Practice Start Test

Thermodynamics

NEET Chemistry · 93 questions · Page 8 of 10 · Click an option or "Show Solution" to reveal answer

Q71
When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?
A q = w = 500 J, Δ\Delta E = 0
B q = Δ\Delta E = 500 J, w = 0
C q = w = 500 J, Δ\Delta E = 0
D Δ\Delta E = 0, q = w = −- 500 J
Correct Answer
Option B
Solution

As,

Δ\Delta

E = q + W Also, W = – p

Δ\Delta

V As

Δ\Delta

V = 0 So, W = 0 ⇒\Rightarrow

Δ\Delta

E = q Now, q = 500 J Thus,

Δ\Delta

E = q = 500 J

Q72
PbO 2 → \to PbO; Δ\Delta G 298 < 0 SnO 2 → \to SnO; Δ\Delta G 298 > 0 Most probable oxidation state of Pb and Sn will be
A Pb 4+ , Sn 4+
B Pb 4+ , Sn 2+
C Pb 2+ , Sn 2+
D Pb 2+ , Sn 4+
Correct Answer
Option D
Solution

PbO 2 → PbO Pb has +4 oxidation state In PbO 2 . Pb has +2 oxidation state In PbO. Here

Δ\Delta

G is negative that is why reaction is spontaneous and Pb 4+ reduces Pb 2+ thus, Pb 2+ is most probable oxidation state of Pb.

SnO 2 →\to SnO; Sn has +4 oxidation state In SnO 2 .

Sn has +2 oxidation state In SnO.

Δ\Delta

G is positive that is why reaction is not spontaneous thus, +4 oxidation state of Sn is more stable as it does not change to +2 oxidation state spontaneously.

Q73
For the reaction, C 2 H 5 OH (l) + 3O 2(g) → \to 2CO 2(g) + 3H 2 O (l) which one is true
A Δ\Delta H = Δ\Delta E −-RT
B Δ\Delta H = Δ\Delta E + RT
C Δ\Delta H = Δ\Delta E + 2RT
D Δ\Delta H = Δ\Delta E −- 2RT
Correct Answer
Option A
Solution

As we know,

Δ\Delta

H =

Δ\Delta

E +

Δ\Delta

n g RT Now,

Δ\Delta

n g = Number of gaseous moles of products – number of gaseous moles of reactions = 2 – 3 = – 1 ⇒\Rightarrow

Δ\Delta

H =

Δ\Delta

E + (–1) RT

Δ\Delta

H =

Δ\Delta

E – RT

Q74
2Zn + O 2 → \to 2ZnO; Δ\Delta G o = −- 616 J 2Zn + S 2 → \to 2ZnS; Δ\Delta G o = −- 293 J S 2 + 2O 2 → \to 2SO 2 ; Δ\Delta G o = −-408 J Δ\Delta G o for the following reaction 2ZnS + 3O 2 → \to 2ZnO + 2SO 2 is
A −- 731 J
B −- 1317 J
C −- 501 J
D + 731 J
Correct Answer
Option A
Solution

2Zn + O 2 →\to 2ZnO;

Δ\Delta

G o = −- 616 J ....(1) 2ZnS →\to 2Zn + O 2 ;

Δ\Delta

G o = ++ 293 J.....(2) S 2 + 2O 2 →\to 2SO 2 ;

Δ\Delta

G o = −-408 J .....(3)

Δ\Delta

G o for the reaction can be obtained by adding (1), (2) and (3) ∴\therefore

Δ\Delta

G o = 293 - 616 - 408 = -731 J

Q75
The entropy change in the fusion of one mole of a solid melting at 27 o C (latent heat of fusion is 2930 J mol –1 ) is :
A 9.77 J/mol-K
B 10.77 J/mol-K
C 9.07 J/mol-K
D 0.977 J/mol-K
Correct Answer
Option A
Solution
ΔS=ΔHT\Delta S = {{\Delta H} \over T}

=

2930300{{2930} \over {300}}

= 9.77 J/mol-K

Q76
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
A Δ\Delta H < 0 and Δ\Delta S > 0
B Δ\Delta H < 0 and Δ\Delta S < 0
C Δ\Delta H < 0 and Δ\Delta S = 0
D Δ\Delta H > 0 and Δ\Delta S < 0
Correct Answer
Option A
Solution
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

If

Δ\Delta

H < 0 and

Δ\Delta

S > 0

ΔG=(−ve)−T(+ve)\Delta G = (-ve) - T (+ve)

then at all temperatures,

ΔG\Delta G

= -ve, spontaneous reaction. If

Δ\Delta

H < 0 and

Δ\Delta

S = 0

ΔG=(−ve)−T(0)\Delta G = (-ve) - T (0)

then at all temperatures,

ΔG\Delta G

= -ve at all temperatures.

Q77
The difference between Δ\Delta H and Δ\Delta U (Δ\Delta H – Δ\Delta U), when the combustion of one mole of heptane(l) is carried out at a temperature T, is equal to :
A – 4 RT
B 3 RT
C – 3 RT
D 4 RT
Correct Answer
Option A
Solution

We know,

Δ\Delta

H -

Δ\Delta

U =

Δ\Delta

ngRT C7H16(

ll

) + 11O2(g) →\to 7CO2(g) + 8H2O(

ll

) Here

Δ\Delta

ng = 7 - 11 = - 4 ∴\therefore

Δ\Delta

H -

Δ\Delta

U = - 4RT

Q78
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, calculate Δ\Delta U and Δ\Delta pV for this process. (R = 8.0 JK–1 mol–1]
A Δ\Delta U = 14 kJ; Δ\Delta (pV) = 4 kJ
B Δ\Delta U = 2.8 kJ; Δ\Delta (pV) = 0.8 kJ
C Δ\Delta U = 14 kJ; Δ\Delta (pV) = 18 kJ
D Δ\Delta U = 14 kJ; Δ\Delta (pV) = 0.8 kJ
Correct Answer
Option A
Solution

For ideal gas,

Δ\Delta

U = nCv

Δ\Delta

T = 5 ×\times 28 ×\times (200 - 100) = 14 kJ We know, PV = nRT ∴\therefore

Δ\Delta

(PV) = nR

Δ\Delta

T = 5 ×\times 8 ×\times 100 = 4 kJ

Q79
The enthalpy change on freezing of 1 mol of water at 5oC to ice at −5oC is : (Given Δ\Delta fusH = 6 kJ mol−-1 at 0oC, Cp(H2O, ℓ\ell = 75.3J mol−-1 K−-1) Cp(H2O s) =36.8 J mol−-1 K−-1)
A 5.44 kJ mol−-1
B 5.81 kJ mol−-1
C 6.56 kJ mol−-1
D 6.00 kJ mol−-1
Correct Answer
Option C
Solution

(1) From 5

∘^\circ

C to 0

∘^\circ

C :

H1=CP×(ΔT){H_1} = {C_P} \times (\Delta T)
=75.3×(278−273)= 75.3 \times (278 - 273)

= 377 J/mol = 0.377 KJ/mol (2) Phase change (0

∘^\circ

C to 0

∘^\circ

C) :

H2=−ΔHfusion{H_2} = - \Delta {H_{fusion}}

= −- 6 KJ/mol (3) 0

∘^\circ

C to −-5

∘^\circ

C (Ice phase) :

H3=−CP(s)×ΔT{H_3} = - {C_{P(s)}} \times \Delta T
=−36.8×(273−268)= - 36.8 \times (273 - 268)

= −-0.184 KJ/mol ∴\therefore

HTotal=H1+H2+H3{H_{Total}} = {H_1} + {H_2} + {H_3}
=0.377−6−0.184= 0.377 - 6 - 0.184

= −-5.837 KJ/mol

Q80
On the basis of the following thermochemical data : (ΔfGoH(aq)+\Delta _fG^oH^+_{(aq)} = 0) H2O(l) →\to H+(aq) + OH-(aq); ΔH\Delta H = 57.32 kJ H2(g) + 12O2(g)→{1 \over 2} O_2(g) \to H2O(l); ΔH\Delta H = -286.20 kJ The value of enthalpy of formation of OH− ion at 25oC is :
A -22.88 kJ
B -228.88 kJ
C +228.88 kJ
D -343.52 kJ
Correct Answer
Option B
Solution

Given, for reaction

(i)(i)
      H2O(ℓ)→H+(aq.)+OH−(aq.);\,\,\,\,\,\,{H_2}O\left( \ell \right) \to {H^ + }\left( {aq.} \right) + O{H^ - }\left( {aq.} \right);
                          ΔHr=57.32 kJ\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = 57.32\,kJ
(ii)(ii)
      H2(g)+12O2(g)→H2O(ℓ);\,\,\,\,\,\,{H_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right) \to {H_2}O\left( \ell \right);
                          ΔHr=−286.20 kJ\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H_r} = - 286.20\,kJ

For reaction

(i)(i)
ΔHr=ΔH∘f(H+.aq)+\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H^ + }.aq} \right) +
                          ΔH∘f(OH−.aq)−ΔH∘f(H2O,ℓ)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)
57.32=0+ΔH∘f(OH−1,aq)−57.32 = 0 + \Delta {H^ \circ }_f\left( {O{H^{ - 1}},aq} \right) -
                          ΔH∘f(H2O,ℓ)      ...(iii)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)\,\,\,\,\,\,...\left( {iii} \right)

For reaction

(ii)(ii)
ΔHr=ΔH∘f(H2O,ℓ)−ΔH∘f(H2,g)−\Delta {H_r} = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right) - \Delta {H^ \circ }_f\left( {{H_2},g} \right) -
                          12ΔH∘f(O2,g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1 \over 2}\Delta {H^ \circ }_f\left( {{O_2},g} \right)
−286.20=ΔH∘f(H2O,ℓ)- 286.20 = \Delta {H^ \circ }_f\left( {{H_2}O,\ell } \right)

On replacing this value in equ.

(iii)(iii)

we have

57.32=ΔH∘f(OH−.aq)−(−286.20)57.32 = \Delta {H^ \circ }_f\left( {O{H^ - }.aq} \right) - \left( { - 286.20} \right)
ΔH∘f=−286.20+57.32\Delta {H^ \circ }_f = - 286.20 + 57.32
=−228.88 kJ= - 228.88\,kJ
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