JEE MAIN - Chemistry (2020 - 7th January Evening Slot - No. 15)
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 :
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 :
Kb[N2][H2O]
Kb[N2][H2O]2/[H2]
Kb[N2][H2O]2/[NO]
Kb[N2][H2O]2
Explanation
Keq = $${{{k_f}} \over {{k_b}}}$$ = $${{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {{{\left[ {{H_2}} \right]}^2}{{\left[ {NO} \right]}^2}}}$$
$${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2} = {k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
At equilibrium rf = rb
Given rf = $${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2}$$
$$ \therefore $$ rb = $${k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
$${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2} = {k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
At equilibrium rf = rb
Given rf = $${k_f}\left[ {{H_2}} \right]{\left[ {NO} \right]^2}$$
$$ \therefore $$ rb = $${k_b}{{\left[ {{N_2}} \right]{{\left[ {{H_2}O} \right]}^2}} \over {\left[ {{H_2}} \right]}}$$
Comments (0)
