Practice Start Test

Thermodynamics

JEE Chemistry · 103 questions · Page 5 of 11 · Click an option or "Show Solution" to reveal answer

Q41
Assuming that water vapour is an ideal gas, the internal energy change (ΔU)\left( {\Delta U} \right) when 11 mol of water is vapourised at 11 bar pressure and 100∘C{100^ \circ }C (Given : molar enthalpy of vapourisation of water at 11 bar and 373373 KK =41 kJ mol−1  = 41\,kJ\,mo{l^{ - 1}}\, and R=8.3 J mol−1 K−1R = 8.3\,J\,mo{l^{ - 1}}\,{K^{ - 1}} )
A 41.00 kJ mol−141.00\,kJ\,mo{l^{ - 1}}
B 4.100 kJ mol−14.100\,kJ\,mo{l^{ - 1}}
C 3.7904 kJ mol−13.7904\,kJ\,mo{l^{ - 1}}
D 37.904 kJ mol−137.904\,kJ\,mo{l^{ - 1}}
Correct Answer
Option D
Solution

Given

ΔH=41 kJ mol−1\Delta H = 41\,kJ\,mo{l^{ - 1}}
=41000 J mol−1= 41000\,J\,mo{l^{ - 1}}
T=100∘C=273+100T = {100^ \circ }C = 273 + 100
=373 K,n=1= 373\,K,n = 1
ΔU=ΔH−ΔnRT\Delta U = \Delta H - \Delta nRT
=41000−(2×8.314×373)= 41000 - \left( {2 \times 8.314 \times 373} \right)
=37898.88 J mol−1= 37898.88\,J\,mo{l^{ - 1}}
=37.9 kJ mol−1= 37.9\,kJ\,mo{l^{ - 1}}
Q42
Identify the correct statement regarding a spontaneous process :
A For a spontaneous process in an isolated system, the change in entropy is positive
B Endothermic processes are never spontaneous
C Exothermic processes are always spontaneous
D Lowering of energy in the reaction process is the only criterion for spontaneity
Correct Answer
Option A
Solution

Spontaneity of reaction depends on tendency to acquire minimum energy state and maximum randomness.

For a spontaneous process in an isolated system the change in entropy is positive.

Q43
Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below: 12Cl2(g){1 \over 2}C{l_2}(g) 12ΔdissHΘ⟶\overset{{{1}2}{\Delta _{diss}}{H^\Theta }} \over \longrightarrow Cl(g)Cl(g) ⟶ΔegHΘ\overset{{{\Delta _{eg}}{H^\Theta }}}\longrightarrow Cl−(g)C{l^ - }(g) ⟶ΔHydHΘ\overset{{{\Delta _{Hyd}}{H^\Theta }}}\longrightarrow Cl−(aq)C{l^ - }(aq) (Using the data, ΔdissHCl2Θ{\Delta _{diss}}H_{C{l_2}}^\Theta = 240 kJ/mol, ΔegHClΘ{\Delta _{eg}}H_{Cl}^\Theta = -349 kJ/mol, ΔhydHCl−Θ{\Delta _{hyd}}H_{C{l^ - }}^\Theta = - 381 kJ/mol) will be :
A +152 kJ mol−1
B −610 kJ mol−1
C −850 kJ mol−1
D +120 kJ mol−1
Correct Answer
Option B
Solution

The energy involved in the conversion of

12C2(g)  {1 \over 2}C{{\rm}_2}\left( g \right)\,\,

to

   Cl−1(aq){\mkern 1mu} {\mkern 1mu} {\mkern 1mu} C{l^{ - 1}}\left( {aq} \right)

is given by

 ΔH=12ΔdissHCl2(−)+ΔegHCl(−)+ΔhyIHCl(−){\mkern 1mu} \Delta H = {1 \over 2}{\Delta _{diss}}H_{C{l_2}}^{\left( - \right)} + {\Delta _{eg}}H_{Cl}^{\left( - \right)} + {\Delta _{hy{\rm{I}}}}H_{Cl}^{\left( - \right)}

Substituting various values from given data, we get

ΔH=(12×240)+(−349)+(−381)kJ mol−1\Delta H = \left( {{1 \over 2} \times 240} \right) + \left( { - 349} \right) + \left( { - 381} \right)kJ\,mo{l^{ - 1}}
=(120−349−381)kJmol−1= \left( {120 - 349 - 381} \right)kJmo{l^{ - 1}}
=−610 kJ mol−1= - 610\,kJ\,mo{l^{ - 1}}

i.e., the correct answer is

(b)(b)
Q44
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK−1 mol−1 , respectively. For the reaction, 12X2{1 \over 2} X_2 + 32Y2→{3 \over 2} Y_2 \to XY3, ΔH\Delta H = -30 kJ, to be at equilibrium, the temperature will be :
A 1250 K
B 500 K
C 750 K
D 1000 K
Correct Answer
Option C
Solution

For a reaction to be at equilibrium

ΔG=0.\Delta G = 0.

Since

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

so at equilibrium

ΔH−TΔS=0\Delta H - T\Delta S = 0

or

    ΔH=TΔS\,\,\,\,\Delta H = T\Delta S

For the reaction

12X2+32Y2→XY3;   {1 \over 2}{X_2} + {3 \over 2}{Y_2} \to X{Y_3};\,\,\,
ΔH=−30kJ  (given)\Delta H = - 30kJ\,\,\left( {given} \right)

Calculating

ΔS\Delta S

for the above reaction, we get

ΔS=50−[12×60+32×40]JK−1\Delta S = 50 - \left[ {{1 \over 2} \times 60 + {3 \over 2} \times 40} \right]J{K^{ - 1}}
=50−(30+60)JK−1= 50 - \left( {30 + 60} \right)J{K^{ - 1}}
=−40 JK−1= - 40\,J{K^{ - 1}}

At equilibrium,

   TΔS=ΔH\,\,\,T\Delta S = \Delta H

[ as

   ΔG=0\,\,\,\Delta G = 0

] ∴\therefore

     \,\,\,\,\,
T×(−40)=−30×1000T \times \left( { - 40} \right) = - 30 \times 1000

[ as

1kJ=1000J1kJ=1000J

] or

     \,\,\,\,\,
T=−30×1000−40T = {{ - 30 \times 1000} \over { - 40}}

or

     \,\,\,\,\,

=

750750
KK
Q45
The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 from its atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N–H bond in NH3 is :
A –964 kJ mol–1
B +352 kJ mol–1
C + 1056 kJ mol–1
D –1102 kJ mol–1
Correct Answer
Option B
Solution
N2+3H2→2NH3 {N_2} + 3{H_2} \to 2N{H_3}\,
 ΔH=2×−46.0  kJ mol−1\,\Delta H = 2 \times - 46.0\,\,kJ\,mo{l^{ - 1}}

Let

xx

be the bond enthalpy of

N−HN-H

bond then [Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive.

]

ΔH=∑ \Delta H = \sum \,

Bond energies of products

                    −∑ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \sum \,

Bond energies of reactants

2×−46=712+3×(436)−6x;2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x;
                     \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
  −92=2020−6x\,\, - 92 = 2020 - 6x
6x=2020+926x = 2020 + 92
⇒6x=2112\Rightarrow 6x = 2112
⇒x=+352 kJ/mol\Rightarrow x = + 352\,kJ/mol
Q46
The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27oC is :
A 35.8 J mol-1 K−1
B 32.3 J mol-1 K−1
C 42.3 J mol-1 K−1
D 38.3 J mol-1 K−1
Correct Answer
Option D
Solution

Entropy change for an isothermal reversible process is given by

ΔS=nR lnV2V1\Delta S = nR{\mkern 1mu} ln{{{V_2}} \over {{V_1}}}
=2×8.314×2.303log 10010= 2 \times 8.314 \times 2.303log\,{{100} \over {10}}
=38.3 J mol−1  K−1= 38.3\,J\,mo{l^{ - 1}}\,\,{K^{ - 1}}
Q47
The incorrect expression among the following is :
A ΔGsystemΔStotal=−T{{\Delta {G_{system}}} \over {\Delta {S_{total}}}} = - T
B In isothermal process wreversible{w_{reversible}} = −nRT ln⁡ VfVi - nRT\,\ln \,{{{V_f}} \over {{V_i}}}
C In K =ΔHo−TΔSoRTK\, = {{\Delta {H^o} - T\Delta {S^o}} \over {RT}}
D K = e−ΔGo/RTK\, = \,{e^{ - \Delta {G^o}/RT}}
Correct Answer
Option C
Solution
ΔG∘=ΔH∘−TΔS∘;\Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ };
         −RT ℓnK=ΔH∘−TΔS∘\,\,\,\,\,\,\,\,\, - RT\,\ell nK = \Delta {H^ \circ } - T\Delta {S^ \circ }
ℓnK=−ΔH∘−TΔS∘RT\ell nK = - {{\Delta {H^ \circ } - T\Delta {S^ \circ }} \over {RT}}
Q48
For complete combustion of ethanol, C2H5OH(l) + 3O2(g) →\to 2CO2(g) + 3H2O(l) the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol–1 at 25oC. Assuming ideality the Enthalpy of combustion, ΔCH\Delta _CH, for the reaction will be : (R = 8.314 kJ mol–1)
A –1460.50 kJ mol–1
B – 1350.50 kJ mol–1
C – 1366.95 kJ mol–1
D – 1361.95 kJ mol–1
Correct Answer
Option C
Solution
C2H5OH(ℓ)+3O2(g){C_2}{H_5}OH\left( \ell \right) + 3{O_2}\left( g \right)
→2CO2(g)+3H2O(ℓ)\to 2C{O_2}\left( g \right) + 3{H_2}O\left( \ell \right)

Bomb calorimeter gives

ΔU\Delta U

of the reaction Given,

ΔU=−1364.47 kJ mol−1\Delta U = - 1364.47\,kJ\,mo{l^{ - 1}}
Δng=−1\Delta {n_g} = - 1
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta {n_g}RT
=−1364.47−1×8.314×2981000= - 1364.47 - {{1 \times 8.314 \times 298} \over {1000}}
=−1366.93 kJ mol−1= - 1366.93\,kJ\,mo{l^{ - 1}}
Q49
The following reaction is performed at 298 K 2NO(g) + O2 (g) ⇋\leftrightharpoons 2NO2 (g) The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (KP = 1.6 × 1012)
A 86600 + R(298) ln(1.6 ×\times 1012)
B 86600 - ln(1.6×1012)R(298)ln (1.6 \times 10^{12}) \over R (298)
C 0.5[2×86,600 – R(298) ln(1.6×1012)]
D R(298) ln(1.6×1012) – 86600
Correct Answer
Option C
Solution
ΔG∘NO(g)\Delta {G^ \circ }_{NO\left( g \right)}
             =86.6kJ/mol=86600J/mol;\,\,\,\,\,\,\,\,\,\,\,\,\, = 86.6kJ/mol = 86600J/mol;
               \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,
G∘NO2(g)=x J/mol{G^ \circ }_{N{O_2}\left( g \right)} = x\,J/mol
T=298, Kp=1.6×1012T = 298,\,{K_p} = 1.6 \times {10^{12}}
ΔG∘=−RT ln⁡ Kp\Delta {G^ \circ } = - RT\,\ln \,{K_p}

Given equation,

2NO(g)+O2(g) ⇌ 2NO2(g)2NO\left( g \right) + {O_2}\left( g \right)\,\rightleftharpoons\,2N{O_2}\left( g \right)
2ΔG∘NO2−2ΔG∘NO2\Delta {G^ \circ }_{N{O_2}} - 2\Delta {G^ \circ }_{NO}
               =−R(298)ln⁡(1.6×1012)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)
2ΔG∘NO2−2×866002\Delta {G^ \circ }_{N{O_2}} - 2 \times 86600
               =−R(298)ln⁡(1.6×1012)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)
2ΔG∘NO22\Delta {G^ \circ }_{N{O_2}}
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,
=2×86600−R(298)ln⁡(1.6×1012)= 2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)
ΔG∘NO2\Delta {G^ \circ }_{N{O_2}}
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,
=12[2×86600−R(298)ln⁡(1.6×1012)]= {1 \over 2}\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]
              \,\,\,\,\,\,\,\,\,\,\,\,\,\,
=0.5[2×86600−R(298)ln⁡(1.6×1012)]= 0.5\left[ {2 \times 86600 - R\left( {298} \right)\ln \left( {1.6 \times {{10}^{12}}} \right)} \right]
Q50
A reaction at 1 bar is non-spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following :
A Both Δ\Delta H and Δ\Delta S are negative.
B Both Δ\Delta H and Δ\Delta S are positive.
C Δ\Delta H is positive while Δ\Delta S is negative.
D Δ\Delta H is negative while Δ\Delta S is positive.
Correct Answer
Option B
Solution

We know that

Δ\Delta

G =

Δ\Delta

H −- T

Δ\Delta

S At low temperature :

Δ\Delta

G is positive (non-spontaneous process), so

Δ\Delta

H is positive and

Δ\Delta

S is positive (T

Δ\Delta

S At high temperature :

Δ\Delta

G is negative (spontaneous process), so

Δ\Delta

H is positive and

Δ\Delta

S is positive (T

Δ\Delta

S >

Δ\Delta

H as T is high).

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