Question

Equimolar concentration of h2 and i2 are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of h2 has reacted at equilibrium if rate constant for both forward and reverse reaction are equal

Answer 1

reaction is ---> $H_2+I_2\leftrightarrow2HI$
if Initially we have 100 % of [H2] then, we can assume that x % get reacted so 2x % of HI will be formed by the stoichiometry.

now, rate constant , k = $\frac{[HI]^2}{[H_2][I_2]}$
we also know, $k=\frac{k_f}{k_b}$
where $k_f$ is forward rate constant and $k_b$ backward rate constant.

a/c to question, $k_f=k_b$
so, k = 1 = $\frac{[HI]^2}{[H_2][I_2]}$

and also given, [H2] = [I2]
so, k = 1 = $\frac{[HI]^2}{[H_2]^2}$
1 = (2x)²/(100- x)²
2x = (100 - x)
x = 33.33 %

hence, answer is 33.33%