The mechanism for the reaction described by
2N2O5(g) ---> 4NO2(g) + O2(g)
is suggested to be
(1) N2O5(g) (k1)--->(K-1) NO2(g) + NO3(g)
(2) NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g)
(3) NO(g) + N2O5(g) --->(K3) 3NO2(g)
Assuming that [NO3] is governed by steady-state conditions, derive the rate law for the production of O2(g) and enter it in the space below.
Rate of reaction= ∆[O2]/∆t = ???
k1k2[N2O5] / k2+k-1
*the numbers following the k's are the subscripts
Hmm, let me put on my thinking cap. Well, looking at the suggested mechanism, we can see that the production of O2(g) only involves the second step:
(2) NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g)
Therefore, the rate law for the production of O2(g) would depend on the concentrations of NO2(g) and NO3(g). Let's assume the rate constant for this step is k2.
Rate of reaction = k2[NO2][NO3]
So, the rate law for the production of O2(g) is given by:
Rate of reaction = k2[NO2][NO3]
I hope that answers your question without making you go "NO" to my attempt at humor!
To derive the rate law for the production of O2(g) based on the given mechanism, we need to determine the rate-determining step. The rate-determining step is the slowest step in a reaction, and it determines the overall rate of the reaction.
Based on the suggested mechanism, the rate-determining step is the second step:
NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g) (eq. 2)
According to the steady-state approximation, the concentration of the intermediate species NO3 remains constant during the reaction. Therefore, we can assume that the rate of formation of NO3 is equal to the rate of its consumption:
Rate formation of NO3 = Rate consumption of NO3
From equation 2, we see that NO2(g) is both formed and consumed. Thus:
Rate formation of NO3 = k1[N2O5]
Rate consumption of NO3 = k2[NO2][NO3]
where k1 and k2 are the rate constants for the respective reactions.
Setting the two rates equal to each other:
k1[N2O5] = k2[NO2][NO3]
Now, we want to find the rate of reaction for O2(g) formation, which is given by:
Rate of reaction = ∆[O2]/∆t
From equation 2, we see that the rate of reaction is equal to the rate of formation of O2(g):
Rate of reaction = k2[NO2][NO3]
Therefore, the rate law for the production of O2(g) based on the given mechanism is:
Rate of reaction = k2[NO2][NO3]
To derive the rate law for the production of O2(g), we need to determine the rate-determining step, which is the slowest step in the reaction mechanism. In this case, it is given by equation (2):
NO2(g) + NO3(g) --->(K2) NO2(g) + O2(g) + NO(g)
Let's assume that the rate law for this step is:
Rate = k2 [NO2]x [NO3]y
Since [NO3] is governed by steady-state conditions, we can express it in terms of the other species involved in the mechanism. Notice that equation (1) involves the formation of NO3. So, we can set up an equation for the rate of the forward reaction (1) and the rate of the backward reaction (1) to determine the steady-state concentration of NO3.
Rate forward = k1 [N2O5]
Rate backward = (K-1) [NO2] [NO3]
At steady-state, the rate of the forward reaction is equal to the rate of the backward reaction. Therefore:
k1 [N2O5] = (K-1) [NO2] [NO3]
Simplifying the equation:
[NO3] = (k1[N2O5]) / ((K-1)[NO2])
Now, substitute this expression for [NO3] back into the rate law for equation (2):
Rate = k2 [NO2]x [(k1[N2O5]) / ((K-1)[NO2])]y
Simplifying further:
Rate = k2 k1^y/N2O5^y-1 (1/(K-1)^y) [NO2]x+y
The rate law for the production of O2(g) is given by the exponent of [O2] in the rate equation. Thus, the rate law for the production of O2(g) is:
Rate = k2 k1^y/N2O5^y-1 (1/(K-1)^y) [NO2]^1
Since the stoichiometric coefficient in front of NO2 is 1, the rate law simplifies to:
Rate = k1^y (k2/N2O5^y-1) (1/(K-1)^y) [NO2]
Please note that the specific values for the rate constants (k1, k2, K-1) and the reaction order (x, y) would need to be provided in order to obtain the numerical expression for the rate law.