Question

2 h₂(g) + 2 no(g) → n₂(g) + 2 h₂o(g)
the experimental rate law for the reaction represented above is rate = kh₂no². which of the following proposed mechanisms is consistent with the rate law?
a
step 1: 2 no → n₂o₂ fast
step 2: h₂ + n₂o₂ ⇌ n₂o + h₂o fast
step 3: h₂ + n₂o → n₂ + h₂o slow
b
step 1: 2 no ⇌ n₂o₂ fast
step 2: h₂ + n₂o₂ → n₂o + h₂o slow
step 3: h₂ + n₂o → n₂ + h₂o fast
c
step 1: 2 no → n₂ + o₂ slow
step 2: h₂ + o₂ → h₂o₂ fast
step 3: h₂ + h₂o₂ → 2 h₂o fast

Explanation:

Step1: Recall Rate Law from Mechanism

The rate law of a reaction is determined by the slow (rate - determining) step. For a mechanism with a fast equilibrium step before the slow step, we can use the equilibrium expression to substitute for the intermediate.

Step2: Analyze Mechanism A

• Step 3 (slow): $\ce{H_{2} + N_{2}O -> N_{2} + H_{2}O}$, rate law would be $rate = k_3[\ce{H_2}][\ce{N_2O}]$. But we need to find $[\ce{N_2O}]$ from previous steps. Step 2 is a fast equilibrium: $\ce{H_{2} + N_{2}O_{2} <=> N_{2}O + H_{2}O}$, $K_2=\frac{[\ce{N_2O}][\ce{H_2O}]}{[\ce{H_2}][\ce{N_2O_2}]}$, so $[\ce{N_2O}]=\frac{K_2[\ce{H_2}][\ce{N_2O_2}]}{[\ce{H_2O}]}$. Step 1 is fast: $\ce{2NO -> N_{2}O_{2}}$, but it's a one - way fast step, not an equilibrium. So we can't easily substitute $[\ce{N_2O_2}]$ in terms of $[\ce{NO}]$ (since it's not an equilibrium, we can't use $K$ for it). So the rate law will be complicated and not match $rate = k[\ce{H_2}][\ce{NO}]^2$.

Step3: Analyze Mechanism B

• Step 2 (slow): $\ce{H_{2} + N_{2}O_{2} -> N_{2}O + H_{2}O}$, rate law from slow step: $rate = k_2[\ce{H_2}][\ce{N_2O_2}]$.
• Step 1 is a fast equilibrium: $\ce{2NO <=> N_{2}O_{2}}$, the equilibrium constant $K_1=\frac{[\ce{N_2O_2}]}{[\ce{NO}]^2}$, so $[\ce{N_2O_2}]=K_1[\ce{NO}]^2$.
• Substitute $[\ce{N_2O_2}]$ into the rate law from step 2: $rate = k_2K_1[\ce{H_2}][\ce{NO}]^2=k[\ce{H_2}][\ce{NO}]^2$ (where $k = k_2K_1$), which matches the given rate law $rate = k[\ce{H_2}][\ce{NO}]^2$.

Step4: Analyze Mechanism C

• Step 1 (slow): $\ce{2NO -> N_{2} + O_{2}}$, rate law would be $rate = k_1[\ce{NO}]^2$, which does not match the given rate law that also has $[\ce{H_2}]$ dependence.

Answer:

B. Step 1: $\boldsymbol{2\ce{NO}
ightleftarrows \ce{N_{2}O_{2}}}$ (fast); Step 2: $\boldsymbol{\ce{H_{2} + N_{2}O_{2} -> N_{2}O + H_{2}O}}$ (slow); Step 3: $\boldsymbol{\ce{H_{2} + N_{2}O -> N_{2} + H_{2}O}}$ (fast)