Question

At 500K, the equilibrium constant, K_eq, is 155 for the reaction H₂(g) + I₂(g) ↔ 2HI(g). Calculate the equilibrium constant for the reaction  2H₂(g) + 2I₂(g) ↔ 4HI(g), 1/2H₂(g) + 1/2I₂(g) ↔ HI(g), and 2HI(g)  ↔ H₂(g) + I₂(g).

Answer 1

Ans;- The given reaction is

H₂ + I₂ 2HI

The rate expression for the given reaction can be written as

Here K= 155 is the equilibrium constant for the above reaction

Now,

a) For the reaction

2H₂ + 2I₂ 4HI

The rate expression will be

K₁ = (K)²

      =(155)²

      =24025 is the equilibrium constant

b) For the reaction

1/2 H₂ + 1/2 I₂ HI

The rate expression will be

K₂ = (K)^1/2

       =(155)^1/2

   = 12.449 is the equilibrium constant

c) For the reaction

2HI H₂ + I₂

The rate expression will be

K₃ = 1/K

      =1/155

      = 0.00645 is the equilibrium constant