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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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451 Medium
451 of 500
Application +4 | -1
ΔfH∘\Delta_{\mathrm{f}} H^{\circ} for NF3(g)\mathrm{NF}_3(\mathrm{g}) is −113 kJ/mol.-113 \mathrm{~kJ} / \mathrm{mol}. Bond energy for the N−F\mathrm{N}-\mathrm{F} bond is 273.5 kJ/mol.273.5 \mathrm{~kJ} / \mathrm{mol}. The bond energies of N2\mathrm{N}_2 and F2,\mathrm{F}_2, if their magnitudes are in the ratio 6:1,6: 1, are respectively.
View answer and explanation
Correct option: C
943.33,157.22 kJ/mol943.33,157.22 \mathrm{~kJ} / \mathrm{mol}
Option C (943.33, 157.22 kJ/mol) is correct because it accurately calculates the bond energies of N2 and F2 using the given enthalpy of formation and bond energy of N-F.
452 Medium
452 of 500
Remembering +4 | -1
The term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure is known as
View answer and explanation
Correct option: A
Gibb's function
Option A (Gibb's function) is correct because the term −TΔStotal-T \Delta S_{\text {total}} at constant temperature and pressure defines the change in Gibb's free energy.
453 Easy
453 of 500
Understanding +4 | -1
well stoppered thermos flask contains some ice cubes. This is an example of
View answer and explanation
Correct option: C
Isolated system
Option C (Isolated system) is correct because a well-stoppered thermos flask prevents both matter and energy exchange with the surroundings.
454 Medium
454 of 500
Analyzing +4 | -1
Assertion: The enthalpy of atomization increases with an increase in atomic number in a period.

Reason: The increase in atomic number results in an increase in the number of protons and electrons in the atom.
View answer and explanation
Correct option: A
Both Assertion and Reason are true and Reason is the correct explanation for Assertion
Option A is correct. The enthalpy of atomization generally increases across a period due to increasing nuclear charge and decreasing atomic size, which strengthens the metallic bond. The reason correctly explains this trend.
455 Medium
455 of 500
Understanding +4 | -1
Internal energy and pressure of a gas of unit volume are related as:
View answer and explanation
Correct option: A
P=23UP = \frac{2}{3}U
Option A (P=23UP = \frac{2}{3}U) is correct because the pressure of a gas is related to its internal energy by the equation P = (2/3)U for a monoatomic gas.
456 Medium
456 of 500
Understanding +4 | -1
Which compound will absorb the maximum amount of heat when dissolved in the same amount of water? (Integral heats of solution at 250c{25^0}c in kcal/mol of each solute are given in brackets):
View answer and explanation
Correct option: C
NH4NO3(ΔH=+16.08)\mathrm{N}{\mathrm{H}_4}\mathrm{N}{\mathrm{O}_3}\left( {\Delta H = + 16.08} \right)
Option C (NH4NO3(ΔH=+16.08)\mathrm{N}{{\mathrm{H}}_4}{\mathrm{N}}{{\mathrm{O}}_3}(\Delta H = + 16.08)) is correct because the compound with the largest positive enthalpy of solution will absorb the maximum amount of heat when dissolved.
457 Hard
457 of 500
Application +4 | -1
The reaction between H2{H}_2 and O2{O}_2 to produce water vapour is given by the equation : 2H2( g)+O2( g)⟶2H2O(g)2 {H}_2({~g})+{O}_2({~g}) \longrightarrow 2 {H}_2 {O}({g})
The work done at STP is :
View answer and explanation
Correct option: C
−2.27 kJ-2.27 {~kJ}
Option C (-2.27 kJ) is correct because the work done is calculated using the formula w = -PΔV, considering the change in the number of moles of gas.
458 Medium
458 of 500
Application +4 | -1
If the bond energies of C=C\mathrm{C}=\mathrm{C} and C−C\mathrm{C}-\mathrm{C} at 298 K298 \mathrm{~K} are 590590 and 331 kJ mol−1331 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively then enthalpy of polymerization per mole of ethylene is:
View answer and explanation
Correct option: A
−72 kJ-72 \mathrm{~kJ}
Option A (-72 kJ) is correct because the enthalpy of polymerization is calculated by subtracting the energy of two C-C bonds from the energy of one C=C bond.
459 Medium
459 of 500
Analyzing +4 | -1
Enthalpies of combustion of CH4,C2H6\mathrm{CH}_4, \mathrm{C}_2 \mathrm{H}_6 and C3H8\mathrm{C}_3 \mathrm{H}_8 are −210,−370-210,-370 and −526 kcal/mol,-526 \mathrm{~kcal} / \mathrm{mol}, respectively. Enthalpy of combustion of hexane can be predicted as
View answer and explanation
Correct option: C
−1000 kcal mol−1-1000 \mathrm{~kcal} \mathrm{~mol}^{-1}
Option C (-1000 kcal mol⁻¹) is correct because the enthalpy of combustion increases by approximately 158 kcal/mol for each additional CH₂ group, allowing for prediction of hexane's enthalpy of combustion.
460 Medium
460 of 500
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.
461 Medium
461 of 500
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.
462 Medium
462 of 500
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.
463 Medium
463 of 500
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.
464 Hard
464 of 500
Application +4 | -1
The enthalpy of vapourisation of water = 45.953 kJ/mol. At 1 atm pressure the boiling point of water is 373 K. What is the boiling point of water when pressure equal to 1/2 atm in Kelvin in nearest possible integers? Given: log⁡2=0.3\log 2=0.3
View answer and explanation
Correct option: A
356
Option A (356) is correct because it is calculated using the Clausius-Clapeyron equation with the given enthalpy of vaporization and pressure change.
465 Medium
465 of 500
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.
466 Easy
466 of 500
Application +4 | -1
For the reaction;   C2H5OH(l)+3O2(g)→  2CO2(g)+3H2O(l)\;{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}\left( l \right) + 3{\mathrm{O}_2}\left( \mathrm{g} \right) \to \;2C{O_2}(g) + 3{H_2}O(l) which one is true?
View answer and explanation
Correct option: A
∆H= ∆U-RT
Option D (∆H= ∆U-RT) is correct because ∆H and ∆U are related by the equation ∆H = ∆U + ∆nRT, where ∆n is the change in the number of moles of gas.
467 Medium
467 of 500
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.
468 Medium
468 of 500
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.
469 Medium
469 of 500
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.
470 Hard
470 of 500
Analyzing +4 | -1
One mole of an ideal gas undergoes a reversible process in which the entropy of the gas changes with absolute temperature TT as S=aT+Cv,mS=a T+C_{\mathrm{v}, \mathrm{m}} ln⁡T,\ln T, where aa is a positive constant. If T=T0T=T_0 at V=V0,V=V_0, then the volume dependence of the gas on temperature in this process is
View answer and explanation
Correct option: C
T=T0+Ra⋅ln⁡VV0T=T_{0}+\frac{R}{a} \cdot \ln \frac{V}{V_{0}}
Option C is correct. The volume dependence of the gas on temperature is derived by using the given entropy equation and the ideal gas law, leading to the expression T=T0+Ra⋅ln⁡VV0T=T_{0}+\frac{R}{a} \cdot \ln \frac{V}{V_{0}}.
471 Hard
471 of 500
Application +4 | -1
Heat of combustion of glucose is −2520 kJ/mol-2520 \mathrm{~kJ} / \mathrm{mol} then calculate the moles of required O2\mathrm{O}_2 when amount of heat evolved is 504 kJ504 \mathrm{~kJ} ?
View answer and explanation
Correct option: D
1.2 mol1.2 \mathrm{~mol}
Option D (1.2 mol1.2 \mathrm{~mol}) is correct because the moles of oxygen required are calculated based on the stoichiometry of the combustion reaction and the amount of heat evolved.
472 Hard
472 of 500
Application +4 | -1
An athlete in the weight room lifts a 50 kg50 \mathrm{~kg} mass through a vertical distance of 2.0 m.2.0 \mathrm{~m}. The mass is allowed to fall through the 2.0 m2.0 \mathrm{~m} distance while coupled to an electrical generator. The electrical generator produces an equal amount of electrical work, which is used to produce aluminium by Hall electrolytic process.

Al2O3(Solution)+3C(Graphite)→2Al(l)+3CO(g);ΔG∘=600 kJ\begin{aligned} & {Al}_2 {O}_3 \text {(Solution)}+3 {C}(\text {Graphite}) \rightarrow 2 {Al}({l})+ 3 {CO}(\mathrm{g}) ; \Delta G^{\circ}=600 \mathrm{~kJ} \end{aligned}

How many times must the athlete lift the 50 kg50 \mathrm{~kg} mass to provide sufficient Gibbs energy to produce 27 g Al?27 \mathrm{~g} \, {Al}? (g=10 m/s2)\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)
View answer and explanation
Correct option: A
300
Option A (300) is correct because it is the number of lifts required to provide the necessary Gibbs energy to produce 27g of Al.
473 Easy
473 of 500
Understanding +4 | -1
Hess’s law is based on
View answer and explanation
Correct option: B
Law of conservation of energy
Option B (Law of conservation of energy) is correct because Hess's law is a direct consequence of the law of conservation of energy, stating that the total enthalpy change is independent of the pathway.
474 Medium
474 of 500
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.
475 Medium
475 of 500
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.
476 Easy
476 of 500
Remembering +4 | -1
Gibbs energy G, enthalpy H and entropy S are related by:
View answer and explanation
Correct option: B
G=H-TS
Option B (G=H-TS) is correct because it represents the fundamental relationship between Gibbs free energy, enthalpy, temperature, and entropy.
477 Easy
477 of 500
Understanding +4 | -1
The enthalpy changes of formation of the gaseous oxides of nitrogen (  N2OandNO)(\;{N_2}OandNO) are positive because of:
View answer and explanation
Correct option: A
The high bond energy of the nitrogen molecule
Option A (The high bond energy of the nitrogen molecule) is correct because the high bond energy of N2N_2 requires a significant amount of energy to break the bond, making the formation of nitrogen oxides endothermic.
478 Medium
478 of 500
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.
479 Hard
479 of 500
Application +4 | -1
Given that
(i) CH3CHO+52O2→2CO2+2H2O\mathrm{CH}_3 \mathrm{CHO}+\frac{5}{2} \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}; ΔH=−1168 kJ mol−1\Delta \mathrm{H}=-1168 \mathrm{~kJ} \mathrm{~mol}^{-1}
(ii) CH3COOH+2O2→2CO2+2H2O\mathrm{CH}_3 \mathrm{COOH}+2 \mathrm{O}_2 \rightarrow 2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O} ; ΔH=−876 kJ mol−1\Delta \mathrm{H}=-876 \mathrm{~kJ} \mathrm{~mol}^{-1}
ΔH\Delta \mathrm{H} for the reaction, CH3CHO+12O2→CH3COOH\mathrm{CH}_3 \mathrm{CHO}+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{CH}_3 \mathrm{COOH}, is
View answer and explanation
Correct option: A
−292 kJ mol−1-292 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (−292 kJ mol−1-292 \mathrm{~kJ} \mathrm{~mol}^{-1}) is correct because it is the result of applying Hess's Law to the given reactions.
480 Medium
480 of 500
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).
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