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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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211 Medium
211 of 329
Application +4 | -1
For a perfectly crystalline solid Cp,m=aT3C_{p, m}=a T^3, where aa is constant. If Cp,mC_{p, m} is 0.42 J/K mol at10 K0.42 \mathrm{~J} / \mathrm{K} \mathrm{~mol} \mathrm{~at} 10 \mathrm{~K}, molar entropy at 20 K20 \mathrm{~K} is :
View answer and explanation
Correct option: C
1.12 J/K mol1.12 \mathrm{~J} / \mathrm{K} \mathrm{~mol}
Option C (1.12 J/K mol1.12 \mathrm{~J} / \mathrm{K} \mathrm{~mol}) is correct because it is calculated by integrating Cp,m/TC_{p,m}/T from 0 to 20 K, using the given relationship Cp,m=aT3C_{p, m}=a T^3.
212 Medium
212 of 329
Analyzing +4 | -1
Consider the following conversions:

(i) O(g)+e−→O−(g),ΔH1\mathrm{O}_{(g)}+e^{-} \rightarrow O^{-}(g), \Delta H_1

(ii) F(g)+e−→F−(g),ΔH2\mathrm{F}_{(g)}+e^{-} \rightarrow F^{-}(g), \Delta H_2

(iii) Cl(g)+e−→Cl−(g),ΔH3\mathrm{Cl}_{(\mathrm{g})}+\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}(\mathrm{g}), \Delta \mathrm{H}_3

(iv) O(g)−+e−→O2−(g),ΔH4\mathrm{O}_{(g)}^{-}+e^{-} \rightarrow O^{2-}(g), \Delta H_4

That according to given information the incorrect statements is:
View answer and explanation
Correct option: D
ΔH1\Delta H_1 and ΔH3\Delta H_3 are negative whereas ΔH2\Delta H_2 and ΔH4\Delta H_4 are positive
Option D is correct because the first electron gain enthalpy for O, F, and Cl are negative, but the second electron gain enthalpy for oxygen is positive due to electron-electron repulsion.
213 Medium
213 of 329
Analyzing +4 | -1
An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If Ti{T_i} is the initial temperature and Tf{T_f} is the final temperature, which of the following statements is correct?
View answer and explanation
Correct option: A
(Tf)irrev>(Tf)rev{\left( {{T_f}} \right)_{\mathrm{irrev}}} > {\left( {{T_f}} \right)_{\mathrm{rev}}}
Option A ((Tf)irrev > (Tf)rev) is correct because in an irreversible adiabatic expansion, less work is done, resulting in a smaller decrease in internal energy and a higher final temperature compared to a reversible process.
214 Medium
214 of 329
Application +4 | -1
A gas is allowed to expand in an insulated container against a constant external pressure of 4 atm from the initial volume of 1.8 L to a final volume of 5.2 L. The change in internal energy of the gas is
View answer and explanation
Correct option: C
–1.38 kJ
Option C (–1.38 kJ) is correct because the change in internal energy is equal to the work done in an adiabatic process, calculated as -Pext * ΔV.
215 Medium
215 of 329
Application +4 | -1
16 gm O216 \mathrm{~gm} \, {O}_2 is expanded reversibly at 300 K300 \mathrm{~K} temperature from 5 dm35 \mathrm{~dm}^3 to 25 dm325 \mathrm{~dm}^3 then find out W{W} and q{q} (in cal)
View answer and explanation
Correct option: A
−483.63 cal,483.63 cal-483.63 \text{ cal}, 483.63 \text{ cal}
Option A (-483.63 cal, 483.63 cal) is correct because the process is isothermal and reversible, allowing for the calculation of work and heat using the appropriate formulas.
216 Medium
216 of 329
Application +4 | -1
Two moles of an ideal gas undergo reversible isothermal expansion from 2.303 L2.303 \mathrm{~L} to 23.03 L23.03 \mathrm{~L} at 27∘C27^{\circ} \mathrm{C}. The change in entropy is: [R=8.314 J K−1\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1}\right. mol−1\mathrm{mol}^{-1}]
View answer and explanation
Correct option: D
38.3 J K−138.3 \mathrm{~J} \mathrm{~K}^{-1}
Option D (38.3 J K−138.3 \mathrm{~J} \mathrm{~K}^{-1}) is correct because it is calculated using the formula for entropy change in a reversible isothermal expansion: ΔS=nRln⁡V2V1\Delta S = nR \ln{\frac{V_2}{V_1}}.
217 Medium
217 of 329
Understanding +4 | -1
The enthalpy change is negative for :
View answer and explanation
Correct option: A
Cl−(g)+aq→Cl−(aq)\mathrm{C}{\mathrm{l}^ - }\left( \mathrm{g} \right) + \mathrm{aq} \to \mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)
Option A is correct because the hydration of ions is an exothermic process, releasing energy and resulting in a negative enthalpy change.
218 Medium
218 of 329
Analyzing +4 | -1
Assertion :- Catalyst changes Gibbs free energy of system.

Reason :- Catalyst changes preexpotential factor of a chemical reaction.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
Option D (If both Assertion & Reason are False) is correct because a catalyst does not change the Gibbs free energy of the system, and it changes the activation energy, not the pre-exponential factor.
219 Medium
219 of 329
Remembering +4 | -1
In the exothermic reaction, the enthalpy of
reaction is always:
View answer and explanation
Correct option: C
negative
Option C (negative) is correct because exothermic reactions release heat, meaning the enthalpy change (ΔH) is negative.
220 Medium
220 of 329
Application +4 | -1
10 moles of an ideal gas compressed from 100  L100\, \mathrm{~L} to 10  L10 \,\mathrm{~L} reversibly at constant temperature of 27∘C27^{\circ} \mathrm{C}. The entropy change during the process is ( R=R= Universal gas constant)
View answer and explanation
Correct option: D
−23.03 R-23.03\, {R}
Option D (−23.03 R-23.03\, {R}) is correct because the entropy change for an isothermal reversible compression is calculated using ΔS = nRln(V2/V1).
221 Medium
221 of 329
Understanding +4 | -1
Identify the correct statement for change of Gibb’s energy for a system (ΔGsystem\Delta {G_{\mathrm{system}}}) at constant temperature and pressure:
View answer and explanation
Correct option: D
If ΔGsystem\Delta {G_{\mathrm{system}}}=0, the system has attained equilibrium
Option D is correct because at equilibrium, the change in Gibb's free energy is zero, indicating no further net change in the system.
222 Medium
222 of 329
Analyzing +4 | -1
When 1 g1 \mathrm{~g} equivalent of strong acid reacts with strong base, the heat released is 13.5 kcal.13.5 \mathrm{~kcal}. When 1 g1 \mathrm{~g} equivalent H2A\mathrm{H}_2 \mathrm{A} is completely neutralized against strong base, 13 kcal13 \mathrm{~kcal} is released. When 1 g1 \mathrm{~g} equivalent B(OH)2\mathrm{B}(\mathrm{OH})_2 is completely neutralized against strong acid, 10 kcal10 \mathrm{~kcal} heat is released. What is the enthalpy change when 1 mole of H2A\mathrm{H}_2 \mathrm{A} is completely neutralized by B(OH)2?\mathrm{B}(\mathrm{OH})_2?
View answer and explanation
Correct option: D
−19 kcal-19 \mathrm{~kcal}
The correct answer is D (-19 kcal) because the enthalpy change is calculated by considering the enthalpy of ionization for both the acid and the base and subtracting them from the total heat released during neutralization.
223 Medium
223 of 329
Analyzing +4 | -1
If H2+1/2O2⟶H2O,ΔH=−68 kcalK+H2O+water⟶KOH(aq)+1/2H2,ΔH=−48 kcalKOH+water⟶KOH(aq),ΔH=−14 kcal\begin{aligned} & \text {If } \mathrm{H}_2+1 / 2 \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}, \Delta \mathrm{H}=-68 \mathrm{~kcal} \\ & \mathrm{K}+\mathrm{H}_2 \mathrm{O}+\text {water} \longrightarrow \mathrm{KOH}(\mathrm{aq})+1 / 2 \mathrm{H}_2, \Delta \mathrm{H}=-48 \mathrm{~kcal} \\ & \mathrm{KOH}+\text {water} \longrightarrow \mathrm{KOH}(\mathrm{aq}), \Delta \mathrm{H}=-14 \mathrm{~kcal} \end{aligned}

Find the heat of formation of KOH\mathrm{KOH}.
View answer and explanation
Correct option: A
-102 k cal
Option A (-102 k cal) is correct because it is calculated by manipulating the given equations to arrive at the formation of KOH.
224 Medium
224 of 329
Application +4 | -1
The bond dissociation energies for Cl2,I2Cl_2, I_2 and
IClICl are 242.3, 151 and 211.3 kJ/mol
respectively. The enthalpy of sublimation of
iodine is 62.8 kJ/mol. What is the standard
enthalpy of formation of ICl(g)?ICl \text{(g)}?
View answer and explanation
Correct option: C
-16.8 kJ/mol
Option C (-16.8 kJ/mol) is correct because it is the standard enthalpy of formation calculated using the given bond dissociation energies and enthalpy of sublimation.
225 Medium
225 of 329
Understanding +4 | -1
Which of the following statements is incorrect?
View answer and explanation
Correct option: C
The standard state of solid iron at 298 K298 \mathrm{~K} is equal to pure iron at 1 bar
Option D (Standard conditions are denoted by adding the superscript 'o' to the symbol ΔH\Delta H) is incorrect because standard conditions are denoted by adding the superscript 'θ\theta' (nought) to the symbol ΔH\Delta H, not 'o'.
226 Medium
226 of 329
Understanding +4 | -1
Which one among the following pairs does not
represent example for intensive property?
View answer and explanation
Correct option: D
Heat capacity and enthalpy
Option D (Heat capacity and enthalpy) is correct because heat capacity and enthalpy are extensive properties, while the others are intensive.
227 Medium
227 of 329
Remembering +4 | -1
The change of enthalpy of a chemical reaction is the same whether the reaction takes place in one step or in several steps". This law was presented by
View answer and explanation
Correct option: A
Hess
Option A (Hess) is correct because Hess's Law states that the enthalpy change of a reaction is independent of the pathway.
228 Medium
228 of 329
Remembering +4 | -1
Which one among the following is the correct option for right relationship between CpandCv{C_p}\mathrm{ and }{C_v} for one mole of ideal gas?
View answer and explanation
Correct option: D
Cp−Cv=R{C_p} - {C_v} = R
Option D (Cp−Cv=R{C_p} - {C_v} = R) is correct because the relationship between Cp and Cv for one mole of an ideal gas is given by Mayer's formula: Cp - Cv = R.
229 Medium
229 of 329
Application +4 | -1
Given that ΔHf(H)=218 kJ/mol\Delta H_f(H)=218 \mathrm{~kJ} / \mathrm{mol}, express the H−H\mathrm{H}-\mathrm{H} bond energy in k cal/mol\mathrm{k ~cal} / \mathrm{mol}.
View answer and explanation
Correct option: C
104
Option C (104) is correct because the H-H bond energy is equivalent to the enthalpy of formation of 2 moles of H atoms from H2, and 436 kJ/mol is approximately 104 kcal/mol.
230 Medium
230 of 329
Analyzing +4 | -1
For which among the following reactions,
change in entropy is less than zero
View answer and explanation
Correct option: B
Dissociation of hydrogen
Option C (Formation of water) is correct because the formation of water from its gaseous elements results in a decrease in entropy.
231 Medium
231 of 329
Application +4 | -1
In a chemical reaction ∆H=150 kJ and ∆S=100 JK−1\mathrm{J}{\mathrm{K}^{ - 1}} at 300 K. The ∆G for the reaction is:
View answer and explanation
Correct option: D
120 kJ
Option D (120 kJ) is correct because it is calculated using the Gibbs free energy equation: ∆G = ∆H - T∆S.
232 Medium
232 of 329
Application +4 | -1
If one mole of an ideal gas (Cp,m=52R)\left(C_{p, m}=\frac{5}{2} R\right) is expanded isothermally at 300 K300 \mathrm{~K} until its volume is tripled then Δsys S\Delta_{\text {sys }} S is
View answer and explanation
Correct option: D
Rln3R ln 3
Option D (Rln3R ln 3) is correct because the change in entropy for an isothermal expansion is given by ΔS=nRln(V2V1)\Delta S = nR ln(\frac{{V_2}}{{V_1}}).
233 Medium
233 of 329
Application +4 | -1
What is the final temperature (in kelvin) of 0.100.10 mole monoatomic ideal gas that performs 75 cal of work adiabatically if the initial temperature is 227∘C227^{\circ} \mathrm{C}? (use R=2 cal/K−mol\mathrm{R}=2 \mathrm{~cal} / \mathrm{K}-\mathrm{mol})
View answer and explanation
Correct option: A
250
Option A (250) is correct because it is calculated using the adiabatic process equation and the given work done.
234 Medium
234 of 329
Application +4 | -1
When a certain amount of ethylene was burnt 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of O2{{\mathrm{O}_2}} (at NTP) that entered into the reaction is:
View answer and explanation
Correct option: B
296.5 litre
Option B (296.5 litre) is correct because the volume of O2 is calculated using the stoichiometry of the combustion reaction and the given heat evolved.
235 Medium
235 of 329
Application +4 | -1
Using bond energy data, calculate heat of formation of isoprene. Given BE\mathrm{BE} of C−H=98.8 kcal,H−H=104 kcal,C−C=83 kcal,C=C=147 kcal\mathrm{C}-\mathrm{H}=98.8 \mathrm{~kcal}, \mathrm{H}-\mathrm{H}=104 \mathrm{~kcal}, \mathrm{C}-\mathrm{C}=83 \mathrm{~kcal}, \mathrm{C}=\mathrm{C}=147 \mathrm{~kcal} and C(s)⟶C(g)=171 kcal\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{C}(\mathrm{g})=171 \mathrm{~kcal}.
View answer and explanation
Correct option: A
20.6 kcal20.6 \mathrm{~kcal}
Option A (20.6 kcal20.6 \mathrm{~kcal}) is correct because it is the calculated heat of formation of isoprene using the given bond energies.
236 Medium
236 of 329
Application +4 | -1
FeCO3(s){FeCO}_3(\mathrm{s}) decomposes at constant pressure as FeO(s)+CO2(g){FeO}(\mathrm{s})+{CO}_2(\mathrm{g})

FeCO3(s)⟶ΔFeO(s)+CO2(g){FeCO}_3(\mathrm{s}) \stackrel{\Delta}{\longrightarrow} {FeO}(\mathrm{s})+{CO}_2(\mathrm{g})

at 25∘C25^{\circ} \mathrm{C}, the heat absorbed during the reaction is 80 kJ80 \mathrm{~kJ}.

Calculate ΔH\Delta H & ΔU\Delta U for the reaction
View answer and explanation
Correct option: A
80 kJ; 77.522 kJ
Option A (80 kJ; 77.522 kJ) is correct because Δ H is equal to the heat absorbed at constant pressure (80 kJ), and Δ U is calculated using the relation Δ H = Δ U + Δ n_g RT.
237 Medium
237 of 329
Understanding +4 | -1
Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction?
View answer and explanation
Correct option: C
Exothermic and increasing disorder
Option C is correct because an exothermic reaction (ΔH < 0) with increasing disorder (ΔS > 0) will always result in a negative ΔG, making the reaction spontaneous at all temperatures.
238 Medium
238 of 329
Understanding +4 | -1
If ΔH\Delta \mathrm{H} is the enthalpy change and ΔE\Delta \mathrm{E} the change in internal energy accompanying a gaseous reaction then
View answer and explanation
Correct option: D
ΔH\Delta \mathrm{H} is less than ΔE\Delta \mathrm{E} if the number of moles of gaseous products is less than the number of moles of gaseous reactants.
The correct answer is D (ΔH\Delta \mathrm{H} is less than ΔE\Delta \mathrm{E} if the number of moles of gaseous products is less than the number of moles of gaseous reactants) because the relationship between enthalpy and internal energy change depends on the change in the number of gaseous moles.
239 Medium
239 of 329
Understanding +4 | -1
The standard heat of combustion of a hydrocarbon compound is an/a-
View answer and explanation
Correct option: C
Intensive property
Option A (Extensive property) is incorrect. The standard heat of combustion is an intensive property because it is defined per mole of the substance.
240 Medium
240 of 329
Application +4 | -1
ΔH\Delta H for the reaction 2C(s)+3H2(g)→C2H6(g)2 \mathrm{C}(\mathrm{s})+3 \mathrm{H}_2(\mathrm{g})\rightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{g}) is −20.24 kcal/mol.-20.24 \mathrm{~kcal} / \mathrm{mol}. The bond energies (in kcal/mol\mathrm{kcal} / \mathrm{mol}) of C−C,C−H\mathrm{C}-\mathrm{C}, \mathrm{C}-\mathrm{H} and H−H\mathrm{H}-\mathrm{H} are 63,85.663 , 85.6 and 102.6,102.6 , respectively. The enthalpy of sublimation of C(s)\mathrm{C}(\mathrm{s}) is
View answer and explanation
Correct option: A
124.3 kcal/mol124.3 \mathrm{~kcal} / \mathrm{mol}
The correct answer is 124.3 kcal/mol124.3 \mathrm{~kcal} / \mathrm{mol} because it is calculated using the given enthalpy change, bond energies, and applying Hess's law.
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