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

Reverse the 2nd reaction,
CO2 + H2 -------> CO + H2O---- i

Now new equilibrium constant will be,
K' = 1/K2 = 10^-5

Adding i and last equation you'll get,
CO2 (g) + 3 H2 (g)----->CH3OH (g) + H2O (g)
K1 = K' x K3 = 1.4 x 10^2 

Answer 2

Answer:

K1 = 1.4*10^2

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,

$K1 = \frac{1}{K2}*K3= \frac{1}{1.0*10^{5}}*1.4*10^{7}=1.4*10^{2}$