Skip to main content
Back to THERMODYNAMICS

Your Progress

0 / 329 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

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.

Take a chapter quiz →
Showing 241-270 of 329 questions

Preparing practice and saved progress…

241 Medium
241 of 329
Understanding +4 | -1
On increasing temperature, the equilibrium constant of exothermic and endothermic reactions repectively
View answer and explanation
Correct option: B
Decreases and increases
Option B (Decreases and increases) is correct because increasing temperature favors the endothermic reaction, shifting the equilibrium and increasing its constant, while it disfavors the exothermic reaction, decreasing its constant.
242 Medium
242 of 329
Application +4 | -1
1 Mole of ideal gas expanded reversibly and isothermally from 1 lit to 10 lit.
then find out the entropy change in cal/Kelvin.
View answer and explanation
Correct option: D
4.606
Option D (4.606) is correct because the entropy change for an isothermal reversible expansion is calculated using the formula ΔS=2.303nRlog⁡(V2/V1)\Delta S = 2.303 nR \log(V_2/V_1).
243 Medium
243 of 329
Application +4 | -1
For a certain reaction the change in enthalpy and change in entropy are 40.63 kJ mol−140.63 \mathrm{~kJ} \mathrm{~mol}^{-1} and 100 JK−1100 \mathrm{~JK}^{-1} respectively. What is the value of ΔG\Delta G at 27∘C?27^{\circ} \mathrm{C}?
View answer and explanation
Correct option: A
10630 J10630 \mathrm{~J}
Option A (10630 J10630 \mathrm{~J}) is correct because it is the calculated value of Δ G using the Gibbs free energy equation.
244 Medium
244 of 329
Analyzing +4 | -1
Which statement(s) is/are correct?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given statements regarding thermodynamic properties are correct and follow standard thermodynamic relationships.
245 Medium
245 of 329
Remembering +4 | -1
Standard molar enthalpy of formation at 298 K298 \mathrm{~K} is zero for
View answer and explanation
Correct option: A
Br2(l)\mathrm{Br}_2(\mathrm{l})
Option A (Br₂(l)) is correct because the standard molar enthalpy of formation is zero for an element in its standard state at a specified temperature (298 K). Bromine exists as a liquid at 298K.
246 Medium
246 of 329
Analyzing +4 | -1
Given,

2Fe2O3(s)⟶4Fe(s)+3O2(g);ΔrG1∘=+1487 kJ mol−16CO(g)+3O2(g)⟶CO2(g);ΔrG2∘=−1543.2 kJ mol−1\begin{aligned} & 2 {Fe}_2 {O}_{3(s)} \longrightarrow 4 {Fe}_{(s)}+3 {O}_{2(g)} ; \\ & \Delta_r G_1^{\circ}=+1487 {~kJ} {~mol}^{-1} \\ & 6 {CO}_{(g)}+3 {O}_{2(g)} \longrightarrow {CO}_{2(g)} ; \\ & \Delta_r G_2^{\circ}=-1543.2 {~kJ} {~mol}^{-1} \end{aligned}

Select the correct statement :
View answer and explanation
Correct option: B
Fe2O3{Fe}_2 {O}_3 can be reduced by CO{CO} spontaneously
Option B (Fe2O3{Fe}_2 {O}_3 can be reduced by CO{CO} spontaneously) is correct because the overall Gibbs free energy change is negative, indicating spontaneity.
247 Medium
247 of 329
Application +4 | -1
1 g1 \mathrm{~g} ice absorbs 335 J335 \mathrm{~J} of heat to melt at 0∘C0^{\circ} \mathrm{C}. The entropy change will be
View answer and explanation
Correct option: C
22.1JK−1 mol−122.1 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
Option C (22.1JK−1 mol−122.1 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}) is correct because it is the entropy change calculated using the formula ΔS=ΔHT\Delta S = \frac{\Delta H}{T} for the melting of ice.
248 Medium
248 of 329
Analyzing +4 | -1
Statement I: Heat evolved in the neutralization of either 1 mole of HCl{HCl} or 1 mole of H2SO4{H}_2 {SO}_4 with NaOH{NaOH} is same.

Statement II: Both HCl{HCl} and H2SO4{H}_2 {SO}_4 are strong acids.
View answer and explanation
Correct option: D
If Statement I is INCORRECT, but Statement II is CORRECT
Option D is correct because the heat evolved in the neutralization of 1 mole of H2SO4{H}_2 {SO}_4 with NaOH{NaOH} is different from that of 1 mole of HCl{HCl} because H2SO4{H}_2 {SO}_4 is a dibasic acid, while HCl{HCl} is a monobasic acid.
249 Medium
249 of 329
Understanding +4 | -1
Among the following the set of parameters that represents path functions, is
(A) q+W          q + W\,\,\,\,\,\,\,\,\,\, (B) q          q\,\,\,\,\,\,\,\,\,\, (C) W          W\,\,\,\,\,\,\,\,\,\, (D) H−TSH - TS
View answer and explanation
Correct option: D
(B) and (C)
Option D ((B) and (C)) is correct because heat (qq) and work (WW) are path functions, meaning their values depend on the path taken during a process.
250 Medium
250 of 329
Remembering +4 | -1
Identify the correct statement regarding entropy
View answer and explanation
Correct option: C
At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero
Option C (At absolute zero of temperature, entropy of a perfectly crystalline substance is taken to be zero) is correct because the Third Law of Thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero (0 K) is zero.
251 Medium
251 of 329
Application +4 | -1
The heat of combustion of CH4(g),C(graphite){CH}_{4({g})}, {C}_{(\mathrm{graphite})} and H2(g){H}_{2({g})} are 20 kcal,−40 kcal20 \, \mathrm{kcal},-40 \, \mathrm{kcal} and −10 kcal-10 \, \mathrm{kcal} respectively. The heat of formation of methane is
View answer and explanation
Correct option: C
−80 kcal-80 \, \mathrm{kcal}
Option C (-80 kcal) is correct because it correctly applies Hess's Law to calculate the heat of formation.
252 Medium
252 of 329
Application +4 | -1
The entropy change for vaporisation of liquid water to steam at 100℃ is…JK−1mol−1\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}Given that heat of vaporisation is 40.8kJmol−1.40.8kJmo{l^{ - 1}}.
View answer and explanation
Correct option: A
109.38
Option A (109.38) is correct because the entropy change is calculated by dividing the heat of vaporization by the temperature in Kelvin.
253 Medium
253 of 329
Remembering +4 | -1
Specific heat is defined as:
View answer and explanation
Correct option: A
Heat capacity/g
Option A (Heat capacity/g) is correct because specific heat is defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
254 Medium
254 of 329
Remembering +4 | -1
The enthalpy of a monoatomic gas at TT Kelvin is:
View answer and explanation
Correct option: D
52RT\frac{5}{2} \mathrm{RT}
Option D (52RT\frac{5}{2} \mathrm{RT}) is correct because the enthalpy of a monoatomic gas is given by H = E + PV = 32RT+RT=52RT\frac{3}{2}RT + RT = \frac{5}{2}RT.
255 Medium
255 of 329
Application +4 | -1
At 25∘C25^{\circ} \mathrm{C} 1 mole of butane upon heating forms CO2\mathrm{CO}_2 and H2O\mathrm{H}_2 \mathrm{O}. The work done is
View answer and explanation
Correct option: B
85.6 L85.6 \mathrm{~L} atm
Option B (85.6 L atm85.6 \mathrm{~L} \mathrm{~atm}) is correct because it involves calculating the work done using the change in the number of moles of gas and the ideal gas constant.
256 Medium
256 of 329
Application +4 | -1
The enthalpy of neutralisation of NH4OHNH_4 OH and CH3COOHCH_3 COOH is −10.5k cal mol−1-10.5 \mathrm{k\,cal\,} \mathrm{mol}^{-1} and enthalpy of neutralisation of CH3COOHCH_3COOH with strong base is −12.5k cal mol−1-12.5 \mathrm{k\,cal\,} \mathrm{mol}^{-1}. The enthalpy of ionisation of NH4OHNH_4OH will be
View answer and explanation
Correct option: C
2.0k cal mol−12.0 \mathrm{k\,cal}\, \mathrm{mol}^{-1}
Option C (2.0k cal}mol−12.0 \mathrm{k\,cal\} \mathrm{mol}^{-1}}) is correct because it is calculated by considering the enthalpy of neutralization of a weak acid with a weak base and a weak acid with a strong base.
257 Medium
257 of 329
Application +4 | -1
Given that N2( g)+3H2( g)→2NH3( g)\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{~g});
ΔrH∘=−92 kJ\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-92 \mathrm{~kJ}, the standard molar enthalpy of formation in kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} of NH3( g)\mathrm{NH}_3(\mathrm{~g}) is
View answer and explanation
Correct option: D
-46
Option D (-46) is correct because the standard molar enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed.
258 Medium
258 of 329
Analyzing +4 | -1
Butane exists in various conformations in nature. At any given instant, the probability that a given butane molecule is in anti, gauche, eclipsed and fully eclipsed conformation is 0.7,0.2,0.060.7,0.2,0.06 and 0.04,0.04 , respectively. If the molar enthalpy of combustion of natural butane is −690 kcal/mol-690 \mathrm{~kcal} / \mathrm{mol} at 25∘C,25^{\circ} \mathrm{C}, then calculate the enthalpy of combustion of butane if all the butane molecules are in gauche conformation. The potential energy of gauche form is 2 kcal/mol2 \mathrm{~kcal} / \mathrm{mol} higher than anti form but it is 3 kcal/mol3 \mathrm{~kcal} / \mathrm{mol} less than eclipsed form and 5.5 kcal/mol5.5 \mathrm{~kcal} / \mathrm{mol} less than fully eclipsed form.
View answer and explanation
Correct option: C
−691 kcal/mol-691 \mathrm{~kcal} / \mathrm{mol}
Option C (-691 kcal/mol) is correct because it calculates the enthalpy of combustion for the gauche conformation based on the given probabilities and energy differences.
259 Medium
259 of 329
Understanding +4 | -1
According to the first law of thermodynamics which of the following quantities represents change in a state function?
View answer and explanation
Correct option: D
qrev+Wrev{q}_{{rev}}+{W}_{{rev}}
Option D (qrev+Wrev{q}_{{rev}}+{W}_{{rev}}) is correct because, according to the first law of thermodynamics, the change in internal energy, a state function, is equal to the sum of heat and work.
260 Medium
260 of 329
Application +4 | -1
1 mol of an ideal gas expanded reversibly from 10 L to 100 L at 27∘C.27^{\circ} \mathrm{C}. Work done involved in the process is
View answer and explanation
Correct option: B
-5.74 kJ
Option B (-5.74 kJ) is correct because the work done in a reversible isothermal expansion is calculated using the formula W = -2.303 nRT log(V2/V1).
261 Medium
261 of 329
Remembering +4 | -1
A process in which no heat change takes place is called.
View answer and explanation
Correct option: B
An adiabatic process
Option B (An adiabatic process) is correct because an adiabatic process is defined as a thermodynamic process in which there is no heat exchange between the system and its surroundings (ΔQ = 0).
262 Medium
262 of 329
Application +4 | -1
The enthalpy changes for the following processes are listed below :

Cl2(g)→2Cl(g),242.3 kJ mol−1I2(g)→2I(g),151.0 kJ mol−1ICl(g)→I(g)+Cl(g),211.3 kJ mol−1I2(s)→I2(g),62.76 kJ mol−1\begin{array}{ll}\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{Cl}(g), & 242.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(g) \rightarrow 2 \mathrm{I}(g), & 151.0 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{ICl}(g) \rightarrow \mathrm{I}(g)+\mathrm{Cl}(g), & 211.3 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{I}_2(s) \rightarrow \mathrm{I}_2(g), & 62.76 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{array}

Given that the standard states for iodine and chlorine are I2(s)\mathrm{I}_2(s) and Cl2(g)\mathrm{Cl}_2(g), the standard enthalpy of formation for ICl⁡(g)\operatorname{ICl}(g) is :
View answer and explanation
Correct option: A
+16.8 kJ mol−1+16.8 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (+16.8 kJ mol⁻¹) is correct because it is calculated using Hess's Law and the given enthalpy changes.
263 Medium
263 of 329
Application +4 | -1
The heat of formations of CO(g)\mathrm{CO(g}) and CO2(g)\mathrm{CO_2(g}) are −26.4 K cal-26.4 \mathrm{~K} \mathrm{~cal} and −94.0 kcal-94.0 \mathrm{~kcal} respectively. The heat of combustion of carbon monoxide will be
View answer and explanation
Correct option: D
−67.6 kcal-67.6 \mathrm{~kcal}
Option D (-67.6 kcal) is correct because it is the result of subtracting the heat of formation of CO from that of CO2.
264 Medium
264 of 329
Application +4 | -1
Consider a reversible isontropic expansion of 1.0 mole of an ideal monoatomic gas from 25∘C25^{\circ} \mathrm{C} to 75∘C75^{\circ} \mathrm{C}. If the initial pressure was 1.0 bar1.0 \mathrm{~bar}, final pressure is
View answer and explanation
Correct option: A
1.474 bar
Option A (1.474 bar) is correct because it is calculated using the isentropic process formula relating pressure and temperature.
265 Medium
265 of 329
Application +4 | -1
If the bond energies of H2,Br2\mathrm{H}_2, \mathrm{Br}_2 and H−Br\mathrm{H}-\mathrm{Br} are 433,192 and 364 kJ/364 \mathrm{~kJ} / mol respectively the ΔH∘\Delta \mathrm{H}^{\circ} for the reaction H2(g)+Br2(g)⟶2HBr(g)\mathrm{H}_2(\mathrm{g})+\mathrm{Br}_2(\mathrm{g}) \longrightarrow 2 \mathrm{HBr}(\mathrm{g}) is:
View answer and explanation
Correct option: C
−103 kJ-103 \mathrm{~kJ}
Option C (-103 kJ) is correct because it calculates the enthalpy change of the reaction using bond energies of reactants and products.
266 Medium
266 of 329
Understanding +4 | -1
Which of the following on combustion gives
maximum energy?
View answer and explanation
Correct option: D
Butane
Option D (Butane) is correct because, for alkanes, the heat of combustion increases with the number of carbon atoms.
267 Medium
267 of 329
Understanding +4 | -1
A process which is unfavourable with respect to enthalpy,
but favourable with respect to entropy
View answer and explanation
Correct option: B
could occur at high temperatures, but not at low
temperatures
Option B is correct because a process unfavorable in enthalpy but favorable in entropy can occur at high temperatures where the entropy term dominates the Gibbs free energy equation.
268 Medium
268 of 329
Understanding +4 | -1
Heat absorbed at constant volume is measured in a calorimeter in which steel vessel is immersed to ensure that
View answer and explanation
Correct option: C
No heat is lost to the surrounding
Option C is correct because a calorimeter is designed to prevent heat exchange with the surroundings, ensuring accurate measurement of heat absorbed at constant volume.
269 Medium
269 of 329
Analyzing +4 | -1
Assertion (A): The dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl} in water is endothermic but it dissolves in water.
Reason (R): On dissolution of NH4Cl\mathrm{NH}_4 \mathrm{Cl}, entropy increases.
View answer and explanation
Correct option: B
Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion
Option B (Both Assertion and Reason are correct; Reason is not the correct explanation of Assertion) is correct because while NH4Cl dissolution is endothermic and dissolves in water, the driving force is entropy increase, not a direct consequence of endothermicity.
270 Medium
270 of 329
Remembering +4 | -1
The molar heat capacity, Cv\mathrm{C}_{\mathrm{v}} of an ideal gas whose energy is that of translational motion only is
View answer and explanation
Correct option: B
12.47 J deg−1 mol−112.47 \mathrm{~J} \mathrm{~deg}^{-1} \mathrm{~mol}^{-1}
Option B (12.47 J deg−1 mol−112.47 \mathrm{~J} \mathrm{~deg}^{-1} \mathrm{~mol}^{-1}) is correct because for an ideal gas with only translational motion, Cv = (3/2)R, which equals 12.47 J/mol·K.
Revision notes and key takeaways

Review THERMODYNAMICS in your textbook or class notes, then use this set to check your understanding.

Showing 241-270 of 329 questions

Your practice

0 / 329 answered · 329 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.