Reaction orders and rate constants

Question

The following data was collected for the reaction below

H₂(g) + Br₂(g) --> 2 HBr (g)

Expiriment [H₂] [Br₂] Rate [M s^-1]

1 0.25 0.0012 1.20 x 10^-4

2 0.50 0.0012 4.80 x 10^-4

3 0.50 0.0048 4.80 x 10^-4

a. write the rate equation and calculate the value of the rate constant (k)

b. what is the overall order of the reaction?

Answer 1

a). In order to write the rate equation (with specificity), we need to first find out the order of the reaction with respect to H₂ and Br₂. We do this by comparing 2 experiments where 1 of them changes and the other remains constant. We look at what happens to the rate and then determine the order with respect to the one that was changing.

So, for H₂, we can compare experiments 1 and 2 because [H₂] doubles from 0.25 to 0.50 while [Br2] remains constant at 0.0012. The rate doubles from 1.20 to 4.80 so this tells us the reaction is FIRST ORDER in [H₂]

Next, for Br_2, we compare experiments 2 and 3 where [Br₂] increases 4 times from 0.0012 to 0.0048 while [H₂] remains fixed at 0.50. The rate doesn't change at all and stays at 4.80x10^-4, so this tells us that the reaction is ZERO ORDER in Br₂.

Now, we can write the proper rate equation:

rate = k[A]¹[B]⁰ = k[A]

To calculate the value of k, we simply choose any one of the experiments and plug in the values for concentrations and rate and solve for k. I will use experiment 3:

rate = k[A]

4.8x10^-4 Ms^-1 = k(0.5 M)

k = 4.8x10^-4 Ms^-1 / 0.5 M

k = 9.6x10^-4 s^-1

b) The overall order of the reaction is the sum of the exponents which is 1 + 0 = 1 so the overall order is FIRST ORDER