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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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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 181-210 of 329 questions

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181 Medium
181 of 329
Understanding +4 | -1
Identify the correct statement from the
following is a chemical reaction.
View answer and explanation
Correct option: B
The change in entropy along with suitable
change in enthalpy decides the rate of
reaction
The correct answer is B because the spontaneity and rate of a chemical reaction are determined by the Gibbs free energy change, which depends on both enthalpy and entropy changes.
182 Medium
182 of 329
Analyzing +4 | -1
In biological cells that have a plentiful supply of O2,\mathrm{O}_2, glucose is oxidized completely to CO2\mathrm{CO}_2 and H2O\mathrm{H}_2 \mathrm{O} by a process called aerobic oxidation. Muscle cells may be deprived of O2\mathrm{O}_2 during vigorous exercise and in that case, one molecule of glucose is converted to two molecules of lactic acid, CH3CH(OH)COOH,\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}, by a process called anaerobic glycolysis.

C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l);ΔH∘=−2880 kJ/mol\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{s})+6 \mathrm{O}_2(\mathrm{g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}({l}) ; \Delta H^{\circ}=-2880 \mathrm{~kJ} / \mathrm{mol}

C6H12O6(s)→2CH3CH(OH)COOH(s);ΔH∘=+2530 kJ/mol\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{s}) \rightarrow 2 \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s}) ; \Delta H^{\circ}=+2530 \mathrm{~kJ} / \mathrm{mol}

Which of the following statements is true regarding aerobic oxidation and anaerobic glycolysis with respect to energy change as heat?
View answer and explanation
Correct option: A
Aerobic oxidation has biological advantage over anaerobic glycolysis by 5410 kJ/mol5410 \mathrm{~kJ} / \mathrm{mol}
Option A (Aerobic oxidation has biological advantage over anaerobic glycolysis by 5410 kJ/mol5410 \mathrm{~kJ} / \mathrm{mol}) is correct because aerobic oxidation releases more energy than anaerobic glycolysis.
183 Medium
183 of 329
Analyzing +4 | -1
Following reaction occurs in an automobile 2C8H18(g)+25 O2(g)→16 CO2(g)+18H2O(g)2{C_8}{H_{18}}(g) + 25\,{O_2}(g) \to 16\,C{O_2}(g) + 18{H_2}O(g) The sign of ΔH, ΔS and  ΔG\Delta H,\,\Delta S\mathrm{ }\,\mathrm{and}\,\,\mathrm{ }\Delta G would be
View answer and explanation
Correct option: B
−,+,−- , + , -
Option A (Both A\mathbf{A} and R\mathbf{R} are true and R\mathbf{R} is the correct explanation of A\mathbf{A}) is correct because the equivalent mass of N₂H₄ is indeed 8, and this is because it is oxidized to N₂ with a loss of four electrons per N₂H₄ unit.
184 Medium
184 of 329
Understanding +4 | -1
In which of the following changes entropy
decreases ?
View answer and explanation
Correct option: A
Crystallization of sucrose from solution
Option A (Crystallization of sucrose from solution) is correct because the process of crystallization involves a transition from a more disordered state (solution) to a more ordered state (crystal), resulting in a decrease in entropy.
185 Medium
185 of 329
Application +4 | -1
If heat of dissociation of CHCl2COOH{CHCl}_2{COOH} is 0.7 Kcal/mol0.7 \mathrm{~Kcal} / \mathrm{mol} then ΔH\Delta {H} for the reaction:

CHCl2COOH+KOH→CHCl2COOK+H2O{CHCl}_2 {COOH}+{KOH} \rightarrow {CHCl}_2 {COOK}+{H}_2 {O}
View answer and explanation
Correct option: A
−13 Kcal-13 \mathrm{~Kcal}
Option A (-13 Kcal) is correct because it is calculated by subtracting the heat of dissociation from the standard enthalpy change of neutralization.
186 Medium
186 of 329
Application +4 | -1
The enthalpy of vaporization of substance is 840 J−mol−1840 \mathrm{~J-} \mathrm{mol}^{-1} and its boiling point is −173∘C-173^{\circ} \mathrm{C}. Its entropy of vaporization is:
View answer and explanation
Correct option: D
8.4 J mol−1 K−18.4 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}
Option D (8.4 J mol−1 K−18.4 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}) is correct because it uses the formula ΔS=ΔHT\Delta S = \frac{{\Delta H}}{{T}} with correct temperature conversion to Kelvin.
187 Medium
187 of 329
Application +4 | -1
What is the entropy change (inJK−1mol−1)(in J{K^{ - 1}}mo{l^{ - 1}}) when one mole of ice is converted into water at 0℃? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol(-1) at 0℃)
View answer and explanation
Correct option: D
21.98
Option D (21.98) is correct because it's the calculated entropy change using the formula ΔS = ΔH/T.
188 Medium
188 of 329
Understanding +4 | -1
Assertion : The mass and volume of a substance are the extensive properties and are proportional to each other.

Reason : The ratio of mass of a sample to its volume is an intensive property.
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and reason are correct, but the reason is not a correct explanation for the assertion. Mass and volume are extensive properties, and their ratio (density) is an intensive property, but density doesn't *cause* mass and volume to be extensive.
189 Medium
189 of 329
Remembering +4 | -1
Cv\mathrm{C}_{\mathrm{v}} for monatomic gas is
View answer and explanation
Correct option: B
32R\frac{3}{2} \mathrm{R}
Option B (32R\frac{3}{2} \mathrm{R}) is correct because for a monatomic gas, the only degrees of freedom are translational, leading to a Cv value of 3/2 * R.
190 Medium
190 of 329
Understanding +4 | -1
Read the following statements carefully and choose the correct option

(i) The state of the system is specified by state functions or state variables.

(ii) Variables like P,V\mathrm{P}, \mathrm{V} and T\mathrm{T} are called state variables or state functions

(iii) Their values depend only on the state of the system and not on how it is reached.
View answer and explanation
Correct option: C
(i), (ii) and (iii) are correct
Option C ((i), (ii) and (iii) are correct) is correct because all three statements accurately describe state functions and state variables.
191 Medium
191 of 329
Analyzing +4 | -1
The heat of neutralization of NaOH{NaOH} with HCl{HCl} is 57.3 kJ57.3 {~kJ} and with HCN{HCN} is 12.1 kJ12.1 {~kJ}. The heat of ionization of HCN{HCN} is
View answer and explanation
Correct option: B
+45.2 kJ+45.2 {~kJ}
Option B (+45.2 {~kJ}) is correct because the heat of ionization of HCN is the difference between the heat of neutralization of NaOH with HCl and the heat of neutralization of NaOH with HCN.
192 Medium
192 of 329
Remembering +4 | -1
Which selection correctly shows units for specific heat capacity?
View answer and explanation
Correct option: C
J/(g⋅∘C)J /\left(g \cdot^{\circ} C\right)
Option C (J/(g⋅∘C)J /\left(g \cdot^{\circ} C\right)) is correct because specific heat capacity is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius, hence the units are Joules per gram per degree Celsius.
193 Medium
193 of 329
Application +4 | -1
Given, H2O(l)→H2O(g);ΔH1=+43.7 kJ\mathrm{H}_2 \mathrm{O}(l) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \Delta \mathrm{H}_1=+43.7 \mathrm{~kJ} H2O(s)→H2O(l);ΔH2=+6.05 kJ\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(l) ; \Delta \mathrm{H}_2=+6.05 \mathrm{~kJ} Calculate the enthalpy of sublimation of ice.
View answer and explanation
Correct option: A
49.75 kJ mol−149.75 \mathrm{~kJ} \mathrm{~mol}^{-1}
Answer A (49.75 kJ mol-1) is correct because the enthalpy of sublimation is the sum of the enthalpy of fusion and the enthalpy of vaporization.
194 Medium
194 of 329
Analyzing +4 | -1
Assertion : Heat absorbed in a reaction at constant temperature and constant volume is −ΔG-\Delta G.

Reason : ΔG\Delta G should be negative for the reaction to be spontaneous.
View answer and explanation
Correct option: D
If assertion is false but reason is true.
Option D is correct. The assertion is false because heat absorbed at constant volume is ΔE\Delta E, not ΔG\Delta G. The reason is true; a negative ΔG\Delta G indicates spontaneity.
195 Medium
195 of 329
Application +4 | -1
At 27∘C27^{\circ} \mathrm{C} latent heat of fusion of a compound is 2930 J/mol2930 \mathrm{~J} / \mathrm{mol}. Entropy change is
View answer and explanation
Correct option: A
9.77 J/mol⋅K9.77 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}
Option A (9.77 J/mol⋅K9.77 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}) is correct because the entropy change is calculated by dividing the latent heat of fusion by the temperature in Kelvin.
196 Medium
196 of 329
Application +4 | -1
During isothermal expansion of one mole of an ideal gas from 10 atm10 {~atm} to 1 atm1 {~atm} at 273 K273 {~K}, thework done is (gas constant =2=2 ) :
View answer and explanation
Correct option: C
−1257.43cal-1257.43 {cal}
Option C (-1257.43 cal) is correct because it's the work done calculated using the formula for isothermal expansion of an ideal gas.
197 Medium
197 of 329
Application +4 | -1
A gas expands against a variable pressure given by P=20VP=\frac{20}{V} (where PP in atm and VV in L\mathrm{L} ). During expansion from volume of 1 litre to 10 litre, the gas undergoes a change in internal energy of 400 J400 \mathrm{~J}. How much heat is absorbed by the gas during expansion?
View answer and explanation
Correct option: C
5065.8 J5065.8 \mathrm{~J}
Option C (5065.8 J) is correct because the heat absorbed is calculated using the first law of thermodynamics, where q = ΔU - w, and the work done is found by integrating P dV.
198 Medium
198 of 329
Application +4 | -1
At 298 K298 \mathrm{~K} what will be the change in standard internal energy change for the given reaction OF2( g)+H2O(g)→O2( g)+2HF(g);\mathrm{OF}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{O}_2(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g}) ;
View answer and explanation
Correct option: C
−310 kJ-310 \mathrm{~kJ}
Option C (-310 kJ) is correct because it is calculated using the formula ΔU = ΔH - Δn_gRT, accounting for the change in the number of moles of gas.
199 Medium
199 of 329
Remembering +4 | -1
Work done in reversible adiabatic process is given by:
View answer and explanation
Correct option: B
nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)
Option B (nR(γ−1)(T2−T1)\frac{{nR}}{{\left( {\gamma - 1} \right)}}\left( {{T_2} - {T_1}} \right)) is correct because it represents the work done in a reversible adiabatic process in terms of the number of moles, gas constant, adiabatic index, and temperature change.
200 Medium
200 of 329
Understanding +4 | -1
For an isothermal reversible expansion process, the value of qq can be calculated by the expression
View answer and explanation
Correct option: A
q=2.303  nRTlog⁡V2V1q = 2.303\,\,nRT\log \frac{{{V_2}}}{{{V_1}}}
Option A (q=2.303  nRTlog⁡V2V1q = 2.303\,\,nRT\log \frac{{{V_2}}}{{{V_1}}}) is correct because it represents the heat absorbed during an isothermal reversible expansion.
201 Medium
201 of 329
Understanding +4 | -1
At 27∘C27^{\circ} \mathrm{C} the reaction, C6H6(l)+15/2O2(g)→6CO2(g)+3H2O(l){C}_6 {H}_{6}({l})+15 / 2 {O}_{2}({g}) \rightarrow 6 {CO}_{2}({g})+3 {H}_2 {O}{({l})} proceeds spontaneously because the magnitude of
View answer and explanation
Correct option: B
ΔH>TΔS\Delta {H} > {T} \Delta {S}
Option B (ΔH>TΔS\Delta H > T \Delta S) is correct because for a spontaneous reaction, the change in Gibbs free energy (ΔG\Delta G) must be negative, and ΔG=ΔH−TΔS\Delta G = \Delta H - T \Delta S. Therefore, ΔH\Delta H must be less than TΔST \Delta S for the reaction to be spontaneous.
202 Medium
202 of 329
Understanding +4 | -1
In an isobaric process, when temperature change from T1\mathrm{T}_1 to T2,ΔS\mathrm{T}_2, \Delta \mathrm{S} is equal to:
View answer and explanation
Correct option: A
2.303Cplog⁡(T2/T1)2.303 \mathrm{C}_{\mathrm{p}} \log \left(\mathrm{T}_2 / \mathrm{T}_1\right)
Option A (2.303Cplog⁡(T2/T1)2.303 \mathrm{C}_{\mathrm{p}} \log (\mathrm{T}_2 / \mathrm{T}_1)) is correct because it represents the entropy change in an isobaric process derived from the general entropy change formula.
203 Medium
203 of 329
Understanding +4 | -1
An ideal monoatomic gas initially at 300 K300 \mathrm{~K} expands adiabatically into vacuum to double its volume. The final temperature of gas is
View answer and explanation
Correct option: A
300 K300 \mathrm{~K}
Option A (300 K300 \mathrm{~K}) is correct because adiabatic expansion into a vacuum is a free expansion, which is an isothermal process for an ideal gas.
204 Medium
204 of 329
Application +4 | -1
Choose the reaction in which ΔH\Delta {H} is not equal to ΔU\Delta {U} ?
View answer and explanation
Correct option: B
C2H4( g)+H2( g)→C2H6( g){C}_2 {H}_4({~g})+{H}_2({~g}) \rightarrow {C}_2 {H}_6({~g})
Option B (C2H4( g)+H2( g)→C2H6( g){C}_2 {H}_4({~g})+{H}_2({~g}) \rightarrow {C}_2 {H}_6({~g})) is correct because the change in the number of gaseous moles (Δng\Delta n_g) is not zero, making ΔH\Delta H not equal to ΔU\Delta U.
205 Medium
205 of 329
Remembering +4 | -1
Which of the following represents the first law
of thermodynamics
View answer and explanation
Correct option: A
ΔU=q+W\Delta U=q+W
Option A (ΔU=q+W\Delta U=q+W) is correct because it represents the first law of thermodynamics, stating that the change in internal energy equals heat plus work.
206 Medium
206 of 329
Understanding +4 | -1
For the reaction 2Cl(g)→Cl2(g),2Cl\left( g \right) \to C{l_2}\left( g \right), the signs of ΔH\Delta H and ΔS\Delta S respectively are
View answer and explanation
Correct option: C
−,  −- ,\,\, -
Option C (−,−-, -) is correct because bond formation releases energy (negative Δ H) and decreases randomness (negative Δ S).
207 Medium
207 of 329
Application +4 | -1
1 g H21 \mathrm{~g~H}_2 gas at S.T.P. is expanded so that volume is doubled. Find the work done in Joules. (1 litre atm = 101.3 J101.3 \mathrm{~J})
View answer and explanation
Correct option: A
1134.56 J1134.56 \mathrm{~J}
Option A (1134.56 J) is correct because the work done during the expansion is calculated using the formula W = P(V2 - V1), where P is the pressure and V is the volume.
208 Medium
208 of 329
Understanding +4 | -1
Considering entropy(s) as the thermodynamic parameter, the criterion for the spontaneity of any process is:
View answer and explanation
Correct option: A
ΔSsys +ΔSsurr >0\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0
Option A (ΔSsys +ΔSsurr >0\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0) is correct because the spontaneity of a process is determined by the total entropy change of the system and surroundings, which must be greater than zero for a spontaneous process.
209 Medium
209 of 329
Analyzing +4 | -1
The standard molar enthalpies of formations of H2O(l)\mathrm{H}_2 \mathrm{O}({l}) and H2O2(l)\mathrm{H}_2 \mathrm{O}_2({l}) are −286-286 and −188 kJ/mol,-188 \mathrm{~kJ} / \mathrm{mol}, respectively. Molar enthalpies of vaporization of H2O\mathrm{H}_2 \mathrm{O} and H2O2\mathrm{H}_2 \mathrm{O}_2 are 4444 and 53 kJ,53 \mathrm{~kJ}, respectively. The bond dissociation enthalpy of O2(g)\mathrm{O}_2(\mathrm{g}) is 498 kJ/mol.498 \mathrm{~kJ} / \mathrm{mol}. Calculate the bond dissociation enthalpy of O−O\mathrm{O}-\mathrm{O} bond in H2O2,\mathrm{H}_2 \mathrm{O}_2, assuming that the bond dissociation enthalpy of O−H\mathrm{O}-\mathrm{H} bond is same in both H2O\mathrm{H}_2 \mathrm{O} and H2O2.\mathrm{H}_2 \mathrm{O}_2.
View answer and explanation
Correct option: A
142 kJ/mol142 \mathrm{~kJ} / \mathrm{mol}
Option A (142 kJ/mol) is correct because by using Hess's Law and the given enthalpies of formation and vaporization, along with the bond dissociation enthalpy of O2, we can calculate the O-O bond dissociation enthalpy.
210 Medium
210 of 329
Application +4 | -1
A gas, while expanding absorbs 25 J25 \mathrm{~J} of heat and does 243 J243 \mathrm{~J} of work. Thus, ΔE\Delta E (change in internal energy) for the gas is
View answer and explanation
Correct option: D
−218 J-218 \mathrm{~J}
Option D (−218 J-218 \mathrm{~J}) is correct because the change in internal energy is calculated using the first law of thermodynamics, considering the signs of heat absorbed and work done.
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