Question

For the reaction

$2H_2\left(g\right)+2NO\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(g\right)$

the observed rate expression is,

rate = $k_f\left[NO{]}^2\right[H_2]$. The rate expression for the reverse reaction is :

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• A. $k_b\left[N_2\right][H_{2}O{]}^2$
• B. $k_b\left[N_2\right]\left[H_{2}O\right]$
• C. $k_b\frac{\left[N_2\right][H_{2}O{]}^2}{\left[H_2\right]}$
• D. $k_b\frac{\left[N_2\right][H_{2}O{]}^2}{[NO{]}^2}$

Answer 1

The correct option is C $k_b\frac{\left[N_2\right][H_{2}O{]}^2}{\left[H_2\right]}$

$k_{eq}=\frac{k_f}{k_b}=\frac{\left[N_2\right][H_{2}O{]}^2}{\left[H_2{]}^2\right[NO{]}^2}$

Rearranging

$k_f\left[NO{]}^2\right[H_2]=\frac{k_b\left[N_2\right][H_{2}O{]}^2}{\left[H_2\right]}$

on comparing $R_f$ and $R_b$ at equilibrium,

$R_b=k_b\frac{\left[N_2\right][H_{2}O{]}^2}{\left[H_2\right]}$