At 700 K, equilibrium constant for the reaction: H 2 (g) + I 2 (g) ⇌ 2HI (g) is 54.8. If 0.5 mol L –1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H 2 (g) and I 2 (g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

Question

Philoid · Accepted Answer

For the reaction,

H₂ (g) + I₂ (g) ⇌2HI (g)

Equilibrium constant is given to be 54.8.

i.e, K_c =

Let x be equilibrium concentration of H₂ at equilibrium. Then equilibrium concentration of I₂ will also be x. Since, initially there was only HI present, equal amount of H₂ and I₂ will be produced.

At equilibrium:

K_c =

= = 54.8

x =

= 0.0675

Therefore, concentration of H₂ and I₂ is 0.0675 mol L^-1

At 700 K, equilibrium constant for the reaction:H2 (g) + I2 (g) ⇌2HI (g)is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?

At 700 K, equilibrium constant for the reaction:
H₂ (g) + I₂ (g) ⇌2HI (g)

is 54.8. If 0.5 mol L^–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?