Question 14.6.4: A kinetic study is done to investigate the following reactio......
A kinetic study is done to investigate the following reaction:
2 H_{2}(g) + 2 NO(g) → N_{2}(g) + 2 H_{2}O(g)
A proposed mechanism is:
Step 1. 2 NO(g) \rightleftarrows N_{2}O_{2}(g) Fast in both directions
Step 2. N_{2}O_{2}(g) + H_{2}(g) → N_{2}O(g) + H_{2}O(g) Slow
Step 3. N_{2}O(g) + H_{2}(g) → N_{2}(g) + H_{2}O(g) Fast
a. Does this mechanism account for the overall reaction?
b. What experimental rate law would be observed if this mechanism is correct?
You are asked to determine whether a given mechanism is possible for a reaction and to predict the experimental rate law for the reaction.
You are given a balanced chemical equation and a proposed mechanism for the reaction.
a. The equation for the overall reaction may be obtained by adding the equations for the elementary steps and simplifying.
2 NO(g) \rightleftarrows N_{2}O_{2}(g)
N_{2}O_{2}(g) + H_{2}(g) → N_{2}O(g) + H_{2}O(g)
N_{2}O(g) + H_{2}(g) → N_{2}(g) + H_{2}O(g)
2\text{ NO(g)}+\cancel{\text{ N}_{2}\text{O}_{2}\text{(g)}} +\cancel{\text{ N}_{2}\text{O}\text{(g)}}+ 2\text{ H}_{2}\text{(g)} \longrightarrow \cancel{\text{ N}_{2}\text{O}_{2}\text{(g)}}+\cancel{\text{ N}_{2}\text{O(g)}}+\text{N}_{2}\text{(g)}+2\text{ H}_{2}\text{O(g)} 2\text{ NO(g)}+ 2\text{ H}_{2}\text{(g)} \longrightarrow \text{N}_{2}\text{(g)}+2\text{ H}_{2}\text{O(g)}
The sum of the proposed mechanism steps is the same as the equation for the reaction being studied.
b. First write the rate law from the slow rate-determining Step 2, where k_{2} is the rate constant for Step 2.
rate = k_{2}[N_{2}O_{2}][H_{2}]
To replace the intermediate (N_{2}O_{2}) in the rate law, assume that the first step comes to equilibrium, equate the forward (k_{1}) and backward (k_{-1}) rates, and solve for [N_{2}O_{2}] in terms of [NO].
k_{1}[NO]² = k_{-1}[N_{2}O_{2}]
[N_{2}O_{2}] = (k_{1}/k_{-1})[NO]²
Finally, replace [N_{2}O_{2}] in the rate equation.
Rate = k_{2}(k_{1}/k_{-1})[NO]²[H_{2}] = k[NO]²[H_{2}], where k = k_{2}(k_{1}/k_{-1})