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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Mixed difficulty

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361 Hard
361 of 500
Application +4 | -1
The carbon dioxide exhaled in the breath of astronaut is often removed from the spacecrafts by reaction with lithium hydroxide.

2LiOH(s)+CO2(g)→Li2CO3(s)+H2O(l)2 \mathrm{LiOH}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightarrow \mathrm{Li}_2 \mathrm{CO}_3(\mathrm{s})+\mathrm{H}_2 \mathrm{O}({l})

Assume that each astronaut requires 2100 kcal2100 \mathrm{~kcal} of energy per day. Further assume that this energy is obtained only from the combustion of glucose into CO2(g)\mathrm{CO}_2(\mathrm{g}) and H2O(l).\mathrm{H}_2 \mathrm{O}(l). The standard enthalpies of formation of glucose, CO2\mathrm{CO}_2 and water are −1280,−395-1280,-395 and −285 kJ/mol,-285 \mathrm{~kJ} / \mathrm{mol}, respectively. The minimum mass of LiOH\mathrm{LiOH} required per astronaut per day to react completely with all the CO2\mathrm{CO}_2 produced is
View answer and explanation
Correct option: A
216 gm216 \mathrm{~gm}
Option A (216 gm216 \mathrm{~gm}) is correct because the calculated minimum mass of LiOH required per astronaut per day is approximately 216 gm.
362 Easy
362 of 500
Understanding +4 | -1
Which of the following statement is incorrect?
View answer and explanation
Correct option: C
The standard state of solid iron at 298 K298 {~K} is pure iron at 1 bar
Option D (Standard conditions are denoted by adding the superscript ⦵ to the symbol ΔH e.g., -ΔH⦵) is correct because the standard state of solid iron at a specified temperature is its pure form at 1 bar, and the example given in the question is correct.
363 Medium
363 of 500
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.
364 Medium
364 of 500
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}}).
365 Hard
365 of 500
Application +4 | -1
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28  JK−1mol−1,{C_V} = 28\mathrm{ }\,\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}}, calculate ΔU\Delta U\mathrm{ } and ΔpV\Delta pV for this process, (R=8.0 JK−1mol−1)(R = 8.0\mathrm{ }\,J{K^{ - 1}}\mathrm{ }mo{l^{ - 1}})
View answer and explanation
Correct option: C
ΔU=14 kJ;Δ(pV)=4 kJ\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ
The correct answer is ΔU=14 kJ;Δ(pV)=4 kJ\Delta U = 14\mathrm{ }\,kJ;\Delta (pV) = 4\mathrm{ }\,kJ. The change in internal energy and the change in pV are calculated using the given values of CVC_V, R, and the temperature change.
366 Medium
366 of 500
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.
367 Medium
367 of 500
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.
368 Medium
368 of 500
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.
369 Medium
369 of 500
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.
370 Medium
370 of 500
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.
371 Hard
371 of 500
Application +4 | -1
A sample of argon gas at 1 atm1 \mathrm{~atm} pressure and 27∘C27^{\circ} \mathrm{C} expands reversibly and adiabatically from 1.25 dm31.25 \mathrm{~dm}^3 to 2.50 dm3.2.50 \mathrm{~dm}^3. Calculate the enthalpy change in this process. Cv,m\mathrm{C}_{\mathrm{v}, \mathrm{m}} for argon is 12.48 JK−1 mol−112.48 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}
View answer and explanation
Correct option: A
−116 J-116 \text{ J}
Option A (-116 J) is correct because the enthalpy change is calculated using the adiabatic expansion formula and the ideal gas law.
372 Medium
372 of 500
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.
373 Medium
373 of 500
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.
374 Medium
374 of 500
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.
375 Medium
375 of 500
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.
376 Hard
376 of 500
Application +4 | -1
Using the data provided, calculate the multiple bond energy (kJ mol−1)(kJ\,mo{l^{ - 1}})of a C&equiv;C bond in C2H2.{\mathrm{C}_2}{\mathrm{H}_2}. Given that the heat of formation of C2H2{\mathrm{C}_2}{\mathrm{H}_2} is 225kJmol−1225kJmo{l^{ - 1}} (take the bond energy of C─H bond as 350  kJ  mol−1.):350\,\,kJ\,\,mo{l^{ - 1}}.):
2C(s)→2C(g)ΔH=1410  kJ  mol−12\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
2C(s)→2C(g)ΔH=1410  kJ  mol−12\mathrm{C}\left( s \right) \to 2\mathrm{C}\left( \mathrm{g} \right)\Delta H = 1410\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
H2(g)→2H(g)ΔH=330  kJ  mol−1{\mathrm{H}_2}\left( \mathrm{g} \right) \to 2\mathrm{H}\left( \mathrm{g} \right)\Delta H = 330\,\,\mathrm{kJ\,\,mo}{\mathrm{l}^{ - 1}}
View answer and explanation
Correct option: D
815
The correct answer is D (815) because it's the calculated C≡C bond energy using the given heat of formation and bond energies.
377 Medium
377 of 500
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).
378 Medium
378 of 500
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.
379 Easy
379 of 500
Remembering +4 | -1
The criterion for a spontaneous process is
View answer and explanation
Correct option: B
ΔG<0\Delta \mathrm{G}<0
Option B (ΔG<0\Delta \mathrm{G}<0) is correct because a spontaneous process is characterized by a decrease in Gibbs free energy.
380 Hard
380 of 500
Application +4 | -1
When one mole of monoatomic ideal gas at T temperature undergoes adiabatic change under a constant external pressure of 1 atm change in volume is from 1 L to 2 L, the final temperature in Kelvin would be
View answer and explanation
Correct option: A
T2(2/3)\frac{T}{{{2^{\left( {2/3} \right)}}}}
Option A (T2(2/3))\frac{T}{{{{2}^{{\left( {2/3} \right)}}}}}) is correct because the final temperature is calculated using the adiabatic process equation and the given volume change.
381 Easy
381 of 500
Understanding +4 | -1
Two moles of an ideal gas expand spontaneously into vacuum. The work is :-
View answer and explanation
Correct option: A
Zero
Option A (Zero) is correct because the work done during the expansion of an ideal gas into a vacuum is zero, as the external pressure is zero.
382 Hard
382 of 500
Application +4 | -1
The value of ΔH∘\Delta H^\circ for the reaction 2A(g)+B(g)⇄A2B(g)2 A_{(g)}+B_{(g)} \rightleftarrows A_2 B_{(g)} for which Kp=1.0×10−10 atm−2K_p=1.0 \times 10^{-10} \mathrm{~atm}^{-2} and ΔS∘=5 JK−1\Delta S^\circ=5 \text{ JK}^{-1} and T=300 KT=300 \text{ K} is
View answer and explanation
Correct option: B
58.94 kJ58.94 \mathrm{~kJ}
Option B (58.94 kJ58.94 \mathrm{~kJ}) is correct because it is calculated using the relationship between Gibbs free energy, enthalpy, entropy, and the equilibrium constant.
383 Hard
383 of 500
Application +4 | -1
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)C{H_3}OH\left( l \right) + \frac{3}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right) + 2{H_2}O\left( l \right) At 298 K, standard Gibb&rsquo;s energies of formation for CH3OH(l), H2O(l)  and  CO2(g)C{H_3}OH\left( l \right),\,{H_2}O\left( l \right)\,\,and\,\,C{O_2}\left( g \right) are -166.2, -237.2 and −394.4kJmol−1- 394.4\mathrm{ }kJ\mathrm{ }mo{l^{ - 1}} respectively. If standard enthalpy of combustion of methanol is −726kJmol−1- 726kJ\mathrm{ }mo{l^{ - 1}} efficiency of the fuel cell will be:
View answer and explanation
Correct option: C
97%
Option C (97%) is correct because the efficiency of the fuel cell is calculated by comparing the change in Gibbs free energy to the standard enthalpy of combustion.
384 Hard
384 of 500
Application +4 | -1
2.56 g2.56 \mathrm{~g} of sulphur is burnt in a constant volume calorimeter with excess of O2(g)\mathrm{O}_2(g). Temperature increases from 21.25∘C21.25^{\circ} \mathrm{C} to 26.72∘C26.72^{\circ} \mathrm{C}. The bomb has a heat capacity of 923 J K−1923 \mathrm{~J} \mathrm{~K}^{-1}. Calorimeter contains 815 g815 \mathrm{~g} of water. Thus, change in internal energy per mole of SO2\mathrm{SO}_2 formed for the reaction is

S8(s)+8O2(g)⟶8SO2(g)\mathrm{S}_8(s)+8 \mathrm{O}_2(g) \longrightarrow 8 \mathrm{SO}_2(g)

(Specific heat of water is 4.184 J K−1 g−14.184 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~g}^{-1} )
View answer and explanation
Correct option: A
−296.27 kJ-296.27 \mathrm{~kJ}
Option A (-296.27 kJ) is correct because the change in internal energy per mole of SO2 formed is calculated using the heat capacity of the calorimeter and the specific heat of water.
385 Medium
385 of 500
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.
386 Easy
386 of 500
Understanding +4 | -1
Decomposition of ozone is exothermic reaction
View answer and explanation
Correct option: C
Spontaneous at all temperatures
The correct answer is C (Spontaneous at all temperatures) because the decomposition of ozone has a negative enthalpy and positive entropy change, making Gibbs free energy negative at all temperatures.
387 Medium
387 of 500
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.
388 Medium
388 of 500
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.
389 Medium
389 of 500
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.
390 Hard
390 of 500
Application +4 | -1
For the reaction, 2A(g)+B(g)⟶2D(g);2 A(g)+B(g) \longrightarrow 2 D(g) ;
ΔU∘=−10.5 kJ and ΔS∘=−44.1 J K−1.\Delta U^{\circ}=-10.5 \mathrm{~kJ} \text { and } \Delta S^{\circ}=-44.1 \mathrm{~J~K}^{-1} .

The ΔG∘\Delta G^{\circ} for the reaction will be, also predict whether the reaction may occur spontaneously.

(R=8.314×10−3 kJ K−1 mol−1 and T=298 K)\text {(} R=8.314 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \text { and } T=298 \mathrm{~K} \text {)}
View answer and explanation
Correct option: A
0.165 kJ mol−10.165 \mathrm{~kJ} \mathrm{~mol}^{-1}, No
Option A (0.165 kJ/mol, No) is correct because the Gibbs free energy is positive, indicating a non-spontaneous reaction.
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