Question

15 moles of H2 and 5.2 moles of I2 are mixed and then allowed to attain equilibrium at 500 degree centigrade.at equilibrium the concentration of HI is found to be 10 moles .
The equilibrium constant for the formation of HI is?

Answer 1

i think k = 3.7 approximately

Answer 2

Answer: The equilibrium constant for the formation of HI is 50.

Explanation:- $H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

Initially:   15   5.2           0

At eq'm  15-x   5.2-x       2x

Given: 2x =10,

x=5 moles

$[H_2]=\frac{15-x}{V}=\frac{10}{V}$

$[I_2]=\frac{5.2-x}{V}=\frac{0.2}{V}$

$[HI]=\frac{2x}{V}=\frac{10}{V}$

Equilibrium constant is the ratio of product of concentration of products to the product of the concentration of reactants each term raised to their stochiometric coefficients.

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

$K_c=\frac{(\frac{10}{V})^2}{(\frac{0.2}{V})(\frac{10}{V})}$

$K_c=50$