For the reaction, `2NO + Cl_(2) rarr 2NOCl` at 300 K, following data are obtained

The specific rate constant will be

Options

A. `1.2 xx 10^(2) mol^(-2)L^(2)s^(-1)`
B. `2.5 xx 10^(2) mol^(-2)L^(2)s^(-1)`
C. `3.2 xx 10^(2) mol^(-2)L^(2)s^(-1)`
D. `4.2 xx 10^(2) mol^(-2)L^(2)s^(-1)`

Answer

A

Explanation

Let the rate law for the reaction be rate `= k[NO]^(x)[Cl_(2)]^(y)`
From Experiment 1, we have `1.2 xx 10^(-4) = k[0.010]^(x)[0.010]^(y)` …(1)
From experiment 2, we have `2.4 xx 10^(-4) = k[0.010]^(x)[0.020]^(y)` …….(2)
Dividing Eq. (2) by Eq. (1), we get `(2.4 xx 10^(-4))/(1.2 xx 10^(-4)) = ((0.020)^(y))/([0.010]^(y)) rArr 2 = 2^(y) rArr y = 1`
from experiment 3, we have `9.6 xx 10^(-4) = k[0.020]^(x)[0.020]^(t)` ............(3)
Dividing Eq. (3) by Eq. (2), we get `(9.6 xx 10^(-4))/(2.4 xx 10^(-4))= ((0.020)^(x))/([0.010]^(x)) rArr 4 = 2^(x) rArr x = 2`
Hence, the order of the reaction `= x + y = 2 + 1 = 3`
Rate law for the reaction is Rate `= k[NO]^(2)[Cl_(2)]`
Considering Eq. (1) again, we get `1.2 xx 10^(-4) = k[0.010]^(2)[0.010]`
`k = (1.2 xx 10^(-4))/([0.010]^(3)) = 1.2 xx 10^(2) mol^(-2)L^(2) s^(-1)`