Question

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

H° = -45.6 kJ and S° = -125.7 J/K

The equilibrium constant for this reaction at 262.0 K is

Answer 1

Answer : The value of equilibrium constant for this reaction at 262.0 K is
`3.35* 10^(2)`

Explanation :

As we know that,

`\Delta G^o=\Delta H^o-T\Delta S^o`

where,

`\Delta G^o` = standard Gibbs free energy = ?

`\Delta H^o` = standard enthalpy = -45.6 kJ = -45600 J

`\Delta S^o` = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

`\Delta G^o=(-45600J)-(262.0K* -125.7J/K)`

`\Delta G^o=-12666.6J=-12.7kJ`

The relation between the equilibrium constant and standard Gibbs free energy is:

`\Delta G^o=-RT* \ln k`

where,

`\Delta G^o` = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

`-12666.6J=-(8.314J/K.mol)* (262.0K)* \ln k`

`k=3.35* 10^(2)`

Therefore, the value of equilibrium constant for this reaction at 262.0 K is
`3.35* 10^(2)`