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11th Standard English Medium Chemistry Subject Thermodynamics Creative 5 Mark Questions with Solution Part - I

Published on: 25/06/2021

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Std: 11th Standard - Chemistry

chapter: - Creative Questions

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Questions:

Generic 5

5 x 5 = 25

The entropy change in the conversion of water to ice at 272 k for the system is -22.88 Jk^-1mol^-1 and that of surrounding is + 24.85 Jk^-1 mol-¹, State whether the process is spontaneous or not?

Calculate the entropy change during the melting of one mole of ice into water at 0°C and 1 atm pressure. Enthalpy of fusion of ice is 6008 J mol^-1.

Calculate the standard entropy change for the following reaction (Δs_f⁰), given the standard entropies of CO_2(g) , C_(s) 'O_2(g) as 213.6, 5.740, and 205 JK^-1 respectively.

Calculate the entropy change in the engine that receives 957.5 kJ of heat reversibly at 110°C temperature.

If an automobile engine burns petrol at a temperature of 816° C and if the surrounding temperature is 21° C, calculate its maximum possible efficiency

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Answers Key:

Generic 5

5 x 5 = 25

\(\Delta{S}_{univ}=\Delta{S}_{sys}+\Delta{S}_{surr}\)
=-22.88+(+24.85)
=1.97 Jk^-1mol^-1
\(\therefore \underline { \Delta } { S }_{ Univ }>0\) at 272K
∴ The process of freezing of water is spontaneous

ΔH_fusion = 6008 J mol-1
Tf=0⁰C = 273K
\({ H }_{ 2 }O_{ (S) }\quad \overset { 273K }{ \longrightarrow } \quad { H }_{ 2 }O_{ (I) }\)
ΔS_(fusion) = \(\frac { \triangle H_{ fusion } }{ { T }_{ f } } \)
ΔS_(fusion) = \(\frac { 6008 }{ 273 } \)
ΔS_(fusion) = 22.007 JK^-1 mole^-1

C(g) + O2(g) ⟶ CO2(g)
ΔS_r⁰ = \({ \sum { S } }_{ (Products) }^{ 0 }-{ \sum { S } }_{ (reactants) }^{ 0 }\)
ΔS_r⁰ = \(\left\{ { S }_{ CO2 }^{ 0 } \right\} -\left\{ { S }_{ C }^{ 0 }+{ S }_{ O2 }^{ 0 } \right\} \)
ΔS_r⁰ = 213.6 - [5.74+205]
ΔS_r⁰ = 213.6 - [210.74]
ΔS_r⁰ = 2.86 JK^-1

q = 957.7 kJ
T = 110+273 = 383 K
ΔS= \(\frac { { q }_{ rev } }{ T } \)
ΔS = \(\frac { 957.7 }{ 383 } \)
ΔS =2.5 kJ K^-1

% Efficiency =\(\left[ \frac { { T }_{ 1 }-{ T }_{ 2 } }{ { T }_{ 1 } } \right] \times 100\)
Here
T₁ = 816+273=1089 K;
T₂ = 21+273 = 294 K
% Efficiency= \(\left( \frac { 1089-294 }{ 1089 } \right) \times 100\)
% Efficiency = 73%