The reaction (nitrogen monoxide reacts with hydrogen gas to form nitrogen gas and water vapor) is believed to take place in three steps, the first of which is the fast reversable dimerization of NO to form N2O2, and the last of which is the reaction N2O + H2 -> N2 + H2O. (i) what is the (slow) second step? (ii) Show, using the stead-state approximation, that the mechanism is consistant with the obseved rate law. (iii) Why is it only approximatily true that the reaction is first-order in H2?
(i) To determine the slow second step of the reaction, we need to consider the information given in the question. We know that the first step is the fast reversible dimerization of NO to form N2O2, and the last step is the reaction N2O + H2 -> N2 + H2O.
Since the first step is fast, it can be assumed that the concentration of the dimer N2O2 is at steady-state, meaning the rate of its formation equals the rate of its consumption. Therefore, the second step must involve the consumption of N2O2.
Hence, the slow second step can be represented as:
N2O2 -> N2O + O2
(ii) To show that the mechanism is consistent with the observed rate law, we first need to determine the rate law for the overall reaction.
The rate law is given in the question as approximately first-order in H2 (hydrogen gas). This means that the rate of the reaction is directly proportional to the concentration of H2.
Let's assume the rate law for the overall reaction is: Rate = k[H2]^x
To use the steady-state approximation, we consider the intermediate species (N2O2) and assume that its rate of formation is equal to its rate of consumption. This allows us to express the rate of the overall reaction in terms of the concentration of N2O2.
Assuming the rate of formation of N2O2 in the first step is k1[NO]^2 and the rate of consumption of N2O2 in the second step is k2[N2O2], we have:
Rate = k2[N2O2] = k1[NO]^2
Now, we use the steady-state approximation, which states that the rate of change of an intermediate species concentration is approximately zero.
Therefore, d[N2O2]/dt = 0
From this, we can write:
k1[NO]^2 - k2[N2O2] = 0
Rearranging the equation, we get:
[N2O2] = (k1/k2) [NO]^2
Substituting this expression for [N2O2] in the rate law, we have:
Rate = k[H2]^x = k2 (k1/k2) [NO]^2
Simplifying further:
Rate = k1k2'/k2 [NO]^2
Where k2' = k[H2]^x
Comparing this rate law with the observed rate law (Rate = k[H2]^x), we can see that the mechanism is consistent with the observed rate law.
(iii) It is only approximately true that the reaction is first-order in H2 because the rate law, as given, assumes that the reaction is first-order with respect to H2. However, based on the mechanism provided, we see that the second step involves the consumption of N2O2, not H2 directly.
So, while the rate law is written as first-order in H2, it is an approximation that assumes the concentration of N2O2 is at steady-state. In reality, the actual reaction order with respect to H2 may be different, but the steady-state approximation simplifies the math and provides a reasonable approximation when comparing the mechanism with experimental observations.