Question

The following reaction is first order in N2O5:
N2O5(g)?NO3(g)+NO2(g)

The rate constant for the reaction at a certain temperature is 0.053/s.

A)

Calculate the rate of the reaction when [N2O5]= 5.4

Answer 1

We know that,

For a first order reaction the rate is,

rate = k [N2O5]

where k is rate constant

and, [N2O5] = concentration at any given time

A) given k = 0.053 /s

[N2O5] = 5.4 x 10^-2 M

Feeding the k and concentration values in the above relation we get,

rate = 0.053 X 0.054 = 2.862 X 10^-3 M/s

Thus, the rate of the reaction when [N2O5] is 5.4 x 10^-2 M is 2.86 x 10^-3 M/s

B) Now If the reaction is second order, then rate equation will be,

rate = k [N2O5]^2

Now the value of rate constant k for the second order reaction would be,

k = 0.053 M-1.s-1

Feeding this value into the above rate euqation,

rate = 0.053 X (0.054)^2

       = 1.55 X 10^-4 M/s

Therefore, the rate of this reaction when it is second order is 1.55 x 10^-4 M/s

C) Now If the reaction is zero order then the rate equation will be,

rate = k [N2O5]^0

       = k X 1

       = k

Now we have the rate constant k for zero order = 0.053 M/s

Thus, rate = 0.053 M/s

Therefore, rate for a zero order reaction will be 0.053 M/s.