Match List - I with List - II. .tg .tg List - I (Partial Derivatives) List - II (Thermodynamic Quantity) (A) $\left(\frac{\partial \mathrm{G}}{\partial \mathrm{T}}\right)_{\mathrm{P}}$ (I) Cp (B

We want to match each partial derivative in List–I to the appropriate thermodynamic quantity in List–II: 1. $\displaystyle\left(\frac{\partial G}{\partial T}\right)_{P}$ A well-known thermodynamic identity is: $ \left(\frac{\partial G}{\partial T}\right)_{P} \;=\; -S. $ Hence, $ (A) \quad \longrightarrow \quad -S \quad (\text{II}). $ 2. $\displaystyle\left(\frac{\partial H}{\partial T}\right)_{P}$ By definition of the heat capacity at constant pressure, $ \left(\frac{\partial H}{\partial T}\righ

Thermodynamics — Page 10 | JEE Chemistry

Q91

The hydration energies of

K^+

and

Cl^-

are

-x

and

-y

kJ/mol respectively. If the lattice energy of KCl is

-z

kJ/mol, then the heat of solution of KCl is :

A

x + y + z

B

z - (x + y)

C

-z - (x + y)

D

x - y - z

Correct Answer

Option B

Solution

$\mathrm{KCl}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{O} \xrightarrow{\Delta \mathrm{H} \text{ sol. }} \mathrm{K}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{\text{(aq) }}^{-1}$

\begin{aligned} \Delta \mathrm{H}_{\mathrm{Sol}^{\mathrm{n}} .} & =\mathrm{L} \cdot \mathrm{E} \cdot+(\mathrm{H} \cdot \mathrm{E})_{\mathrm{K}_{(\mathrm{g})}^{+}}+(\mathrm{HE})_{\mathrm{Cl}_{(\mathrm{g})}^{-1}} \\ & =\mathrm{Z}-\mathrm{x}-\mathrm{y} \\ & =\mathrm{z}-(\mathrm{x}+\mathrm{y}) \end{aligned}

Q92

The correct statement amongst the following is :

A The standard state of a pure gas is the pure gas at a pressure of 1 bar and temperature 273 K.

B

\Delta_f H^{\circ}_{500}

is zero for

O_2(g)

.

C

\Delta_f H^{\circ}_{298}

is zero for

O(g)

.

D The term 'standard state' implies that the temperature is 0°C.

Correct Answer

Option B

Solution

Option A: It claims that the standard state for a pure gas is defined at 1 bar and 273 K.

In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is 298 K (25°C), not 273 K (0°C).

Thus, Option A is incorrect.

Option B: It states that the standard enthalpy of formation

\Delta_f H^\circ_{500}

is zero for

O_2(g)

.

By convention, the standard enthalpy of formation for an element in its most stable form is defined as zero.

Even though typical tables list

\Delta_f H^\circ_{298}

values, if one were to evaluate the formation enthalpy at any temperature where the element remains in its standard state, the reaction

O_2(g) \rightarrow O_2(g)

has no change in enthalpy. Therefore, by definition,

\Delta_f H^\circ_{500}(O_2(g)) = 0

. This makes Option B correct. Option C: It claims that

\Delta_f H^\circ_{298}

is zero for

O(g)

(atomic oxygen). However, the standard state of oxygen is diatomic

O_2(g)

, not atomic oxygen. Since

O(g)

is not the most stable form of oxygen under standard conditions, its formation enthalpy is not zero.

Hence, Option C is incorrect.

Option D: It asserts that the term "standard state" implies that the temperature is 0°C (273 K).

In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature—commonly 298 K for tabulated thermodynamic data—not necessarily 0°C.

Thus, Option D is also incorrect.

To summarize: The only correct statement is Option B.

Therefore, the correct answer is: Option B.

Q93

Let us consider a reversible reaction at temperature, T. In this reaction, both

\Delta \mathrm{H}

and

\Delta \mathrm{S}

were observed to have positive values. If the equilibrium temperature is Te , then the reaction becomes spontaneous at:

A

\mathrm{Te}>\mathrm{T}

B

\mathrm{T}>\mathrm{Te}

C

\mathrm{T}=\mathrm{Te}

D

\mathrm{Te}=5 \mathrm{~T}

Correct Answer

Option B

Solution

For reaction to be spontaneous according to $2^{\text{nd }}$ law:

\begin{aligned} & \Delta \mathrm{G}\left(\frac{\Delta \mathrm{H}}{\Delta \mathrm{~S}}\right)=\mathrm{T}_{\mathrm{e}} \\ & \Rightarrow \mathrm{~T}>\mathrm{T}_{\mathrm{e}} \end{aligned}

Q94

Match with . }{\partial \mathrm{T}}\right)_{\mathrm{P}}$

List - I		List - II
(B)	$\left(\dfrac{\partial \mathrm{H}}{\partial \mathrm{T}}\right)_{\mathrm{P}}$	(I)	Cp
(C)	$\left(\dfrac{\partial \mathrm{G}}{\partial \mathrm{P}}\right)_{\mathrm{T}}$	(II)	$-$ S
(D)	$\left(\dfrac{\partial \mathrm{U}}{\partial \mathrm{T}}\right)_{\mathrm{V}}$	(III)	Cv
()		(IV)	V

A (A)-(I), (B)-(II), (C)-(IV), (D)-(III)

B (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

C (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

D (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

Correct Answer

Option C

Solution

We want to match each partial derivative in List–I to the appropriate thermodynamic quantity in List–II: 1.

$\displaystyle\left(\dfrac{\partial G}{\partial T}\right)_{P}$ A well-known thermodynamic identity is: $\left(\dfrac{\partial G}{\partial T}\right)_{P} \;=\; -S.$ Hence, $(A) \quad \longrightarrow \quad -S \quad (\text{II}).$ 2.

$\displaystyle\left(\dfrac{\partial H}{\partial T}\right)_{P}$ By definition of the heat capacity at constant pressure, $\left(\dfrac{\partial H}{\partial T}\right)_{P} \;=\; C_P.$ Hence, $(B) \quad \longrightarrow \quad C_P \quad (\text{I}).$ 3.

$\displaystyle\left(\dfrac{\partial G}{\partial P}\right)_{T}$ A standard thermodynamic relation (for a single‐component system) is: $\left(\dfrac{\partial G}{\partial P}\right)_{T} \;=\; V.$ Hence, $(C) \quad \longrightarrow \quad V \quad (\text{IV}).$ 4.

$\displaystyle\left(\dfrac{\partial U}{\partial T}\right)_{V}$ By definition of the heat capacity at constant volume, $\left(\dfrac{\partial U}{\partial T}\right)_{V} \;=\; C_V.$ Hence, $(D) \quad \longrightarrow \quad C_V \quad (\text{III}).$ Final Matching $(A) \to \text{(II)}, \quad (B) \to \text{(I)}, \quad (C) \to \text{(IV)}, \quad (D) \to \text{(III)}.$ Which corresponds to: Option C: (A)-(II), (B)-(I), (C)-(IV), (D)-(III).

Q95

Given below are two statements: Statement I : When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II : At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option C

Solution

The correct choice is Option C: Both Statement I and Statement II are true.

Explanation: Statement I is true because During the ice–water phase change at 0 °C, any heat added goes into the latent heat of fusion.

The temperature remains fixed at 0 °C until all the ice has melted.

Statement II is also true because The heat absorbed at the melting point is used to overcome (break) the hydrogen‐bonding forces between water molecules in ice—in other words, it increases the system’s potential energy.

Since no heat goes into raising the kinetic energy of the molecules, the temperature (which measures average kinetic energy) does not increase during melting.

Mathematically, for a mass m of ice, the heat needed is

Q = m\,L_f

where $L_f$ is the latent heat of fusion of ice.

This heat does not change the temperature but is entirely consumed in changing the phase (i.e., increasing potential energy).

Q96

Match with . } < 0$$

List - I		List - II
(C)	$\Delta {H_{reaction}}$	(III)	Isothermal and isobaric process
(D)	Exothermic Process	(IV)	[Bond energies of molecules in reactants] $-$ [Bond energies of product molecules

A (A) - (III), (B) - (II), (C) - (IV), (D) - (I)

B (A) - (II), (B) - (III), (C) - (IV), (D) - (I)

C (A) - (II), (B) - (III), (C) - (I), (D) - (IV)

D (A) - (II), (B) - (I), (C) - (III), (D) - (IV)

Correct Answer

Option B

Solution

(A) For a spontaneous process

\Delta

GT,P reaction = (

\Sigma

Bond energies of reactants) - (

\Sigma

bond energies of products) (D)

\Delta

H < 0 is for exothermic reaction

Q97

Given (i) 2Fe2O3(s)

\to

4Fe(s) + 3O2(g);

\Delta

rGo = + 1487.0 kJ mol-1 (ii) 2CO(g) + O2(g)

\to

2CO2(g);

\Delta

rGo =

-

514.4 kJ mol-1 Free energy change,

\Delta

rGo for the reaction 2Fe2O3(s) + 6CO(g)

\to

4Fe(s) + 6CO2(g) will be :

A

-

112.4 kJ mol-1

B

-

56.2 kJ mol-1

C

-

168.2 kJ mol-1

D

-

208.0 kJ mol-1

Correct Answer

Option B

Solution

Given (i) 2Fe2O3(s) $\to$ 4Fe(s) + 3O2(g);

\Delta

rGo = + 1487.0 kJ mol-1 (ii) 2CO(g) + O2(g) $\to$ 2CO2(g);

\Delta

rGo = $-$ 514.4 kJ mol-1 Multiply reaction (ii) with 3 and we get, (iii) 6CO(g) + 3O2 (g)

\overset{\,}\longrightarrow

6CO2(g) ;

\Delta

rGo = 3 $\times$ $-$ 514.4 = $-$ 1543.2 kJ mol $-$ 1 Now add reaction (i) with (iii), we get 2Fe2O3(s) + 6CO(g)

\overset{\,}\longrightarrow

4Fe(s) + 6CO2(g)

\therefore\,\,\,

Free energy change of the reaction,

\Delta

rGo = 1487.0 $-$ 1543.2 = $-$ 56.2 kJ/mol

Q98

Given (i) C (graphite) + O2(g)

\to

CO2(g);

\Delta

rH

^\Theta

= x kJ mol

-

1 (ii) C(graphite) +

{1 \over 2}

O2(g)

\to

CO(g);

\Delta

rH

^\Theta

= y kJ mol

-

1 (iii) CO(g) +

{1 \over 2}

O2(g)

\to

CO2(g);

\Delta

rH

^\Theta

= z kJ mol

-

1 Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?

A z = x + y

B x = y + z

C x = y – z

D y = 2z – x

Correct Answer

Option B

Solution

In reaction (i), the product is CO2 If we add reaction (ii) and (iii) we get, C + O2 $\to$ CO2 here also product is CO2 from same reactant C and O2.

So according to hess law, we can say x = y + z

Q99

At 25

^\circ

C and 1 atm pressure, the enthalpies of combustion are as given below : .tg .tg Substance

{H_2}

C (graphite)

{C_2}{H_6}(g)

{{{\Delta _c}{H^\Theta }} \over {kJ\,mo{l^{ - 1}}}}

- 286.0

- 394.0

- 1560.0

The enthalpy of formation of ethane is

A +54.0 kJ mol

-

1

B

-

68.0 kJ mol

-

1

C

-

86.0 kJ mol

-

1

D +97.0 kJ mol

-

1

Correct Answer

Option C

Solution

2 \mathrm{C}

(graphite)

+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})

\Delta \mathrm{H}_{\mathrm{r}}=+1560+2(-394)+3(-286)

=-86.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

Enthalpy of formation of C2H6(g) = –86.0 kJ mol–1

Q100

Given .tg .tg (A)

\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)}

\mathrm{\Delta H_1^0=-x~kJ~mol^{-1}}

(B)

\mathrm{C(graphite)+O_2(g)\to CO_2(g)}

\mathrm{\Delta H_2^0=-y~kJ~mol^{-1}}

\mathrm{\Delta H^0}

for the reaction

\mathrm{C(graphite)+\dfrac{1}{2}O_2(g)\to CO(g)}

is :

A

\dfrac{x-2y}{2}

B

\dfrac{x+2y}{2}

C

2y-x

D

\dfrac{2x-y}{2}

Correct Answer

Option A

Solution

The standard enthalpy change for the reaction

\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}

can be calculated using Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in many steps.

If we modify the given reactions to reflect the formation of CO, we can add the modified reactions together to get the reaction of interest.

Consider the reactions: (A)

\mathrm{2CO(g)+O_2(g)\to 2CO_2(g)} \quad \Delta H_1^0=-x~kJ~mol^{-1}

(B)

\mathrm{C(graphite)+O_2(g)\to CO_2(g)} \quad \Delta H_2^0=-y~kJ~mol^{-1}

First, divide reaction (A) by 2 to get the reaction for 1 mol of CO: (A/2)

\mathrm{CO(g)+\frac{1}{2}O_2(g)\to CO_2(g)} \quad \Delta H_1^0=-\frac{x}{2}~kJ~mol^{-1}

Then, subtract the divided reaction (A) from reaction (B) to get the reaction for the formation of CO:

\mathrm{C(graphite)+\frac{1}{2}O_2(g)\to CO(g)}

\Delta H^0=\Delta H_2^0-\Delta H_1^0=-y-(-\frac{x}{2})= \frac{x}{2} - y~kJ~mol^{-1}

Therefore, the correct answer is Option A:

\frac{x-2y}{2}