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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Medium

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61 Medium
61 of 329
Application +4 | -1
The entropy change (in cal K−1 mol−1\mathrm{K}^{-1} \mathrm{~mol}^{-1} ) when one mole of an ideal gas at 27∘C27^{\circ} \mathrm{C} is expanded reversibly from 1ℓ1 \ell to 10 I10 \mathrm{~I} is
View answer and explanation
Correct option: C
4.6
Option C (4.6) is correct because it is the entropy change calculated using the formula for reversible expansion of an ideal gas.
62 Medium
62 of 329
Application +4 | -1
1 mole of an ideal gas at 25℃ is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:
View answer and explanation
Correct option: A
19.15JK−1mol−119.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}
Option A (19.15JK−1mol−119.15\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because it is the entropy change calculated using the formula for reversible expansion of an ideal gas.
63 Medium
63 of 329
Application +4 | -1
Calculate the Gibbs energy change when 1 mole of NaCl{NaCl} is dissolved in water at 298 K298 {~K}. Given,

(a) Lattice energy of NaCl=778 kJ mol−1{NaCl}=778 {~kJ} {~mol}^{-1}.

(b) Hydration energy of NaCl=774.3 kJ mol−1{NaCl}=774.3 {~kJ} {~mol}^{-1}.

(c) Entropy change at 298 K=43 J mol−1298 {~K}=43 {~J} {~mol}^{-1}.
View answer and explanation
Correct option: A
-9.114
The Gibbs energy change is -9.114 kJ. This is calculated using the formula ΔG = ΔH - TΔS, where ΔH is the enthalpy change and ΔS is the entropy change.
64 Medium
64 of 329
Understanding +4 | -1
In which case of mixing of a strong acid and a base each of 1N concentration, temperature increase is highest?
View answer and explanation
Correct option: C
25 mL acid-25mL alkali
Option C (25 mL acid-25mL alkali) is correct because the highest temperature increase occurs when the acid and base completely neutralize each other, which happens when equal milliequivalents react.
65 Medium
65 of 329
Application +4 | -1
When 100 ml100 \mathrm{~ml} of 1.0 M HCl1.0 \mathrm{~M} \mathrm{~HCl} was mixed with 100 ml100 \mathrm{~ml} of 1.0 M NaOH1.0 \mathrm{~M} \mathrm{~NaOH} in an insulated beaker at constant pressure, a temperature increase of 5.8∘C5.8^{\circ} \mathrm{C} was measured for the beaker and its constants. If the enthalpy of neutralization of a strong acid with a strong base is −58.0 kJ mol−1-58.0 \mathrm{~kJ} \mathrm{~mol}^{-1}. the calorimeter constant in (kJk−1)\left(\mathrm{kJ} \mathrm{k}^{-1}\right) is ___
(Given: Heat capacity of solution is 4.2 J g−1k−14.2 \mathrm{~J} \mathrm{~g}^{-1} \mathrm{k}^{-1} and density of solution is 1 g mL−11 \mathrm{~g} \mathrm{~mL}^{-1})
View answer and explanation
Correct option: D
0.160.16
Option D (0.160.16) is correct because it is calculated by using the enthalpy of neutralization and the temperature increase to find the calorimeter constant.
66 Medium
66 of 329
Application +4 | -1
For the reaction, N2+3H2⇌2NH3;ΔH=\mathrm{N}_2+3 \mathrm{H}_2 \rightleftharpoons 2 \mathrm{NH}_3 ; \Delta \mathrm{H}= ?
View answer and explanation
Correct option: B
△E−2RT\triangle \mathrm{E}-2 \mathrm{RT}
Option B (△E−2RT\triangle \mathrm{E}-2 \mathrm{RT}) is correct because the relationship between ΔH and ΔE is given by ΔH = ΔE + Δn_g RT.
67 Medium
67 of 329
Remembering +4 | -1
Which one of the following is used to remove most of plutonium from spent nuclear fuel?
View answer and explanation
Correct option: B
O2F2{O_2}{F_2}
Option B (O2F2{O_2}{F_2}) is correct because dioxygen difluoride is a strong fluorinating agent used to convert plutonium to plutonium hexafluoride.
68 Medium
68 of 329
Application +4 | -1
Enthalpies of formation of CO(g),   CO2(g),   N2O(g)CO\left( g \right),\,\,\,C{O_2}\left( g \right),\,\,\,{N_2}O\left( g \right) and N2O4(g){N_2}{O_4}\left( g \right) are –110, –393, 81 and 9.7 KJ/mol respectively. Find the value of ΔrH{\Delta _r}H for the reaction N2O4(g)+3CO(g)→N2O(g)+3CO2(g){N_2}{O_4}\left( g \right) + 3CO\left( g \right) \to {N_2}O\left( g \right) + 3C{O_2}\left( g \right)
View answer and explanation
Correct option: A
–778 KJ
Option A (–778 KJ) is correct because it is calculated by subtracting the sum of the enthalpies of formation of the reactants from the sum of the enthalpies of formation of the products.
69 Medium
69 of 329
Application +4 | -1
500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are:

Given: R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}
View answer and explanation
Correct option: A
348 K and 300 J
Option A (348 K and 300 J) is correct because the final temperature is approximately 348 K, and the change in internal energy is 300 J, calculated using the given heat and gas properties.
70 Medium
70 of 329
Understanding +4 | -1
Assertion :- The value of specific heat at constant pressure is more than that at constant
volume.

Reason :- The energy required to raise temperature by a unit at constant pressure is
greater because some amount of heat is used in doing work.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because the specific heat at constant pressure is higher than at constant volume due to the additional energy required for the system to do work against the external pressure.
71 Medium
71 of 329
Application +4 | -1
The standard heat of combustion of propane is −2220.1 kJ/mol.-2220.1 \mathrm{~kJ} / \mathrm{mol}. The standard heat of vaporization of liquid water is 44 kJ/mol.44 \mathrm{~kJ} / \mathrm{mol}. What is the ΔH∘\Delta H^{\circ} of the following reaction?

C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)\mathrm{C}_3 \mathrm{H}_8(\mathrm{g})+5 \mathrm{O}_2(\mathrm{g}) \rightarrow 3 \mathrm{CO}_2(\mathrm{g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g})
View answer and explanation
Correct option: B
−2044.1 kJ-2044.1 \mathrm{~kJ}
Option B (−2044.1 kJ-2044.1 \mathrm{~kJ}) is correct because it accounts for the heat of vaporization of water when converting the heat of combustion to the desired reaction.
72 Medium
72 of 329
Application +4 | -1
The entropy change accompanying the transfer of 12,000 J12,000 \mathrm{~J} of heat from a body A{A} at 327∘C327^{\circ} \mathrm{C} to a body BB at 127∘C127^{\circ} \mathrm{C} is
View answer and explanation
Correct option: B
+10.0 J/K+10.0 \mathrm{~J} / \mathrm{K}
Option B (+10.0 J/K) is correct because the entropy change is calculated by summing the entropy changes of the two bodies, considering heat transfer and temperature.
73 Medium
73 of 329
Application +4 | -1
The ratio KpK_p to KcK_c of a reaction is 24.63 L atm mol−124.63 \mathrm{~L ~atm ~mol}^{-1} at 27∘C27^\circ \mathrm{C}. If heat of reaction at constant pressure is 98.8 kcal, what is the heat of reaction (in kcal) at constant volume?
View answer and explanation
Correct option: A
98.2
Option A (98.2) is correct because it is calculated using the relationship between heat of reaction at constant pressure and constant volume.
74 Medium
74 of 329
Understanding +4 | -1
The least random state of H2O{H_2}O is
View answer and explanation
Correct option: A
Ice
Option A is correct. Ice has the most ordered structure due to the hydrogen bonding network, making it the least random state of water compared to liquid water and steam.
75 Medium
75 of 329
Analyzing +4 | -1
NH4Cl(s)+H2O(aq)⇌NH4Cl(aq)\mathrm{NH}_4 \mathrm{Cl}(s)+\mathrm{H}_2 \mathrm{O}(a q) \rightleftharpoons \mathrm{NH}_4 \mathrm{Cl}(a q)

Due to dissolution of ammonium chloride in H2O\mathrm{H}_2 \mathrm{O}, temperature falls rapidly. Thus,
View answer and explanation
Correct option: C
ΔH>0,ΔS>0,ΔG<0\Delta H>0, \Delta S>0, \Delta G < 0
Option C (ΔH>0,ΔS>0,ΔG<0\Delta H>0, \Delta S>0, \Delta G < 0) is correct because the dissolution of ammonium chloride is endothermic, increases disorder, and is spontaneous.
76 Medium
76 of 329
Analyzing +4 | -1
The correct descending order of the heat liberated (in kJ\mathrm{kJ} ) during the neutralisation of the acids CH3COOH(W),HF(X),HCOOH(Y)\mathrm{CH}_3 \mathrm{COOH}(W), \mathrm{HF}(X), \mathrm{HCOOH}(Y) and HCN(Z)H C N(Z) under indentical conditions (Ka\left(K_a\right. of CH3COOH=1.8×10−5,HCOOH=1.8×10−4\mathrm{CH}_3 \mathrm{COOH}=1.8 \times 10^{-5}, {H C O O H}=1.8 \times 10^{-4}, HCN=4.9×10−10\mathrm{HCN}={4 . 9} \times 10^{-10} and HF=3.2×10−4\mathrm{HF}=3.2 \times 10^{-4} ) is
View answer and explanation
Correct option: B
X >> Y >> W >Z>Z
Option B is correct because the heat liberated during neutralization is directly proportional to the acid strength, and the given acids follow the order X > Y > W > Z based on their Ka values.
77 Medium
77 of 329
Analyzing +4 | -1
The standard enthalpy of formation (ΔfH∘298)\left(\Delta_f {H}^{\circ}{ }_{298}\right) for methane, CH4{CH}_4 is −74.9 kJ mol−1-74.9 {~kJ} {~mol}^{-1}. In order to calculate the average energy given out in the formation of a C−H{C}-{H} bond from this it is necessary to know which one of the following?
View answer and explanation
Correct option: A
The dissociation energy of the hydrogen molecule, H2{H}_2.
Option C (The dissociation energy of H2{H}_2 and enthalpy and sublimation of carbon (graphite)) is correct because these values are needed along with the enthalpy of formation of methane to calculate the average C-H bond energy.
78 Medium
78 of 329
Understanding +4 | -1
Spontaneous adsorption of a gas on solid surface is an
exothermic process because
View answer and explanation
Correct option: C
△S\triangle \mathrm{S} decreases for gas
Option C is correct because gas adsorption onto a solid surface decreases the gas's entropy, and since the process is spontaneous, the enthalpy change must be negative (exothermic).
79 Medium
79 of 329
Analyzing +4 | -1
Assertion : For a reaction

2NH3( g)⟶N2( g)+3H2( g);ΔH>ΔE2 {NH}_3({~g}) \longrightarrow {N}_2({~g})+3 {H}_2({~g}) ; \Delta H>\Delta E

Reason : Enthalpy change is always greater than internal energy change.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion) is correct because the given reaction has ΔH > ΔE due to the increase in the number of moles, and while enthalpy change can be greater than internal energy change, it's not always the case.
80 Medium
80 of 329
Application +4 | -1
In a process, 701 J701 \mathrm{~J} of heat is absorbed by a system and 394 J394 \mathrm{~J} of work is done by the system. What is the change in internal energy for the process?
View answer and explanation
Correct option: A
Increases by 307 J307 \mathrm{~J}
Option A (Increases by 307 J307 \mathrm{~J}) is correct because the change in internal energy is the sum of heat absorbed by the system and work done on the system.
81 Medium
81 of 329
Analyzing +4 | -1
Based on following thermochemical reaction:

H2O(g)+C(s)→CO(g)+H2(g);ΔH=131 KJCO(g)+12O2(g)→CO2(g);ΔH=−282 KJH2(g)+12O2(g)→H2O(g);ΔH=−242 KJC(s)+O2(g)→CO2(g);ΔH=x KJ\begin{aligned} & {H}_2 {O}(\mathrm{g})+{C}(\mathrm{s}) \rightarrow {CO}(\mathrm{g})+{H}_2(\mathrm{g}) ; \Delta {H}=131 \mathrm{~KJ} \\ & {CO}(\mathrm{g})+\frac{1}{2} {O}_2(\mathrm{g}) \rightarrow {CO}_2(\mathrm{g}) ; \Delta {H}=-282 \mathrm{~KJ} \\ & {H}_2(\mathrm{g})+\frac{1}{2} {O}_2(\mathrm{g}) \rightarrow {H}_2 {O}(\mathrm{g}) ; \Delta {H}=-242 \mathrm{~KJ} \\ & {C}(\mathrm{s})+{O}_2(\mathrm{g}) \rightarrow {CO}_2(\mathrm{g}) ; \Delta {H}=x \mathrm{~KJ} \end{aligned}

The value of xx will be
View answer and explanation
Correct option: A
−393 KJ-393 \mathrm{~KJ}
Option A (-393 KJ) is correct because by adding the given thermochemical reactions, we can derive the target reaction and calculate the corresponding enthalpy change.
82 Medium
82 of 329
Application +4 | -1
C+12O2→CO;ΔH=−42 kJ{C}+\frac{1}{2} {O}_2 \rightarrow {CO} ; \Delta {H}=-42 \mathrm{~kJ}

CO+12O2→CO2;ΔH=−24 kJ{CO}+\frac{1}{2} {O}_2 \rightarrow {CO}_2 ; \Delta {H}=-24 \mathrm{~kJ}

The heat of formation of CO2{CO}_2 is
View answer and explanation
Correct option: D
−66 kJ-66 \mathrm{~kJ}
Option D (−66 kJ-66 \mathrm{~kJ}) is correct because the heat of formation of CO2 is calculated by adding the enthalpy changes of the two given reactions.
83 Medium
83 of 329
Application +4 | -1
The heat change taking place during the reaction   H2O(l)\;{\mathrm{H}_2}\mathrm{O}\left( l \right)→H2O(g)\to {\mathrm{H}_2}\mathrm{O(g)} is: [Given,ΔHf\Delta {H_f} of H2O(g){\mathrm{H}_2}\mathrm{O}(g) = -57 kcal, ΔHf=H2O(l)\Delta {H_f} = {\mathrm{H}_2}\mathrm{O}\left( l \right) = -68.3 kcal]
View answer and explanation
Correct option: A
+11.3kcal+ 11.3\mathrm{kcal}
Option A (+11.3 kcal) is correct because the heat change is calculated by subtracting the heat of formation of liquid water from the heat of formation of gaseous water.
84 Medium
84 of 329
Application +4 | -1
0.75 mole of an ideal gas expands at 27∘C27^{\circ} \mathrm{C} from 15 L15 \mathrm{~L} to 25 L25 \mathrm{~L}, then maximum work done is
View answer and explanation
Correct option: C
−956 J-956 \mathrm{~J}
The correct answer is C (-956 J) because it is the maximum work done in an isothermal reversible expansion, calculated using the formula W = -2.303nRTlog(V2/V1).
85 Medium
85 of 329
Application +4 | -1
Calculate the standard enthalpy of solution of AgCl(s)\mathrm{AgCl}(\mathrm{s}) in water ΔHf∘(AgCl,s)=−127.07 kJ mol−1\Delta \mathrm{H}^\circ_{\mathrm{f}}(\mathrm{AgCl}, \mathrm{s})=-127.07 \mathrm{~kJ} \mathrm{~mol}^{-1}, ΔHf∘(Ag+,aq)=105.58 kJ mol−1,ΔHf∘(Cl−,aq)=−167.35 kJ mol−1\Delta \mathrm{H^\circ_f}\left(\mathrm{Ag}^{+}, \mathrm{aq}\right)=105.58 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{H}^\circ_\mathrm{f}\left(\mathrm{Cl}^{-}, \mathrm{aq}\right)=-167.35 \mathrm{~kJ} \mathrm{~mol}^{-1}.
View answer and explanation
Correct option: A
−65.3 kJ mol−1-65.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option A (-65.3 kJ mol⁻¹) is correct because the standard enthalpy of solution is calculated using the standard enthalpies of formation of the products and reactants.
86 Medium
86 of 329
Application +4 | -1
A sample of 100 gH2O100 {~g} {H}_2 \mathrm{O} is slowly heated from 27∘C27^{\circ} {C} to 87∘C87^{\circ} {C}. Calculate the change in entropy during heating (Specific heat of water =4200 J/kg−K=4200 {~J} / {kg}-{K} ).
View answer and explanation
Correct option: A
76.59
Option A (76.59) is correct because the change in entropy is calculated using the formula ΔS = 2.303 * m * Cp * log(T2/T1).
87 Medium
87 of 329
Understanding +4 | -1
Assertion (A): The difference between ΔH\Delta H and ΔU\Delta U is not significant for systems consisting of only solids or liquids.
Reason (R): Solids and liquids do not suffer any significant volume changes upon heating.
View answer and explanation
Correct option: A
Both Assertion and Reason are correct; Reason is the correct explanation of Assertion
The correct answer is A (Both Assertion and Reason are correct; Reason is the correct explanation of Assertion) because the volume change is negligible for solids and liquids, making the difference between ΔH\Delta H and ΔU\Delta U insignificant.
88 Medium
88 of 329
Analyzing +4 | -1
Assertion: When ideal gas expand from P1,V1,T1P_1, V_1, T_1 to P2,V2,T2P_2, V_2, T_2 in multiple steps, work done is high where number of steps are high.
Reason: Work is path function.
View answer and explanation
Correct option: B
If both Assertion and Reason are correct, but
Reason is not the correct explanation of
Assertion.
Option A (Gain of 2 moles of electrons) is correct because in the reaction, one mole of ozone gains two moles of electrons to form water and oxygen.
89 Medium
89 of 329
Application +4 | -1
For a reaction Y2O4(ℓ)⟶2YO2(g){Y}_2{O}_4(\ell) \longrightarrow 2 {YO}_2(\mathrm{g})

ΔE=2.1 KCal,ΔS=20 calK−1\Delta {E}=2.1 \mathrm{~KCal}, \Delta {S}=20 \mathrm{~calK}^{-1} at 300 K,ΔG300 \mathrm{~K}, \Delta {G} will be:
View answer and explanation
Correct option: B
−2.7 Kcal-2.7 \mathrm{~Kcal}
Option B (-2.7 Kcal) is correct because it is the Gibbs free energy change calculated using the given values of internal energy change, entropy change, and temperature.
90 Medium
90 of 329
Application +4 | -1
The enthalpies of formation of C2H2( g)\mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) and C6H6( g)\mathrm{C}_6 \mathrm{H}_6(\mathrm{~g}) at 298 K298 \mathrm{~K} are 230 and 85 kJ/mol85 \mathrm{~kJ} / \mathrm{mol} respectively. The enthalpy change for the reaction is
View answer and explanation
Correct option: C
−605 kJ/mol-605 \mathrm{~kJ} / \mathrm{mol}
Option C (-605 kJ/mol) is correct because the enthalpy change for the reaction is calculated by subtracting the sum of the enthalpies of formation of the reactants from the enthalpy of formation of the products.
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