Question

Consider the chemical equation and equilibrium constant at 25∘C: H2(g)+I2(g)⇌2HI(g) , K=6.2×102 Calculate the equilibrium constant for the following reaction at 25∘C: HI(g)⇌12H2(g)+12I2(g)

Answer 1

The given reaction is

H2 (g) + I2 (g) <------------------> 2 HI (g) .................Keq = 6.2 x 10^2

Let's flip the above reaction.

When a reaction is reversed, we have to take inverse of equilibrium constant.

2 HI (g) <----------------------> H2 (g) + I2 (g) ............. K'eq = 1/ 6.2 x 10^2

Let's divide above reaction by 2 .

When a reaction is divided by 2, we have to square root the equilibrium constant

HI (g) <---------------> 1/2 H2 (g) + 1/2 I2 (g) ...............K'eq = sq rt ( 1/ 6.2 x 10^2 )

This is the reaction for which we are finding equilibrium constant . So let's solve for K'eq.

K'eq = sq rt ( 1/ 6.2 x 10^2)

K'eq = sq rt ( 1.6129 x 10^-3)

K' eq = 0.0402

Equilibrium constant for the required reaction is 0.0402