Question

Initial rate data at 25^o C are listed in the table below for the reaction:

2NO(g) + Cl₂(g) → 2NOCl(g)

[NO]i (M)	[Cl2]i(M)	Initial rate (M/s)
0.13	0.20	1.0 x 10-2
0.26	0.20	4.0 x 10-2
0.13	0.10	5.0 x 10-3

First determine the experimental rate law and calculate the the value of the rate constant, k (no units).
k =

Tries 0/45

Then, calculate the initial rate of this reaction, when the initial concentration of NO is 0.78 M and the inital concentration of Cl₂ is 0.93 M? The units should be M/s.

Answer 1

Answer - We are given, reaction -   2 NO(g) + Cl₂(g) -----> 2 NOCl(g)

We assume rate law is ,rate = k [NO]^x [Cl₂]^y

So, x and y are the order with respect to NO and Cl₂

So the rate law for each experiment is as follow-

Rate1 = k [NO]₁^x [Cl₂]₁^y

Rate2 = k [NO]₂^x [Cl₂]₂^y

Rate 3 = k [NO]₃^x [Cl₂]₃^y

Order with respect to NO -

Rate2/ Rate1 = k [NO]₂^x [Cl₂]₂^y / k [NO]₁^x [Cl₂]₁^y

4.0E-2 / 1.0E-2 = (0.26)^x /(0.13)^x *(0.20)^y /(0.20)^y

4 = (2)^x

So, x = 2

So order with respect to NO is second order

Now order with respect to H₂

Rate3/Rate1 = k [NO]₃^x [Cl₂]₃^y / k [NO]₁^x [Cl₂]₁^y

5.0E-3 / 1.0E-2 = (0.13)² /(0.13)² *(0.10)^y /(0.20)^y

0.5 = (0.2)^y

So, ny= 1

So order with respect to Cl₂ is first

So overall order = 2 + 1 = 3

So rate law, Rate = k [NO]² [Cl₂]

b) Now we need to calculate the rate constant

We know rate law

Rate = k [NO]² [Cl₂]

1.0*10^-2 M/s = k *(0.13 M)² (0.20 M)

k = 1.0*10^-2 M.s^-1/ (0.13 M)² (0.20 M)

= 2.95 M^-2 s^-1

Now we are given, [NO] = 0.78 and for [Cl] = 0.93

We know,

Rate = k [NO]² [Cl₂]

        = 2.95 * (0.78)²(0.93)

         = 1.67 M/s