Skip to main content
Back to THERMODYNAMICS

Your Progress

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

Thermodynamics practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

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 301-329 of 329 questions
301 Medium
301 of 329
Application +4 | -1
2.1 g of Fe combines with S evolving 3.77 kJ. The heat of formation of FeS in kJ/mole is
View answer and explanation
Correct option: B
−100.5- 100.5
Option B (-100.5) is correct because the heat of formation is calculated based on the amount of Fe that combines with S and the evolved heat.
302 Medium
302 of 329
Application +4 | -1
In which of the given reactions, ΔH>ΔU\Delta {H} > \Delta {U} ?
View answer and explanation
Correct option: C
Reduction of ZnO{ZnO} by coke
Option C (Reduction of ZnO{ZnO} by coke) is correct because it is the only reaction where Δng>0\Delta n_g > 0, leading to ΔH>ΔU\Delta H > \Delta U.
303 Medium
303 of 329
Understanding +4 | -1
Condition for spontaneity in an isothermal
process is
View answer and explanation
Correct option: A
ΔA+W<0\Delta \mathrm{A}+\mathrm{W}<0
Option A (ΔA+W<0\Delta \mathrm{A}+\mathrm{W}<0) is correct because for a spontaneous isothermal process, the change in Helmholtz free energy plus the work done must be less than zero.
304 Medium
304 of 329
Understanding +4 | -1
(A): The heat absorbed during isothermal expansion of an ideal gas against vacuum is zero.

(R) : The volume occupied by the molecules is zero.
View answer and explanation
Correct option: A
AA and RR are true, RR explains AA
Option A (AA and RR are true, RR explains AA) is correct because the heat absorbed is zero due to no external pressure, and this is because ideal gas molecules occupy negligible volume.
305 Medium
305 of 329
Analyzing +4 | -1
Reaction

1. C+O2→CO2\mathrm{C}+\mathrm{O}_2 \rightarrow \mathrm{CO}_2 \quad ; ΔG=−380 KJ\Delta \mathrm{G}=-380 \mathrm{~KJ}

 2. 4Al+3O2→2Al2O3;ΔG=−2250 KJ\text { 2. } 4 \mathrm{Al}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3 \quad ; \Delta \mathrm{G}=-2250 \mathrm{~KJ}

Reduction of Al2O3\mathrm{Al}_2 \mathrm{O}_3 in presence of carbon is possible or not ?
View answer and explanation
Correct option: C
+555 KJ+555 \mathrm{~KJ}
The correct answer is C (+555 KJ) because the Gibbs free energy change for the reduction of Al2O3 by carbon is positive, indicating that the reaction is not spontaneous.
306 Medium
306 of 329
Application +4 | -1
Three moles of an ideal gas are expanded isothermally from a volume of 300 cm3300 \mathrm{~cm}^3 to 2.5 L2.5 \mathrm{~L} at 300 K300 \mathrm{~K} against a pressure of 1.9 atm1.9 \mathrm{~atm}. The work done in joules is
View answer and explanation
Correct option: A
−423.56 J-423.56 \mathrm{~J}
Option A (-423.56 J) is correct because the work done is calculated using the formula W = -PΔV, considering the pressure is constant.
307 Medium
307 of 329
Application +4 | -1
For a specific work, on an average a person requires 5616 kJ5616 \mathrm{~kJ} of energy. How many kilograms of glucose must be consumed if all the required energy has to be derived from glucose only? ΔH\Delta H for combustion of glucose is −2808 kJ mol−1.-2808 \mathrm{~kJ} \mathrm{~mol}^{-1}.
View answer and explanation
Correct option: B
0.36 kg0.36 \mathrm{~kg}
Option D (34.2) is incorrect. The correct answer is C (68.4) because isotonic solutions have the same osmotic pressure, allowing for the calculation of the unknown solute's molar mass.
308 Medium
308 of 329
Analyzing +4 | -1
Consider following statements:
(a) The absolute value of internal energy cannot be
evaluated.
(b) Entropy is a state function and extensive
property
(c) Temperature is an extensive property
(d) The standard enthalpy of formation of an element (ΔHf)\left(\Delta \mathrm{H}_{\mathrm{f}}\right) is always considered to be zero

(e) For an adiabatic process ΔS>0\Delta S>0
wrong statements are :
View answer and explanation
Correct option: C
(c) and (e)
Option C ((c) and (e)) is correct because temperature is an intensive property, and for an adiabatic process, the change in entropy is zero, not greater than zero.
309 Medium
309 of 329
Understanding +4 | -1
The heat evolved during neutralisation is maximum in the reaction of :
View answer and explanation
Correct option: D
NaOH  and  HCl\mathrm{NaOH}\,\,\mathrm{and}\,\,\mathrm{HCl}
Option D (NaOH and HCl\mathrm{NaOH}\,\mathrm{and}\,\mathrm{HCl}) is correct because it involves the neutralization of a strong acid (HCl) and a strong base (NaOH), which releases the maximum amount of heat.
310 Medium
310 of 329
Analyzing +4 | -1
Assertion :- At constant temperature 0∘C0^{\circ} \mathrm{C} and 1 atm1 \mathrm{~atm}, the change H2O(s)→H2O(ℓ),ΔH\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\ell), \Delta \mathrm{H} and ΔE\Delta \mathrm{E} both are zero.

Reason :- During isothermal process H\mathrm{H} and E\mathrm{E} both remains constant.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D is correct because during the phase transition from solid to liquid, both enthalpy and internal energy change, and the process is not truly isothermal.
311 Medium
311 of 329
Understanding +4 | -1
The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound
View answer and explanation
Correct option: C
May be positive or negative
Option A (HCl) is correct because it has the lowest boiling point among the given hydrogen halides due to weaker intermolecular forces.
312 Medium
312 of 329
Analyzing +4 | -1
Predict which of the following reactions has a negative entropy change?

I. CH4( g)+2O2( g)→CO2( g)+2H2O(ℓ){CH}_4({~g})+2 {O}_2({~g}) \rightarrow {CO}_2({~g})+2 {H}_2 {O}(\ell)

II. Ag(aq)++Cl(aq)−→AgCl(s){Ag}_{({aq})}^{+}+{Cl}_{({aq})}^{-} \rightarrow {AgCl}({s})

III. H2( g)→2H(g){H}_2({~g}) \rightarrow 2 {H}({g})
View answer and explanation
Correct option: C
I and II
Option C is correct because reactions I and II involve a decrease in the number of gaseous molecules or a change from aqueous ions to a solid, both of which lead to a decrease in entropy (negative entropy change).
313 Medium
313 of 329
Application +4 | -1
Assign the sign of work done (based on SI convention) in
the following chemical changes taking place against
external atmospheric pressure
I. N2(g)+3H2(g)⟶2NH3(g)\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \longrightarrow 2 \mathrm{NH}_3(g)

II. NH4HS(s)⟶NH3(g)+H2 S(g)\mathrm{NH}_4 \mathrm{HS}(s) \longrightarrow \mathrm{NH}_3(g)+\mathrm{H}_2 \mathrm{~S}(g)

III. C(s)+CO2(g)⟶2CO(g)\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(g) \longrightarrow 2 \mathrm{CO}(g)

IV. C2H6(g)+3.5O2(g)⟶2CO2(g)+3H2O(l)\mathrm{C}_2 \mathrm{H}_6(g)+3.5 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}_2(g)+3 \mathrm{H}_2 \mathrm{O}(l)
View answer and explanation
Correct option: A
IIIIIIIV+−−+\begin{array}{cccccccccccccccccccc} {\text{I}}&{{\text{II}}}&{{\text{III}}}&{{\text{IV}}} \\ + & - & - & + \end{array}
Option A (+, -, -, +) is correct because it correctly assigns the sign of work based on the change in the number of moles of gas (Δng\Delta n_g).
314 Medium
314 of 329
Application +4 | -1
At the given condition of CH4−O2CH_4 - O_2 fuel
cell the cell emf is 0.8 V and the
enthalpy of combustion of CH4 (g)C{H_4} \text{ (g)} is
-772 kJ/mol. The maximum efficiency of
the given fuel cell in the given condition
is :
View answer and explanation
Correct option: C
0.8
Option C (80%) is correct because the maximum efficiency is calculated by dividing the Gibbs free energy change by the enthalpy change of combustion.
315 Medium
315 of 329
Remembering +4 | -1
The state of a gas can be described by quoting the relationship between
View answer and explanation
Correct option: D
Pressure, volume, temperature and amount
Option D (Pressure, volume, temperature and amount) is correct because the state of a gas is defined by the relationship between these four variables, as described by the ideal gas law.
316 Medium
316 of 329
Application +4 | -1
One mole of a non-ideal gas undergoes a change of state (2.0 atm,3.0 L,95 K)&rarr;(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
44
Option C (44) is correct because the change in enthalpy is calculated using the formula ∆H = ∆U + (P₂V₂ - P₁V₁), which gives 44 L atm.
317 Medium
317 of 329
Remembering +4 | -1
If water is formed from H+{H^ + }ions andOH−,O{H^ - },the heat of formation of water is :
View answer and explanation
Correct option: A
−13.7kcal- 13.7\mathrm{kcal}
Option A (−13.7kcal- 13.7\mathrm{kcal}) is correct because the heat of formation of water from H+ and OH- is the heat of neutralization, which is -13.7 kcal.
318 Medium
318 of 329
Application +4 | -1
The volume of O2\mathrm{O}_2 required to completely burn 52 g52 \mathrm{~g} at Ethyne (C2H2)\left(\mathrm{C}_2 \mathrm{H}_2\right) of STP is:
View answer and explanation
Correct option: C
112 L112 \mathrm{~L}
Option C (112 L112 \mathrm{~L}) is correct because the balanced equation shows that 1 mole of ethyne requires 2.5 moles of oxygen for complete combustion, and 52g of ethyne is 2 moles, thus requiring 5 moles of oxygen which equals 112L at STP.
319 Medium
319 of 329
Application +4 | -1
The entropy change in the conversion of one mole of liquid water at 373 K373 \mathrm{~K} to vapour at the same temperature is: (Latent heat of vaporization of water, ΔHvap =2.257 kJ/g\Delta H_{\text {vap }}=2.257 \mathrm{~kJ} / \mathrm{g} )
View answer and explanation
Correct option: C
108.9JK−1108.9 \mathrm{JK}^{-1}
Option C (108.9JK−1108.9 \mathrm{JK}^{-1}) is correct because it is calculated by dividing the molar heat of vaporization by the temperature.
320 Medium
320 of 329
Remembering +4 | -1
Which of the following is a highly corrosive salt?
View answer and explanation
Correct option: D
HgCl2{HgCl}_2.
Option D (HgCl2{HgCl}_2) is correct because Mercuric chloride (HgCl2HgCl_2) is known as corrosive sublimate due to its highly corrosive and poisonous nature.
321 Medium
321 of 329
Application +4 | -1
The standard free energy change for a gaseous reaction at 27∘C  is  x k.cal.27^\circ C\,\,\mathrm{ is }\,\,x\,k.cal. If equilibrium constant for a reaction is 100 and R is 2 cal.k−1.mol−1.2\,cal.{k^{ - 1}}.mo{l^{ - 1}}. Then xx is
View answer and explanation
Correct option: A
– 2.7636
Option A (– 2.7636) is correct because it is the standard free energy change calculated using the given equilibrium constant and temperature.
322 Medium
322 of 329
Understanding +4 | -1
To calculate the amount of work done in joules during reversible isothermal expansion of an ideal gas, the volume must be expressed in
View answer and explanation
Correct option: D
Any of these
Option D (Any of these) is correct because the work done in a reversible isothermal expansion involves a ratio of volumes, so the units cancel out, making any volume unit acceptable as long as both volumes are in the same unit.
323 Medium
323 of 329
Understanding +4 | -1
For which reaction from the following, ΔS\Delta {S} will be maximum ?
View answer and explanation
Correct option: B
CaCO3( s)⟶CaO(s)+CO2( g){CaCO}_3(\mathrm{~s}) \longrightarrow {CaO}(\mathrm{s})+{CO}_2(\mathrm{~g})
Option B is correct because the reaction CaCO3( s)⟶CaO( s)+CO2( g){CaCO}_3(\mathrm{~s}) \longrightarrow {CaO}(\mathrm{~s})+{CO}_2(\mathrm{~g}) has the largest increase in the number of gaseous moles (Δng=1\Delta n_g = 1), leading to the maximum increase in entropy (ΔS\Delta S).
324 Medium
324 of 329
Understanding +4 | -1
Which of the following is/are intensive properties?

(I) Temperature

(II) Heat

(III) Density

(IV) Refractive index
View answer and explanation
Correct option: C
(I), (III) and (IV) only
Option C ((I), (III) and (IV) only) is correct because temperature, density, and refractive index are intensive properties, while heat is an extensive property.
325 Medium
325 of 329
Understanding +4 | -1
For the reaction: PCl5(g)→PCl3(g)+Cl2(g)\mathrm{PCl}_{5(g)} \rightarrow \mathrm{PCl}_{3(g)}+\mathrm{Cl}_{2(g)}, which one is true?
View answer and explanation
Correct option: B
ΔH=+ve;ΔS=+ve\Delta H=+\mathrm{ve} ; \Delta S=+\mathrm{ve}
Option B is correct because the decomposition of PCl5(g)\mathrm{PCl}_{5(g)} is endothermic (ΔH>0\Delta H > 0) and increases the number of gas molecules, leading to an increase in entropy (ΔS>0\Delta S > 0).
326 Medium
326 of 329
Analyzing +4 | -1
One mole of CH3COOH\mathrm{CH}_3 \mathrm{COOH} undergoes dimerization in vapour phase at 127∘C127^{\circ} \mathrm{C} as:

2CH3COOH(g)⇌(CH3COOH)2( g)2 \mathrm{CH}_3 \mathrm{COOH}(\mathrm{g}) \rightleftharpoons\left(\mathrm{CH}_3 \mathrm{COOH}\right)_2(\mathrm{~g})

if dimer formation is due to two H\mathrm{H}-bonds involved in dimer, each of 33 kJ33 \mathrm{~kJ} stength and the degree of dimerisation of acetic acid 98.2%98.2 \%. The change in standard entropy during dimerization is
View answer and explanation
Correct option: A
−104JK−1 mol−1-104 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
Option A is correct because the change in standard entropy during dimerization is calculated using the equilibrium constant and enthalpy change, resulting in approximately -104 J/K·mol.
327 Medium
327 of 329
Application +4 | -1
Given, 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∘=−92.4 kJ mol−1\Delta_{\mathrm{r}} \mathrm{H}^{\circ}=-92.4 \mathrm{~kJ} \mathrm{~mol}^{-1}

What is the magnitude of standard enthalpy of formation of NH3\mathrm{NH}_3 gas?
View answer and explanation
Correct option: A
46.2
Option A (46.2) is correct because the standard enthalpy of formation is half of the given reaction enthalpy since 2 moles of NH3 are formed in the reaction.
328 Medium
328 of 329
Understanding +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: B
During complete combustion of 1 mole of butane, 2658 kJ2658 \mathrm{~kJ} of heat is released
Option B (During complete combustion of 1 mole of butane, 2658 kJ2658 \mathrm{~kJ} of heat is released) is correct because the combustion of butane is an exothermic reaction that releases a specific amount of heat.
329 Medium
329 of 329
Understanding +4 | -1
Two mole of an ideal gas is expanded isothermally and reversibly from 1 L to 10 L at 300 K. The enthalpy change (in kJ) for the process is
View answer and explanation
Correct option: C
0
Option C (0) is correct because enthalpy change for an isothermal process involving an ideal gas is zero, as enthalpy depends only on temperature.
Revision notes and key takeaways

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

Showing 301-329 of 329 questions
0 /329
0✓ 0✗
0% Accuracy
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.