Question

1. Rate equations can only be found experimentally. The table below shows the results of an investigation into the following reaction:

2H2(g) + 2NO(g) 2H2O(g) + N2(g) (Temp =973K) Experiment no. [H2] moldm-3 (x10-2) [NO] moldm-3 (x10-2) Rate moldm-3s-1 (x10-6)
1 2.00 2.50 4.80
2 2.00 1.25 1.20
3 1.00 1.25 0.60

Answer 1

The rate law is Rate = k[
`H_2`] [NO
`]^2`

The value of the rate constant is 1.04 *
`10^2`
`mol^2`
`dm^{-6`

Temperature, the existence of catalysts, and the particular reaction mechanism are some of the variables that affect the rate constant. It is specific to a given reaction at a given temperature and gives information about the speed of the reaction.

We have that;

(2.50/1.25
`)^n` = 4.80/1.20

`2^n`= 4

`2^n`=
`2^2`

n = 2

Again;

(2/1
`)^n` = 1.20/0.60

`2^n` =
`2^1`

n = 1

We then have the rate law as;

Rate = k[
`H_2`] [NO
`]^2`

The numerical value;

4.8 *
`10^{-6` = k[2 *
`10^{-2`] [2.5 *
`10^{-2`]

k = [2 *
`10^{-2`] [2.5 *
`10^{-2`]/4.8 *
`10^{-6`

k = 1.04 *
`10^2`
`mol^2`
`dm^{-6`