Now the reaction (2NO+2H2==> 2H20+N2) is believed to take place in three steps, the first of which is the fast reversible dimerization of NO to form N2O2, and the last of which (again fast) is the reaction N20 + H2 --> N2 + H2O. So, what is the slow (second) step? Then show, using the steady-state approximation, that the mechanism is consistent with the observed rate law. Why is it only approximately true that the reaction is first-order in H2?
The slow (second) step in the reaction is the dimerization of N2O2 to form N2 and H2O. The reaction can be represented as follows:
N2O2 ⇌ 2NO (slow)
Using the steady-state approximation, we assume that the rate of formation of N2O2 and its rate of consumption are equal, meaning that the concentration of N2O2 remains constant over time.
Let's denote the rate of dimerization of N2O2 as r1 and the rate of consumption of N2O2 as r2. According to the steady-state approximation, we have:
r1 = r2
The rate of formation of N2O2 (r1) can be expressed as the product of the rate constant (k1) and the concentration of NO squared:
r1 = k1[NO]^2
The rate of consumption of N2O2 (r2) can be expressed as the product of the rate constant (k-1) and the concentration of N2O2:
r2 = k-1[N2O2]
Setting r1 equal to r2 and substituting the expressions for r1 and r2, we get:
k1[NO]^2 = k-1[N2O2]
Rearranging the equation, we have:
[N2O2] = (k1/k-1)[NO]^2
This equation shows that the concentration of N2O2 is proportional to the square of the concentration of NO.
Now, let's consider the rate-determining step (the slowest step), which is the dimerization of NO:
2NO ⇌ N2O2
The rate of this reaction depends on the concentration of NO, and in a typical rate law, the rate would be proportional to the concentration of NO raised to a power (let's call it n). However, since this step is fast, the concentration of NO is much higher than the concentration of N2O2. Therefore, we can assume that the concentration of N2O2 remains approximately constant and the rate is only dependent on the concentration of NO. This leads to a simplified rate law:
rate = k'[NO]^n
Since the concentration of N2O2 remains constant, it does not contribute to the observed rate law, resulting in a first-order dependence on [NO] (n = 1).
To summarize, using the steady-state approximation, we found that the slow (second) step is the dimerization of N2O2 to form N2 and H2O. The mechanism is consistent with the observed rate law because the rate-determining step is fast, and the concentration of N2O2 is assumed to be constant. The approximation of the reaction being first-order in H2 is only approximately true because the rate-determining step involves the dimerization of NO, and the concentration of H2 does not directly affect this step's rate.
To determine the slow (second) step in the given reaction mechanism, we can use the steady-state approximation. According to the proposed mechanism, the overall reaction is believed to occur in three steps.
1. Fast reversible dimerization of NO: 2NO ⇌ N2O2
2. Slow step (unknown intermediate): N2O2 + H2 ⟶ ??
3. Fast reaction: N2O + H2 ⟶ N2 + H2O
The steady-state approximation can be applied to the intermediate species N2O2. According to this approximation, the rate of production of N2O2 is equal to its rate of consumption. By assuming the concentration of the intermediate N2O2 remains constant throughout the reaction, we can now focus on the overall rate of formation of the product, which is given by the last step of the reaction.
Considering the overall reaction: 2NO + 2H2 ⟶ 2H2O + N2
If we express the rate of formation of N2, d[N2]/dt, we can write it as the rate of consumption of the intermediate N2O2, d[N2O2]/dt. This can be expressed as the rate of production of N2O2 in the first step, multiplied by the rate constant for the second slow step, and divided by the rate constant for the first step:
d[N2]/dt = k2 * [N2O2]/k1
This expression represents the rate law for the overall reaction. It suggests that the rate of formation of N2 (the product) is directly proportional to the concentration of the intermediate (N2O2) and the rate constant for the second slow step (k2), while being inversely proportional to the rate constant for the first step (k1).
The reason it is only approximately true that the reaction is first-order in H2 is due to the fast and reversible nature of the first step, where the dimerization of NO to form N2O2 occurs. The concentration of NO is typically much greater than H2, and thus, the concentration of N2O2 primarily depends on the concentration of NO and not H2. Consequently, the rate-limiting step and the observed rate law primarily depend on the concentration of NO, resulting in the reaction being approximately first-order in H2.