Question

A proposed mechanism for the reaction N2O5-->2NO2 +1/2O2

N2O5 k1--> NO2+NO3 <---k_1-- NO2 +NO3 --k2--> NO +O2 +NO2 NO +NO3 - k3-->2NO2

Applying the steady-state approximation to show that the overall reaction rate is -d[N2 O5 ]/dt=k[N2 O5 ] In this process you should evaluate K in terms of K1, K_1, K2 and K3

Answer 1

Hi, the answer to the question is here:

Rate of formation of N2O5 can be written as:

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

Since NO3 is an intermediate, we have to remove it from the rate expression, so we can write it rate expression as:

d[NO3]/dt = k1[N2O5] – k(-1)[NO2][NO3] – k2[NO2][NO3] – k3[NO][NO3]

Let’s write another expression for NO2

d[NO2]/dt = k1[N2O5] – k(-1)[NO2][NO3] – k2[NO2][NO3]

since NO is also an intermediate, we have to write another expression for it

d[NO]/dt = k2[NO2][NO3] – k3[NO][NO3]

if we apply steady state approximation then, d[NO]/dt = 0

0 = k2[NO2][NO3] – k3[NO][NO3]

Rearranging the equation

k3[NO][NO3] = k2[NO2][NO3]

k3[NO] = k2[NO2]

[NO] = k2/k3[NO2] substituting the value of [NO] into the equation for NO3 and applying steady state approximation we get

d[NO3]/dt = k1[N2O5] – k(-1)[NO2][NO3] – k2[NO2][NO3] – k3[NO][NO3]

0= k1[N2O5] – k(-1)[NO2][NO3] – k2[NO2][NO3] – k2[N2][NO3]

0= k1[N2O5] – k(-1)[NO2][NO3] – 2(k2[NO2][NO3])

0= k1[N2O5] – [NO2][NO3](k(-1)+2k2) and

[NO2][NO3] = k1/(k(-1)+2k2) [N2O5]

Now, we have the value of [NO2][NO3]

Let’s put it into our first equation

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

d[N2O5]/dt = k1[N2O5] – k(-1){ k1/(k(-1)+2k2) [N2O5]}

d[N2O5]/dt = [N2O5]x{k1 – k(-1){ k1/(k(-1)+2k2)}

I apologize for the clumsy writing, it was not possible for me to write such a large number of equations on the editor. I am writing final equations in the writer.

or we can write

now we have the final equation that

I hope it helps.