A proposed mechanism for the reaction of NO and H 2 gases is given:2NO (g) ⇌ N 2 O 2 (g) fast equilibriumN 2 O 2 (g) + H 2 (g) ® N 2 O (g) + H 2 O (g) slowN 2 O (g) + H 2 (g) ® N 2 (g) + H 2 O (g) fastDerive the rate law for the overall reaction based on the above mechanism.
Question
A proposed mechanism for the reaction of NO and H 2 gases is given:2NO (g) ⇌ N 2 O 2 (g) fast equilibriumN 2 O 2 (g) + H 2 (g) ® N 2 O (g) + H 2 O (g) slowN 2 O (g) + H 2 (g) ® N 2 (g) + H 2 O (g) fastDerive the rate law for the overall reaction based on the above mechanism.
Solution
The rate law for a reaction is determined by the slowest step in the reaction mechanism, also known as the rate-determining step. In this case, the slow step is the second step: N2O2 (g) + H2 (g) → N2O (g) + H2O (g).
The rate law for this step is: Rate = k[N2O2][H2]
However, N2O2 is an intermediate, and we need to express the rate law in terms of the reactants. From the first step, which is a fast equilibrium, we can write the equilibrium expression: K = [N2O2]/[NO]^2
Solving for [N2O2], we get: [N2O2] = K[NO]^2
Substituting this into the rate law, we get: Rate = k[K[NO]^2][H2] = k' [NO]^2[H2]
So, the rate law for the overall reaction is: Rate = k' [NO]^2[H2]
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