Question

How can I calculate the rates and order in the following reaction? `H_2 + Br_2 -> 2HBr`

Answer 1

First and foremost, you need actual data, and second, you need a plausible mechanism. This is the most plausible one, in the presence of heat or light:

`1) H_2 => cancel(2H*)`
`2) Br_2 => cancel(2Br*)`
`3) cancel(2H*) + cancel(2Br*) => 2HBr`

Since it overall sums to:
`H_2 + Br_2 -> 2HBr`

...it works. This is a radical mechanism as you can see.

Now, according to that mechanism, it involves two molecules participating in the two-part rate-determining steps (slow steps), so it is a bimolecular process.

The first two steps are slow because breaking bonds to form radicals is much harder than waiting for two radicals to react (radicals are very reactive). Since the two participants are `H*` and `Br*`, we can write the rate law temporarily as:

`r(t) = k_(obs)[H_2]^m[Br_2]^n`

where `m` and `n` are unknown so far.

Now, let's pretend we actually have data. This is fake data, but let's say it was this:

`H_2 + Br_2 -> 2HBr` .... `r(t) (M^(2)s^(-1))`
`1M` ... `1M` ..... `2M` ........ `1"x"10^(-5)`
`1M` ... `2M` ..... `2M` ........ `2"x"10^(-5)`
`2M` ... `1M` ..... `2M` ........ `2"x"10^(-5)`

Each row is one trial.

If we look at rows 1 and 2, `[H_2]` did not change, but `[Br_2]` doubled. The rate also doubled. That means doubling the concentration of `Br_2` doubled the rate of reaction, and the reaction is said to be first-order with respect to `Br_2`.

If we look at rows 1 and 3, `[Br_2]` did not change, but `[H_2]` doubled. The rate doubled here too. That means doubling the concentration of `H_2` doubled the rate of reaction, and the reaction is said to be first-order with respect to `H_2`.

So, the rate law can be written as:

`-(d[H_2])/(dt) = -(d[Br_2])/(dt) = 1/2(d[HBr])/(dt) = r(t) = k_(obs)[H_2][Br_2]`

which is good enough for high school chemistry.