For the reaction
2H2(g) + 2NO(g) $\to$ N2(g) + 2H2O(g)
the observed rate expression is, rate = Kf[NO]2[H2]. The rate expression for the reverse reaction is :
Solution
K<sub>eq</sub> = ${{{k_f}} \over {{k_b}}}$ = $${{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {{{\left[ {{H_2}} \right]}^2}{{\left[ {NO} \right]}^2}}}$$
<br><br>$${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2} = {k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
<br><br>At equilibrium r<sub>f</sub> = r<sub>b</sub>
<br><br>Given r<sub>f</sub> = ${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2}$
<br><br>$\therefore$ r<sub>b</sub> = $${k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
This question is part of PrepWiser's free JEE Main question bank. 96 more solved questions on Chemical Kinetics are available — start with the harder ones if your accuracy is >70%.