Decomposition of N2O5: Reaction Mechanisms and Intermediate Concentrations

[SpinzTronics]

Homework Statement

Consider the decomposition of N2O5:
1. N2O5 → NO2 + NO3
2. NO2 + NO3 → NO2 + O2 + NO
3. NO + N2O5 → 3NO2

a) Find the reaction for the decomposition of N2O5 to stable products
b) Find the steady-state concentration of the intermediate(s)
c) Use the previous to show how the rate of reaction depends on the concentration of N2O5

Homework Equations

First I made the following forward and reverse reactions using k1, k2 and k3:
1. N2O5 → NO2 + NO3 using k1 forward
NO2 + NO3→ N2O5 using k-1 reverse
2. NO2 + NO3 → NO2 + O2 + NO with k2
3. NO + N2O5 → 3NO2 with k3

The Attempt at a Solution

a) -d[N2O5]/dt = k1[N2O5] - (k-1)[NO2][NO3]

b)The intermediates are NO and NO3
d[NO]/dt = 0= -k3[NO][N2O5]+k2[NO2][NO3]
=> simplifies to [NO] = (k2/k3)([NO2][NO3]/[N2O5])
d[NO3]/dt = 0= -k2[NO2][NO3]+k1[N2O5]-(k-1)[NO2][NO3]
=> simplifies to [NO3] = (-k1/(-k2-(k-1)))*[N2O5]/[NO2]

c) from the last expression, it is possible to have [NO3][NO2]= (-k1/(-k2-(k-1)))*[N2O5]
So, I substituted this for [NO3][NO2] in the overall reaction decomposition of N2O5 (part a):
R= k1[N2O5] -(k-1)[-k1/(-k2-(k-1))*[N2O5]] = (k1+((K-1)k1)/(-k2-(k-1)))*[N2O5] ~ keff[N2O5]

But isn't k3 supposed to be included in the overall reaction equation? Is this answer correct? Hope someone can help, thanks in advance!

 

[epenguin]

Homework Statement

Consider the decomposition of N2O5:
1. N2O5 → NO2 + NO3
2. NO2 + NO3 → NO2 + O2 + NO
3. NO + N2O5 → 3NO2

a) Find the reaction for the decomposition of N2O5 to stable products
b) Find the steady-state concentration of the intermediate(s)
c) Use the previous to show how the rate of reaction depends on the concentration of N2O5

Homework Equations

First I made the following forward and reverse reactions using k1, k2 and k3:
1. N2O5 → NO2 + NO3 using k1 forward
NO2 + NO3→ N2O5 using k-1 reverse
2. NO2 + NO3 → NO2 + O2 + NO with k2
3. NO + N2O5 → 3NO2 with k3

The Attempt at a Solution

a) -d[N2O5]/dt = k1[N2O5] - (k-1)[NO2][NO3]

b)The intermediates are NO and NO3
d[NO]/dt = 0= -k3[NO][N2O5]+k2[NO2][NO3]
=> simplifies to [NO] = (k2/k3)([NO2][NO3]/[N2O5])
d[NO3]/dt = 0= -k2[NO2][NO3]+k1[N2O5]-(k-1)[NO2][NO3]
=> simplifies to [NO3] = (-k1/(-k2-(k-1)))*[N2O5]/[NO2]

c) from the last expression, it is possible to have [NO3][NO2]= (-k1/(-k2-(k-1)))*[N2O5]
So, I substituted this for [NO3][NO2] in the overall reaction decomposition of N2O5 (part a):
R= k1[N2O5] -(k-1)[-k1/(-k2-(k-1))*[N2O5]] = (k1+((K-1)k1)/(-k2-(k-1)))*[N2O5] ~ keff[N2O5]

But isn't k3 supposed to be included in the overall reaction equation? Is this answer correct? Hope someone can help, thanks in advance!

You have ommited to include a term for reaction 3 in your equation a, and also in yuour equation for R.
I recommend you rewrite the equation for R in terms of the two steps that are irreversibly giving you product.

And yes, if intermediates are present in only small concentration it does follow that the rate Is independent of k₃.

No reverse reaction for the first step (k_-1) is shown in the question scheme, so I believe you would be justified in leaving this out, though there is no harm in including it just to see what difference it would make.
On second thoughts, also because it seems it is causing you confusion you, I would work it out first first for the simpler scheme where k_-1 = 0.

Food for thought - the mechanism seems to give such a simple result that I doubt it has been worked out mainly by kinetics.