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Question: For the water gas reaction \[{\text{C}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}...

For the water gas reaction C(s) + H2O(g) .⇌ CO(g) + H2O(g){\text{C}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right){\text{ }}{\text{.}} \rightleftharpoons {\text{ CO}}\left( {\text{g}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right) the standard Gibbs free energy of reaction (at 1000 K1000{\text{ K}}) is −8.1 kJ/mol - 8.1{\text{ kJ/mol}} . Calculate its equilibrium constant.

Explanation

Solution

To calculate the value of the equilibrium constant for the formation of water gas, use the following reaction.
K=eΔGo−2.303RTK = {e^{\dfrac{{\Delta {{\text{G}}^o}}}{{ - 2.303RT}}}}
Here, ΔGo\Delta {{\text{G}}^o} represents the standard Gibbs free energy change, RR represents the ideal gas constant, TT represents the absolute temperature and KK represents the equilibrium constant .

Complete Step by step answer: Write the reaction for the formation of the water gas as shown below:
C(s) + H2O(g)⇌ CO(g) + H2(g){\text{C}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{g}} \right) \rightleftharpoons {\text{ CO}}\left( {\text{g}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)
In this reaction, carbon solid reacts with water vapors (steam) at 1000 K1000{\text{ K}} to form carbon monoxide gas and hydrogen gas. The mixture of carbon monoxide gas and hydrogen gas is called water gas.
The above reaction is an equilibrium reaction.
Write the relationship between the equilibrium constant and the standard Gibbs free energy change of the reaction as shown below:
ΔGo=−2.303RTlog⁡10K\Delta {{\text{G}}^o} = - 2.303RT{\log _{10}}K
Here, ΔGo\Delta {{\text{G}}^o} represents the standard Gibbs free energy change, RR represents the ideal gas constant, TT represents the absolute temperature and KK represents the equilibrium constant .
Rearrange the above expression:
log⁡10K=ΔGo−2.303RT{\log _{10}}K = \dfrac{{\Delta {{\text{G}}^o}}}{{ - 2.303RT}}
Take antilog on both sides of the reaction
K=eΔGo−2.303RTK = {e^{\dfrac{{\Delta {{\text{G}}^o}}}{{ - 2.303RT}}}}
Substitute 1000 K1000{\text{ K}} for temperature, 8.314 × 10−3 kJ/molK{\text{8}}{\text{.314 }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ kJ/molK}} for the ideal gas constant and −8.1 kJ/mol - 8.1{\text{ kJ/mol}}for the standard Gibbs free energy change in the above expression and calculate the value of the equilibrium constant.
K=e(−8.1 kJ/mol−2.303 × 8.314 × 10−3 kJ/molK × 1000 K)K = {e^{\left( {\dfrac{{ - 8.1{\text{ kJ/mol}}}}{{ - 2.303{\text{ }} \times {\text{ 8}}{\text{.314 }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ kJ/molK }} \times {\text{ }}1000{\text{ K}}}}} \right)}}
⇒K=2.65\Rightarrow K = 2.65

Hence, the value of the equilibrium constant for the formation of water gas is 2.65.

Note: Write an express for the equilibrium constant for the formation of the water gas. K=[CO]×[H2][H2O]K = \dfrac{{\left[ {{\text{CO}}} \right] \times \left[ {{{\text{H}}_{\text{2}}}} \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}
In the above equilibrium constant expression, the concentration of carbon solid does not appear. In an equilibrium constant expression, the concentration of solids and pure liquids does not appear.