Question

Practice 14.2: Predict the equilibrium constant for the first reaction given the equilibrium constants for the second and third reactions: (ans: 1.4 x 102)

CO2 (g) + 3 H2 (g)-><-CH3OH (g) + H2O (g) K1 = ?
CO (g) + H2O (g)-><- CO2 (g) + H2 (g) K2 = 1.0 x 105
CO (g) + 2 H2 (g)-><-CH3OH (g) K3 = 1.4 x 107

Answer 1

K1 = 1.4*10^2

Step-by-step explanation:

The given reactions are:

Reaction 1

`CO (g) + H2O (g)\rightleftharpoons  CO2 (g) + H2 (g).....K2 = 1.0 * 10^(5)`

Reaction 2

`CO (g) + 2 H2 (g)\rightleftharpoons CH3OH (g)....K3 = 1.4 * 10^(7)`

The required reaction is:

`CO2 (g) + 3 H2 (g)\rightleftharpoons CH3OH (g) + H2O (g)... K1 = ?`

This reaction can be obtained by reversing the first reaction and then adding the reactions 1 and 2.

As per convention

1) For a multistep reaction, the net equilibrium constant is the product of Keq of the individual steps

2) if a reaction is reversed the new equilibrium constant will be the inverse of the old equilibrium constant.

Therefore,

`K1 = (1)/(K2)*K3= (1)/(1.0*10^(5))*1.4*10^(7)=1.4*10^(2)`