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

Answer 1

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

If we write the expression of equilbrium constant here,
$K_C=k_f/k_b=\frac{\left[N_2\right][H_{2}O{]}^2}{\left[H_2{]}^2\right[NO{]}^2}$
We know that at equilibrium,
rate of forward reaction $=$ rate of backward reaction
i.e., $k_f\left[NO{]}^2\right[H_2{]}^2=k_b\left[N_2\right][H_{2}O{]}^2$
Thus, rate of backward reaction is $k_b\left[N_2\right][H_{2}O{]}^2$