What is the rate constant k for this reaction ?

Question

For the following reaction the below data were measured.


2NO(g) + 2H2 (g) -----> N2 (g) + 2H2Og

Initial Concentrations (mol/L)
[NO] [H2]
0.10 0.10
0.10 0.20
0.20 0.10

Initital Rate of Formation of H2O (mol/Ls)
0.00123
0.00246
0.00492


What is the rate constant k for this reaction ?

Answer 1

Rate = k * [NO]^m * [H2]^n

0.00123 = k * 0.1^m * 0.1^n -------------------(1)
0.00246 = k * 0.1^m * 0.2^n -------------------(2)
0.00492 = k * 0.2^m * 0.1^n -------------------(3)

Divide (1) by (2)
0.00123/0.00246 = k/k * (0.1^m)/(0.1^m) * (0.1^n)/(0.2^n)
1/2 = (1/2)^n
Therefore n = 1

Divide (1) by (3)
0.00123/0.00492 = k/k * (0.1^m)/(0.2^m) * (0.1^n)/(0.1^n)
1/4 = (1/2)^m
4 = 2^m
Therefore m = 2

Now Substitute the values for m and n to the first (1) equation
0.00123 = k * 0.1^m * 0.1^n
0.00123 = k * 0.1^2 * 0.1^1
0.00123 = k * 0.001
k = 0.00123/0.001
k = 1.23

Units for k = (mol L-1)/{(mol2 L-2) * (mol L-1)}
= mol-2 L2 (litre squared per mol squared)

Thus, k = 1.23 mol-2 L2
=================

Answer 2

When you double the [H2], while keeping [NO2] the same, the rate doubles; so the reaction is directly proportional to [H2]. When you double [NO2] while keeping [H2] the same, the rate becomes 4 times. So the reaction is second order in [NO2].
The rate law is Rate = k [NO2]^2 [H2].

Use one of the sets of values in this equation to get k:
0.00123 = k (0.1)^2*(0.1)
k = 0.00123/(0.1*0.1*0.1) = 1.23 L^2/mol^2.s

Answer 3

Rate equation is r= k[0.1]^2 [0.1]
Note that as you double the first reactant in trial #3 that the rate of the reaction quadruples (=4x) indicating that it is second order with respect to the first reactant.

k = r/([0.1]^2 [0.1]) = 1.23