For the reactions: ${\rm{C}} + {{\rm{O}}_2} \to {\rm{C}}{{\rm{O}}_2};\Delta H = - 393{\rm{kJ}}$ $2{\rm{Zn}} + {{\rm{O}}_2} \to 2{\rm{ZnO}};\Delta H = - 412{\rm{kJ}}$ which one is correct?

Oxidation of carbon is not feasible

Oxidation of Zn is not feasible

One mole of ice is converted into water at 273 K. The entropies of $\;{{\rm{H}}_2}{\rm{O}}$ (S) and $\;{{\rm{H}}_2}{\rm{O}}$ (l)are 38.20 and 60.01 J ${\rm{mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ respectively. The enthalpy change for the conversion is:

$5954{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$59.54{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$595.4{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$320.6{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 500 questions

1 Hard

1 of 500

Analyzing +4 | -1

Consider following reactions
I:

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II:

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III:

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

Select the correct statement

View answer and explanation

Correct option: D

All the above are correct statements

Option D is correct because all the given statements are correct based on Hess's Law and the definitions of heat of formation and combustion.

2 Hard

2 of 500

Application +4 | -1

When

0.18 \mathrm{~g}

of glucose was burnt in a bomb calorimeter, the temperature rose by

4 \mathrm{~K}.

The heat capacity of the calorimeter system is

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

View answer and explanation

Correct option: A

700 \mathrm{~J} / \mathrm{K}

Option A (

700 \mathrm{~J} / \mathrm{K}

) is correct because it is the heat capacity calculated using the heat released by the combustion of glucose and the temperature rise.

3 Medium

3 of 500

Analyzing +4 | -1

Statement-1:

{CsI}

is less soluble in water compared to

{CsCl}

Satement-2: More amount of heat is liberated during the formation of

{CsCl}

than the formation

{Cs} I

from their elements

View answer and explanation

Correct option: A

Both statements 1 and 2 correct

Both statements 1 and 2 are correct. CsI is less soluble than CsCl due to lower hydration enthalpy, and more heat is released during CsCl formation.

4 Medium

4 of 500

Application +4 | -1

Consider the following processes :

\begin{array}{cccccccccccccccccccc} {}&{\Delta H(KJ/mol)} \\ {1/2A \to B}&{ + 150} \\ {3B \to 2C + D}&{ - 125} \\ {E + A \to 2D}&{ + 350} \end{array}

For

{B}+{D} \rightarrow {E}+2 {C}, \Delta {H}

will be :

View answer and explanation

Correct option: B

-175 {~kJ} / {mol}

Option B (

-175 {~kJ} / {mol}

) is correct because it is calculated by manipulating the given equations and their enthalpy changes according to Hess's Law.

5 Medium

5 of 500

Understanding +4 | -1

The difference b/w

\Delta H

and

\Delta u

at constant volume is equal to

View answer and explanation

Correct option: C

V\Delta P

Option C (

V\Delta P

) is correct because the difference between

\Delta H

and

\Delta u

at constant volume is equal to

V\Delta P

based on the enthalpy equation.

6 Medium

6 of 500

Application +4 | -1

The heat evolved in the combustion of methane is given by the following equations

\mathrm{C}{\mathrm{H}_4}\left( \mathrm{g} \right) + 2{\mathrm{O}_2}\left( \mathrm{g} \right) \to \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( l \right);\Delta H = - 890.3\mathrm{kJ}

How many grams of methane would be required to produce 44.15 kJ of heat of combustion?

View answer and explanation

Correct option: B

8 g

Option B (8 g) is correct because it is calculated based on the heat evolved in the combustion of methane and the given enthalpy change.

7 Medium

7 of 500

Understanding +4 | -1

A reaction accompanied with the absorption of energy is :

View answer and explanation

Correct option: C

Electrolysis of water

Option C (Electrolysis of water) is correct because electrolysis requires the absorption of electrical energy to break chemical bonds.

8 Medium

8 of 500

Understanding +4 | -1

\Delta {U}

of combustion of methane is

-{x} \, \mathrm{kJ} \mathrm{~mol}^{-1}

. The value of

\Delta {H}

View answer and explanation

Correct option: C

< \Delta {U}

Option C (< ΔU) is correct because the change in the number of moles of gas is negative, making ΔH less than ΔU.

9 Hard

9 of 500

Application +4 | -1

A sample of oxygen gas expands its volume from

3 \mathrm{~L}

5 \mathrm{~L}

against a constant pressure of

3 \mathrm{~atm}.

If the work done during expansion is used to heat

10 \mathrm{~mol}

of water initially present at

290 \mathrm{~K},

its final temperature will be

(Specific heat capacity of water

=4.18 \mathrm{~JK}^{-1} \mathrm{~g}^{-1}

)

View answer and explanation

Correct option: B

290.8 \mathrm{~K}

Option B (

290.8 \mathrm{~K}

) is correct because it accurately calculates the final temperature of the water after accounting for the work done by the expanding oxygen gas.

10 Medium

10 of 500

Application +4 | -1

A 3 mole sample of a triatomic ideal gas at

300 \mathrm{~K}

is allowed to expand under adiabatic reversible condition from

5 \mathrm{~L}

40 \mathrm{~L}

. The value of

\Delta {H}

View answer and explanation

Correct option: B

-14.965 \mathrm{~kJ}

Option B (-14.965 kJ) is correct because the enthalpy change is calculated using adiabatic reversible expansion and the heat capacity at constant pressure.

11 Medium

11 of 500

Analyzing +4 | -1

48 g of C (diamond) on complete combustion evolves 1584 kJ of heat. The standard heat of formation of gaseous carbon is 725 kJ/mol. The energies required for the process
(i)C (graphite) → C(gas)
(ii)C (diamond) → C(gas)are:

View answer and explanation

Correct option: C

725, 723

Option C (725, 723) is correct because the heat of formation of gaseous carbon from graphite is 725 kJ/mol, and the heat of formation from diamond is slightly less due to diamond's higher energy.

12 Medium

12 of 500

Understanding +4 | -1

When 1 mole of gas is heated at constant volume. Temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct?

View answer and explanation

Correct option: B

q=∆U=500 J,W=0

Option B (q=∆U=500 J,W=0) is correct because at constant volume, no work is done, and all the heat supplied goes into changing the internal energy.

13 Easy

13 of 500

Analyzing +4 | -1

For the reactions:

\mathrm{C} + {\mathrm{O}_2} \to \mathrm{C}{\mathrm{O}_2};\Delta H = - 393\mathrm{kJ}

2\mathrm{Zn} + {\mathrm{O}_2} \to 2\mathrm{ZnO};\Delta H = - 412\mathrm{kJ}

which one is correct?

View answer and explanation

Correct option: A

Carbon can reduce ZnO to Zn

Option D (Zn liberates more heat than carbon during oxidation) is correct because the absolute value of ∆H for the oxidation of Zn is greater than that of carbon.

14 Medium

14 of 500

Analyzing +4 | -1

Heat of combustion of

{CH}_4, {C}_2 {H}_4, {C}_2 {H}_6

are

-890,-1411

and

-1560 \mathrm{~KJ} / \mathrm{mol}

respectively. Which has the lowest fuel value is:

View answer and explanation

Correct option: B

{C}_2{H}_4

Option B (

{C}_2{H}_4

) has the lowest fuel value because, when the heat of combustion is normalized per gram, it yields the smallest magnitude.

15 Medium

15 of 500

Application +4 | -1

If equilibrium constant

K_c

for a reaction is

10^{10}

27^{\circ} \mathrm{C}

, then

\Delta G^0

of the reaction at

27^{\circ} \mathrm{C}

will be

View answer and explanation

Correct option: B

-6909 R

Option B (

-6909 R

) is correct because it accurately calculates the standard Gibbs free energy change using the given equilibrium constant and temperature.

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16 Medium

16 of 500

Analyzing +4 | -1

The heats of neutralization of

C{H_3}COOH,\,\,\,HCOOH,\,\,HCN

and

{H_2}S

are

- 13.2,\, - \,13.4,\, - 2.9

and - 3.8 kcal per equivalent respectively. Arrange the increasing order of strength.

View answer and explanation

Correct option: A

HCCOH > C{H_3}COOH > {H_2}S > HCN

Option A is correct because the strength of an acid is inversely proportional to its heat of neutralization (more negative = stronger acid).

17 Medium

17 of 500

Understanding +4 | -1

The absolute enthalpy of neutralisation of the reaction

\mathrm{MgO}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_2(a q)+\mathrm{H}_2 \mathrm{O}(l) \text { will be}

View answer and explanation

Correct option: C

less than

-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option A (greater than

-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is incorrect. The correct answer is less than

-57.33 \mathrm{~kJ} \mathrm{~mol}^{-1}

because the reaction involves the formation of water and the neutralization of a strong acid with a metal oxide, which releases more heat than the neutralization of a strong acid with a strong base.

18 Medium

18 of 500

Understanding +4 | -1

Consider the following statements about the Hess's law of constant heat summation.
I. If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
II. The enthalpy of an overall reaction

A \rightarrow B

along one route is

\Delta_r H

and

\Delta_r H_1, \Delta_r H_2, \Delta_r H_3 \ldots

representing enthalpies of reactions leading to the same product,

B

along another route, then we have

\Delta_r H=\Delta_r H_1+\Delta_r H_2+\Delta_r H_3 \cdots

Select the correct statement and choose the correct option.

View answer and explanation

Correct option: C

Both I and II

Option C (Both I and II) is correct because both statements accurately describe Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway taken.

19 Medium

19 of 500

Application +4 | -1

One mole of a non-ideal gas undergoes a change of state (2.0 atm,3.0 L,95 K)→(4.0 atm,5.0 L,245 K) with a change in internal energy, ∆U=30.0 L atm. The change in enthalpy (∆H) of the process in L atm is:

View answer and explanation

Correct option: C

Option C (44) is correct because it is the change in enthalpy calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁).

20 Easy

20 of 500

Remembering +4 | -1

Which of the following is an example of extensive property?

View answer and explanation

Correct option: C

Mass

Option C (Mass) is correct because mass is an extensive property, meaning its value depends on the amount of substance.

21 Medium

21 of 500

Understanding +4 | -1

Which of the following statements is true?

View answer and explanation

Correct option: C

The total energy of the universe remains
constant

Option C (The total energy of the universe remains constant) is correct because it is a statement of the first law of thermodynamics.

22 Hard

22 of 500

Application +4 | -1

One mole of ice is converted into water at 273 K. The entropies of

\;{\mathrm{H}_2}\mathrm{O}

(S) and

\;{\mathrm{H}_2}\mathrm{O}

(l)are 38.20 and 60.01 J

\mathrm{mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}

respectively. The enthalpy change for the conversion is:

View answer and explanation

Correct option: B

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

Option B (

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

) is correct because it is the enthalpy change calculated using the formula ∆H=T∆S at equilibrium.

23 Medium

23 of 500

Understanding +4 | -1

Which of the following conditions lead to a
spontaneous process?

View answer and explanation

Correct option: D

Both

\Delta \mathrm{H}

and

\mathrm{T} \Delta \mathrm{S}

are + ve but

\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}

Option D (Both ΔH and TΔS are + ve but TΔS>ΔH) is correct because for a spontaneous process, ΔG must be negative, which occurs when TΔS > ΔH and both are positive.

24 Medium

24 of 500

Application +4 | -1

For vaporization of water at 1 atmospheric pressure, the values of ∆H and ∆S are 40.63 kJ

mo{l^{ - 1}}

and 108.8

J{K^{ - 1}}mo{l^{ - 1}},

respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is:

View answer and explanation

Correct option: C

373.4 K

Option C (373.4 K) is correct because at this temperature, the Gibbs free energy change (∆G) is zero, indicating equilibrium between liquid and gas phases.

25 Medium

25 of 500

Understanding +4 | -1

Assertion (A):

\Delta G

has the unit of energy.
Reason (R):

\Delta H

and

T \Delta S

are the energy dependent terms.

View answer and explanation

Correct option: A

Both Assertion and Reason are correct; Reason is the correct explanation of Assertion

Option A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) is correct because Gibbs free energy change has units of energy, and both enthalpy and entropy terms contribute to the energy change.

26 Hard

26 of 500

Application +4 | -1

In the reaction

\mathrm{CS}_2({l})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{SO}_2(\mathrm{g}) ; \Delta H= -265 \text{ kcal}.

The enthalpies of formation of

\mathrm{CO}_2

and

\mathrm{SO}_2

are both negative and are in the ratio

4: 3.

The enthalpy of formation of

\mathrm{CS}_2

+26 \mathrm{~kcal} \mathrm{~mol}^{-1}.

Calculate the enthalpy of formation of

\mathrm{SO}_2.

View answer and explanation

Correct option: D

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

Option D (

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

) is correct because it is calculated using Hess's law and the given enthalpy of reaction and formation of other compounds.

27 Medium

27 of 500

Application +4 | -1

For a reaction at 25℃ enthalpy change (∆H) and entropy change (∆S) are -11.7×

{10^3}

Jmo{l

- 105Jmo{l^{- 1}}{K^{- 1}}

^{ - 1}}

Jmo{l^{ - 1}}

respectively. The reaction is:

View answer and explanation

Correct option: B

Non-spontaneous

The correct answer is B (Non-spontaneous) because the calculated Gibbs free energy change (∆G) is positive, indicating a non-spontaneous reaction.

28 Easy

28 of 500

Application +4 | -1

The melting point a solid is

27^{\circ} \mathrm{C}

and its latent heat of fusion is 600 calories/mole the entropy change for the fusion of one mole of the solid in

\mathrm{cal/K}

is ___.

View answer and explanation

Correct option: A

Option A (2) is correct because it is calculated using the formula

\Delta S = \Delta H / T

29 Hard

29 of 500

Analyzing +4 | -1

Calculate the enthalpy of formation of

{I}_2 {O}_5(\mathrm{s})

from the following data.

(i)

{I}_2 {O}_5(\mathrm{s})+{H}_2{O}({l}) \rightarrow 2 {HIO}_3\text{(aq)}; \Delta H=+4.0 \mathrm{~kJ}

(ii)

{KI} \text{(aq)} +3 {HClO}\text{(aq)} \rightarrow {HIO}_3 \text{(aq)} +2 {HCl}\text{(aq)}+{KCl}(\mathrm{aq}); \Delta H=-322.0 \mathrm{~kJ}

(iii)

{NaOH}(\mathrm{aq})+{HClO}(\mathrm{aq}) \rightarrow {NaOCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-44.0 \mathrm{~kJ}

(iv)

{NaOH}(\mathrm{aq})+{HCl}(\mathrm{aq}) \rightarrow {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-57.0 \mathrm{~kJ}

(v)

2 {NaOH}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow{NaOCl}(\mathrm{aq})+ {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-100.0 \mathrm{~kJ}

(vi)

2 {KI}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow 2 {KCl}(\mathrm{aq})+{I}_2(\mathrm{s}); \Delta H=-224.0 \mathrm{~kJ}

(vii)

{H}_2(\mathrm{g})+1 / 2 \, {O}_2(\mathrm{g}) \rightarrow {H}_2 {O}({l}); \Delta H=-285.0 \mathrm{~kJ}

(viii)

1 / 2 \,{H}_2(\mathrm{g})+1 / 2 \, {Cl}_2(\mathrm{g}) \rightarrow {HCl}(\mathrm{g}); \Delta H= -92.0 \mathrm{~kJ}

(ix)

{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow {HCl}(\mathrm{aq}); \Delta H=-75.0 \mathrm{~kJ}

View answer and explanation

Correct option: A

-169.0 \mathrm{~kJ}

Option A (-169.0 kJ) is correct because the enthalpy of formation of I2O5(s) is calculated by combining the given reactions using Hess's Law.

30 Medium

30 of 500

Application +4 | -1

The internal energy change (in J) when 90 g of water undergoes complete evaporation at

100^\circ C

is …..,
(Given :

\Delta {H_{\mathrm{vap}}}

for water at

373\,\,K = 41kJ/mol,\,R = 8.314\,\,\,J{K^{ - 1}}mo{l^{ - 1}})

View answer and explanation

Correct option: A

189494

Option A (189494) is correct because it is the internal energy change calculated using the enthalpy of vaporization and the ideal gas constant.

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