At a particular temperature

Question

At a particular temperature, K=6.5×10² for the reaction 2NO(g)+2H2​(g)⇌N2​(g)+2H2​O(g) 

Calculate K for each of the following reactions.

a. 2 N2​(g)+4H2​O(g)⇌4NO(g)+4H2​(g) 

b. NO(g)+H2​(g)⇌1/2 N2​(g)+H2​O(g)

Answer 1

2NO(g) + 2H₂​(g) ⇌ N₂​(g) + 2H_2​O(g) K = 6.5×10² (given)

a. 2 N₂​(g) + 4H₂​O(g) ⇌ 4NO(g) + 4H₂​(g) K_a = ?

The direction of the reaction is reversed ⇒ take the reciprocal of the first equilibrium constant K

All coefficients are doubled ⇒ square the first equilibrium constant K

So K_a = 1/K² = 1/(6.5×10²)² = 2.4×10^–6 

b. NO(g) + H₂​(g) ⇌ 1/2 N₂​(g) + H₂O(g) K_b = ?

All coefficients are halved ⇒ take the square root of the first equilibrium constant K

So K_b = √K = √(6.5×10²) = 25.5