chemical processes) used to calculate the lattice energy of ionic solids, which is difficult to determine experimentally. You can think of BH cycle as a special case of Hess's law which states that the overall energy change in a chemical process can be calculated by breaking down the process into several steps and adding the energy change from each step.

${ \Delta H }_{ r }$ =2

$\times$ 90 + 2

$\times$ 418 + 436 - 2

$\times$ 78 - 2

$\times$ 710

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

$- 124 kJ/mole$

$2$ moles of

$CO$ and

$1$ mole of

$O_{2}$ are taken in a container of volume

$1$ litre to form

$2$ moles of

$CO_{2}$ according to equation

$2CO + O_{2}\rightarrow 2CO_{2}, \ \Delta H = -560KJ$ .

In this process, pressure changes from

$70\ atm$ to

$40\ atm$ . Find out value of

$\Delta U$ at

$500\ K$ .

[Given: gases deviates apprecially from ideal nature.

$1\ atm-litre = 0.1\ KJ$ ].

0%
$-530\ KJ$
0%
$-500\ KJ$
0%
$-557\ KJ$
0%
$-560\ KJ$

The Born Haber cycle below represents the energy changes occurring at 298K when

$KH$ is formed from its elements
v :

${ \Delta H }_{ atomisation }$

$K = 90 kJ/mol$
w :

${ \Delta H }_{ ionisation }$

$K = 418 kJ/mol$
x :

${ \Delta H }_{ dissociation }$

$H = 436 kJ/mol$
y :

${ \Delta H }_{ electron affinity }$

$H = 78 kJ/mol$
z :

${ \Delta H }_{ lattice }$

$KH = 710 kJ/mol$

On complete reaction with water,

$0.1 g$ of

$KH$ gave a solution requiring 25

${ cm }^{ 3 }$ of 0.1M

$HCl$ for neutralisation.Calculate the relative atomic mass of potassium from this information.

0%
$39$
0%
$40$
0%
$41$
0%
None of these

Explanation

Meq. of KH = Meq. of

$HCl$

$\frac { { 0.1 } }{ { E }_{ KH } } \times 1000\quad =\quad 25\quad \times \quad 0.1$
Valency factor (05955) of

$K$ is 1 hence

${ E }_{ K }$ =

${ M }_{ K }$

${ M }_{ K }$ =39

${ E }_{ KH }$ =40

${ E }_{ KH }$ =

${ E }_{ K }$ =
40=

${ E }_{ K }$ +1

${ E }_{ KH }$

${ E }_{ K }$

$\Rightarrow$ 39

Write a balanced equation for the reaction of

$KH$ with water.

0%
$KH + H_2O \rightarrow KOH + H_2O_2$
0%
$KH + H_2O \rightarrow KOH + H_2$
0%
$KH + H_2O \rightarrow K(OH)_2 + H_2$
0%
$KH + H_2O \rightarrow KOH + H_2O$

Explanation

Potassium hydride hydrolyzes in water to form hydroxide.

$KH + H_2O \rightarrow KOH + H_2$

Hence, option B is correct

The amount of heat required to raise the temperature of a substance through 1

$^o$ C is called

0%
thermal energy
0%
calories
0%
heat capacity
0%
specific heat capacity

Explanation

Energy required to raise the temperature of an object by

$1^o C$ is called heat capacity.

Hence C is correct

Hess's law is used to calculate

0%
Enthalpy of reaction
0%
Entropy of reaction
0%
Work done in reaction
0%
All of these

Explanation

Correct Option:

$A$

Explanation:

Hess’s law states that the change in enthalpy in a chemical reaction is the same whether the process is carried out in one step or in multiple steps.

It is the outcome of the first law of thermodynamics. It is used for the calculation of standard reaction enthalpy.

Additional Information:

Hess’s law states that the standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction can be divided, while each occurs at the same temperature.

The enthalpy change for a reaction is independent of the number of ways a product can be obtained if the initial and final conditions are the same.

The negative enthalpy change for a reaction indicates an exothermic process, while positive enthalpy change corresponds to an endothermic process.

Which of the following type of energies are involved in Born Haber`s cycle ?

0%
a. $\Delta_{ sub }H$
0%
b. ionisation energy
0%
c. bond dissociation energy
0%
d. lattice energy

Specific heat capacity of lead is

$120\space J\space kg^{-1}\space K^{-1}$ . When

$7200J$ of heat is supplied to

$5kg$ of lead, the rise in temperature is

0%
$8^{\small\circ}C$
0%
$10^{\small\circ}C$
0%
$12^{\small\circ}C$
0%
$14^{\small\circ}C$

Explanation

$Q = ms\triangle t$

$\triangle t = \displaystyle\frac{Q}{ms} = \displaystyle\frac{7200}{5\times120} = 12^{\small\circ}$

Gibbs- Helmholtz equation is:

0%
$\Delta G = T\Delta H + \Delta S$
0%
$\Delta G = - T\Delta H - \Delta S$
0%
$\Delta G = T\Delta H - \Delta S$
0%
$\Delta G = -T\Delta H + \Delta S$

Explanation

Gibbs-Helmholtz Equation

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

Measure

of heat

in the

reaction

$\blacksquare$ In chemical reactions, reactions absorb

energy to break bonds

$\blacksquare$ Energy is then released when bonds from

between rearranged atoms of the product

A hypothetical reaction

$A \rightarrow 2B$ , proceeds through following sequence of steps:
(i)

$A \rightarrow C; \Delta H = q$
(ii)

$C \rightarrow D; \Delta H = v$
(iii)

$\dfrac{1}{2}D \rightarrow B; \Delta H = x$
Then the heat of reaction is:

0%
$q - v + 2x$
0%
$q + v - 2x$
0%
$q + v + 2x$
0%
$q + 2v - 2x$

Explanation

$A\rightarrow C,\Delta H=q\quad$ ...(1)

$\\ C\rightarrow D,\Delta H=v$ ....(2)

$\\ 1/2\quad D\rightarrow B,\Delta H=x$ ..(3)

From (3) ,

$D\rightarrow 2B,\Delta H=2x$ ...(4)

Now adding equations (1)+(2)+(4) we get:

$\\ A\rightarrow 2B,\Delta H=q+v+2x$

For the reaction,

$X_2Y_4(l) \longrightarrow 2XY_2(g)$ at 300 K, the values of

$\Delta U$ and

$\Delta S$ are 2 kcal and 20 cal

$K^{-1}$ respectively. The value of

$\Delta G$ for the reaction is

0%
- 3400 cal
0%
3400 cal
0%
- 2800 cal
0%
2000 cal

Explanation

$\Delta H=\Delta U+\Delta n_gRT=2000+2\times2\times300=3200\ cal$

$\displaystyle \Delta G = \Delta H - T\Delta S = 3200 - 300 \times 20 = -2800 \: cal$

What wiII be the heat of formation of methane, if the heat of combustion of carbon is

$'-x' kJ$ , heat of formation of water is

$'-y' kJ$ and heat of combustion of methane is

$'z' kJ$ ?

0%
$(-x - y + z) kJ$
0%
$(-z -x + 2y) kJ$
0%
$(-x - 2y - z) kJ$
0%
$(-x -2y + z) kJ$

Explanation

From given data, we have

$C + O_{2} \rightarrow CO_{2} - x\ kJ ...... (i)$

$H_{2} + \dfrac {1}{2} O_{2} \rightarrow H_{2} O - y\ kJ ...... (ii)$

$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O + z\ kJ ..... (iii)$
The required equation is

$C + 2H_{2} \rightarrow CH_{4} + Q$

where Q is heat of combustion of

$CH_4$
To get the required equation, operate

$(i) + 2\times (ii) - (iii)$

$C + 2H_{2} \rightarrow CH_{4} + [(-x) + (-2y) - (-z)]$
Thus, heat of formation of methane is

$(-x -2y + z) kJ$

For the reaction of one mole of zinc dust with one mole of sulphuric acid in a bomb calorimeter,

$\triangle U$ and

$w$ corresponds to:

0%
$\triangle U < 0, w = 0$
0%
$\triangle U < 0, w < 0$
0%
$\triangle U > 0, w = 0$
0%
$\triangle U > 0, w > 0$

Explanation

$Zn + H_{2}SO_{4} = ZnSO_{4} + H_{2}$

In bomb calorimeter, there is no expansion in volume, so, work done will be zero.

This reaction is exothermic. So, some heat will be evolved which will result in lowering of internal energy.

Hence,

$\triangle U < 0$ &

$w = 0$ .

Mark the point where the temperature of

$H_2O$ is changing at

$4.18 J/g^o$ C(or

$1$

$cal/g^oC$ )?

0%
A
0%
B
0%
C
0%
D
0%
E

Explanation

At point (C) the temperature of the water is changing at

$4.18J/g ^ oC$

The specific heat of liquid water is

$4.18J/g ^ oC$ .

Thus the section (C) of the curve corresponds to the region where the temperature of the water is heated from

$0^0 C$ to

$100^0 C$ .

Hence, the correct option is C.

From the following information given, determine

$\Delta H^o$ for the decomposition of sodium chlorate?

$NaClO_3(s)\rightarrow NaCl(s)+\displaystyle\frac{3}{2}O_2(g)$

$(\Delta H^o_f$ values

$:$

$NaClO_3(s)=358J/mol,$

$NaCl(s)=-410j/mol$ ,

$O_2(g)=0$

$kcal/mol)$

0%
$52.0$ J
0%
$-52.0$ J
0%
$768$ J
0%
$-768$ J
0%
$\displaystyle\frac{3}{2}(-768J)$

Explanation

$NaCl{ O }_{ 3 }\left( s \right) \rightarrow NaCl\left( s \right) +\cfrac { 3 }{ 2 } { O }_{ 2 }\left( g \right) \\ \triangle H°=\sum { n\triangle { H° }_{ product } } -\sum { n\triangle { H° }_{ reactants } } \\ Here\\ \triangle { H° }=\left[ \cfrac { 3 }{ 2 } \left( 0 \right) +1\left( -410 \right) \right] -\left[ 1\left( 358 \right) \right] \\ =-410-358\\ =-768$

According to Hess's law, the heat of reaction depends upon

0%
initial condition of reactants
0%
initial and final conditions of reactants
0%
intermediate path of the reaction
0%
end conditions of reactants

Explanation

According to Hess's law, the heat of reaction depends upon Initial and final conditions of reactants. The the heat of reaction is independant of the intermediate path of the reaction. For example, the heat of the reaction for the reaction

$\displaystyle A \rightarrow D$ is equal to the sum of the heats of the reactions

$\displaystyle A \rightarrow B \\ B \rightarrow C \\ C \rightarrow D$

$C+{O}_{2}.\longrightarrow {CO}_{2}+94.2kcal$

${H}_{2}+\cfrac{1}{2}{O}_{2}\longrightarrow {H}_{2}+68.3kcal$

${CH}_{4}+2{O}_{2}\longrightarrow {CO}_{2}+2{H}_{2}O+210.8kcal$
Then, the heat of formation of methane will be

0%
$47.3kcal$
0%
$20.0kcal$
0%
$45.9kcal$
0%
$-47.3kcal$

Explanation

$C+{O}_{2}\longrightarrow {CO}_{2}+94.3kcal.........(i)$

${H}_{2}+\cfrac{1}{2}{O}_{2}\longrightarrow {H}_{2}O+68.3kcal.......(ii)$

On multiplying Eq.

$(ii)$ by

$2$ and then adding in Eq.

$(i)$ , we get

$C+2{H}_{2}+2{O}_{2}\longrightarrow {CO}_{2}+2{H}_{2}O+230.8kcal.....(iii)$

On substracting, adding the following equation from Eq.

$(iii)$

${CH}_{4}+2{O}_{2}\longrightarrow {CO}_{2}+2{H}_{2}O+210.8kcal$

we get

$C+2{H}_{2}\longrightarrow {CH}_{4}+20kcal$

What is the characteristic of a material which undergo spontaneous combustion?

0%
High calorific value
0%
High vapour pressure
0%
Low ignition temperature
0%
All of the above

Statement 1 : Candles can be safely stored at room temperature, even though their reaction with air is spontaneous at room temperature.
Statement 2 : The reaction that takes place when a candle is burned involves a decrease in entropy.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

Explanation

Statement : If equal masses of sodium metal and chlorine gas are allowed to react, some sodium will be left over after all the chlorine is used up.
The molar mass of chlorine is higher than the molar mass of sodium. Hence, for equal masses, the number of moles of chlorine will be smaller than the number of moles of sodium.
Statement 2 : The reaction required twice as many atoms of sodium as the molecules of chlorine.

$\displaystyle 2Na + Cl_2 \rightarrow 2NaCl$
Statement 1 is correct but Statement 2 is not correct.

Which statement is correct about heat?

0%
It is a measure of the transfer of energy due a temperature difference
0%
The heat transfer is always from hot to cold
0%
The heat absorbed by the surroundings equals the heat released by the system
0%
The amount of heat lost or gained during a reaction is directly related to the enthalpy change of the reaction.
0%
All answers are correct

Hess's Law is used

0%
When a directly measured enthalpy change of reaction is not available
0%
To calculate an enthalpy change value through multiple steps
0%
Because enthalpy is a State Function
0%
As an easy way to calculate the enthalpy change of a reaction
0%
All of these options(A,B,C,D) are correct

If 100.0 J of energy is added to 50.0 g of Cu, initially at

$25.0^o C$ , what will be the final temperature? (Specific heat capacity = 0.382 J/ C.g )

0%
$31.2^o C$
0%
$30.24^o C$
0%
$25.2^o C$
0%
$25.19^o C$
0%
$19.8^o C$

Explanation

let T deg C be the final temperature.

$\displaystyle q = mS \Delta T$

$\displaystyle q = 100.0$ J = energy added

$\displaystyle m = 50.0$ g = mass of Cu

$\displaystyle S = 0.382$ J/C.g = the specific heat of Cu

$\displaystyle \Delta T = T - 25.0$ = temperature difference.
Substite values in the above equation.

$\displaystyle 100.0 = 50.0 \times 0.382 \times (T-25.0)$

$\displaystyle 5.235= T-25.0$

$\displaystyle T = 30.24 ^oC$

10 g of liquid at 300K is heated to 350 K. The liquid absorbs 6 kcals. What is the specific heat of the liquid (in

$cal/ \displaystyle { g }^{ \circ }C$ )?

0%
6
0%
12
0%
60
0%
120

Explanation

$Q=nC\triangle T$ where

$C$ =specific heat of the liquid

$6KCal=Q$

$6\times 10^3 Cal=Q$

So,

$6 \times 10^3=10\times C\times (350-300)K$

$\Rightarrow \cfrac {6\times 10^3}{10\times 50}=C$

$\Rightarrow C=12$ .

A spontaneous reaction occurs:

0%
by itself and quickly
0%
with outside intervention and quickly
0%
by itself and slowly
0%
with outside intervention and slowly
0%
by itself but it has now effect on how fast the reaction takes

$10\ g$ of liquid at

$300\ K$ are heated to

$350\ K$ . The liquid absorbs

$6\ kcal$ . What is the specific heat of the liquid (in

$cal/g^{\circ}C)$ ?

0%
$6$
0%
$12$
0%
$60$
0%
$120$
0%
$600$

Explanation

Answer(B)

liquid is heated from 300k to 350k so temperature difference

$=$

$50K$

$C$

$=$

$Q/(m*T)$

$=$

$6000/(10*50)$

$=$

$12cal/gC$

Which set of the conditions given below guarantees that a reaction will be spontaneous?

0%
$\Delta H(+)$ and $\Delta S(-)$
0%
$\Delta H(-)$ and $\Delta S(+)$
0%
$\Delta H(+)$ and $\Delta S(+)$ at low temperature
0%
$\Delta H(-)$ and $\Delta S(-)$ at high temperature
0%
$\Delta G(+)$

Explanation

For a spontaneous reaction,

$\displaystyle \Delta G < 0$
Also,

$\displaystyle \Delta G = \Delta H - T \Delta S$
Hence, when

$\displaystyle \Delta H < 0$ and

$\displaystyle \Delta S > 0$ ,

$\displaystyle \Delta G < 0$ , and the reaction will become spontaneous.

Heat required, to rise one unit mass of a substance by 1 degree Celsius, is known as ......................... .

0%
gibbs free energy
0%
heat of formation
0%
specific heat
0%
Heinsenberg uncertainty principle
0%
heat of vaporization

Explanation

The energy of a system available to do work is called as gibbs free energy.

Heat required to rise one unit mass of a substance by 1 degree Celsius is known as specific heat. Its formula is

$\dfrac{Q}{m\times{\Delta{T}}}$ .

Heat absorbed or released during production of a substance from elements in their standard states is known as heat of formation.

Heinsenberg uncertainty principle states that, "It is not possible to determine simultaneously the position and momentum of a moving microsopic particle (electron) with absolute accuracy".

Heat required to vaporize unit mass of a substance, without changing its temperature at constant pressure, is called heat of vaporization.

Determine the heat of reaction for the combustion of sulphur dioxide:

$\displaystyle 2{ SO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \rightarrow 2{ SO }_{ 3 }\left( g \right)$

Given the following thermochemical equations:

$\displaystyle { S }_{ 8 }\left( s \right) +8{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 2 }\left( g \right) \Delta H=-2374.6kJ$
II.

$\displaystyle { S }_{ 8 }\left( s \right) +12{ O }_{ 2 }\left( g \right) \rightarrow 8{ SO }_{ 3 }\left( g \right) \Delta H=-3165.8kJ$

0%
-5540.4 kJ
0%
-1385.1 kJ
0%
-791.2 kJ
0%
-197.8 kJ
0%
-700 kJ

Explanation

$S_8$

$+$

$8O_2 \to 8SO_2$ ......

$\Delta H$

$=$

$-2374.6kJ$ .....

$(1)$

$S_8$

$+$

$12O_2 \to 8SO_3$ ......

$\Delta H$

$=$

$-3165.8kJ$ .....

$(2)$

Subtracting the two equations, that is

$(1)-(2)$ , we get-

$-4O_2 \to 8SO_2$

$-$

$8SO_3$ ......

$\Delta H$

$=$

$791.2kJ$ ......

$(3)$

Diving the equation

$(3)$ by

$4$ , we get-

$-O_2 \to 2SO_2$

$-$

$2SO_3$ .....

$\Delta H$

$=$

$197.8kJ$

Now, multiplying by a negative sign on both the sides of the equation obtained, we get-

$O_2 \to -$

$2SO_2$

$+$

$2SO_3$ .

The value of

$\Delta H$ will also be multiplied by a negative sign, that is, it will be

$-197.8kJ$ .

Hence,

$O_2$

$+$

$2SO_2 \to 2SO_3$ , will have the value of

$\Delta H$ as

$-197.8kJ$ .

Option

$D$ is the correct answer.

When one mole of sulfur burns to form

$SO_2$ , 1,300 calories are released. When one mole of sulfur burns to form

$SO_3$ , 3,600 calories are released. What is the

$\Delta H$ when one mole of

$SO_2$ is burned to form

$SO_3$ ?

0%
3,900 cal
0%
-1,950 cal
0%
1,000 cal
0%
-500 cal
0%
-2, 300 cal

Explanation

Let's write out our information in a chemical way:

(1)

$S + O_2 \rightarrow SO_2 ΔH = -1.3 kcal$
(2)

$S + (3/2)O_2 \rightarrow SO_3 ΔH = -3.6 kcal$

$SO_2 + (1/2)O_2 \rightarrow SO_3$
To obtain the target equation, what I will do is flip the first equation, remembering to also change the sign on the enthalpy. Here are equations (1) and (2) with (1) flipped:
(1)

$SO_2 \rightarrow S + O_2 ΔH = +1.3 kcal$
(2)

$S + (3/2)O_2 \rightarrow SO_3 ΔH = -3.6 kcal$
When those two equations are added, an O2 will cancel out, leaving (1/2)O2 on the right-hand side. (The S also cancels.) We add the enthalpies:
+1.3 plus -3.6 = -2.3 kcal or -2300 cal

$\displaystyle S\left( s \right) +{ O }_{ 2 }\left( g \right) \rightarrow { SO }_{ 2 }\left( g \right) \quad \Delta { H }^{ \circ }=-297kJ$

$\displaystyle 2{ SO }_{ 2 }\left( g \right) +{ O }_{ 2 }\left( g \right) \rightarrow 2{ SO }_{ 3 }\left( g \right) \quad \Delta { H }^{ \circ }=-198kJ$
Given the above thermochemical reactions, what is the heat of reaction for the formation of

$\displaystyle { SO }_{ 3 }\left( g \right)$ provided below?

$\displaystyle S\left( s \right) +{ 3/2O }_{ 2 }\left( g \right) \rightarrow { SO }_{ 3 }\left( g \right)$

0%
-495 kJ
0%
-396 kJ
0%
-198 kJ
0%
+99
0%
+198 kJ

Explanation

$S$

$+$

$O_2$

$=$

$SO_2$ ,

$\Delta H^o$

$=$

$-297kJ$ .....

$(1)$

$2SO_2$

$+$

$O_2$

$=$

$2SO_3$ ,

$\Delta H^o$

$=$

$-198kJ$ ....

$(2)$

Dividing equation (2) by

$2$ , we get-

$SO_2$

$+$

$\dfrac{1}{2}O_2$

$=$

$SO_3$ ,

$\Delta H^o$

$=$

$-99kJ$ ....

$(3)$

Now, adding equation (1) and (3), we get-

$S$

$+$

$\dfrac{3}{2}O_2$

$+$

$SO_2$

$=$

$SO_2$

$+$

$S O_3$ ......

$(4)$ .

Cancelling

$SO_2$ from both the sides of the equation, we get the equation

$S$

$+$

$\dfrac{3}{2}O_2$

$=$

$SO_3$ .

Thus the value of

$\Delta H^o$ will also be added. Since we are adding the equations (1) and (3), which comes out to be equal to

$[-297+(-99)]kJ$ , that is

$-396kJ$ .

Thus option

$B$ is the correct answer.

Statement I
As ice absorbs heat and begins to melt, its temperature remains constant
Because
Statement II
Changes of state bring about changes in a substances potential energy, not in its kinetic energy

0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1
0%
Both the Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%
Statement 1 is correct but Statement 2 is not correct.
0%
Statement 1 is not correct but Statement 2 is correct.
0%
Both the Statement 1 and Statement 2 are not correct.

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 3 - MCQExams.com

Born-Haber Cycle

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Thermodynamics Quiz 3 - MCQExams.com

Born-Haber Cycle

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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