Question

What is the rate constant of a reaction if rate 1.5 (mol/L)s, [A] is 1 M, [B] is 3 M, m=2, and n=1?

Answer 1

For a reaction's rate law, in general the relevant portion is:

`\mathbf(r(t) = k[A]^m [B]^n)`

where:

• `r(t)` is the rate of the reaction. We know this to be in `"M/s"`, or `"mol"/("L"cdot"s")`, i.e. `"mol"cdot"L"^(-1)"s"^(-1)` or `"M"cdot"s"^(-1)`.
• `k` is the rate constant, in units that are based on the rest of the rate law. We'll need to determine these units.
• `["A"]` and `["B"]` are the concentrations of `"A"` and `"B"`, respectively, in `"M"`. Recall that `"M"` `=` `"mol/L"`.
• `m` is the order of reactant `"A"` and `n` is the order of reactant `"B"`. These contribute to the overall order of this reaction.

We already have all the information we need to solve this.

`"1.5 M"/"s" = k("1 M")^2("3 M")^1`

This bimolecular reaction is observed to be second order with respect to `"A"`, first order with respect to `"B"`, and third order overall.

Naturally, in the original rate law, since we are multiplying the rate constant of units `"???"` by the concentrations of overall units `"M"^3` to get a rate in `"M"/"s"`, we ought to use `1/("M"^2cdot"s")` for the rate constant to cancel out the units properly.

`color(blue)(k) = (("1.5 M")/"s")/(("1 M"^2)("3 M")) = ("1.5" cancel"M")/"s"xx1/("3 M"^(cancel(3)^(2)))`

`= color(blue)(0.5)` `color(blue)("M"^(-2)"s"^(-1))` or `color(blue)(1/("M"^2cdot"s"))`