KVPY 2020-21 SB SX Exam Previous Papers

Question : 106 of 120

Marks: +1, -0

The reaction

{CaCO}_{3}({s}) \⇌{CaO}({s})+{CO}_{2}({g})

is in equilibrium in a closed vessel at

298 {K}

. The partial pressure (in atm) of

{CO}_{2}({g})

in the reaction vessel is closest to:
[Given : The change in Gibbs energies of formation at

298 {K}

and 1 bar for

{CaO}({s})=-603.501 {kJ} {mol}^{-1}

{CO}_{2}({g})=-394.389 {kJ} {mol}^{-1}

{CaCO}_{3}({s})=-1128.79 {kJ} {mol}^{-1}

Gas constant

{R}=8.314 {J} {K}^{-1} {mol}^{-1}]

Solution:

{CaCO}_{3}({s}) \⇌{CaO}({s})+{CO}_{2}({g})

\Δ_{r} G^{0}=\Δ_{f} G^{0}({CaO})+\Δ_{{f}} {G}^{0}({CO}_{2})

-\Δ_{{f}} {G}^{0}({CaCO}_{3})

=-603.501-394.389+1128.79

=130.9 {kJmol}^{-1}

\Δ_{{r}} {G}^{0}=-2.303 {RT} \log {K}_{{p}}

\log {K}_{{p}}={130.9 \× 1000}/{-2.303 \×298 \×8.314}

=-22.94

{K}_{0}=

antilog

(-22.94)=1.13 \×10^{-23}

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