Question

For a reaction whose mechanism is given below with initial rate `"0.0030 M/s"`, `(a)` what is the new rate if `["A"]` is doubled, `(b)` if `["B"]` is doubled, and `(c)` if both `["A"]` and `["B"]` are multiplied by `5`?

Overall reaction:

`A + C -> D`

Mechanism:

`A => B` (`"slow"`)
`B + C => D` (`"fast"`)

Answer 1

Well, we know that the rate-limiting step is unimolecular with `A` as the only reactant.

Single mechanistic steps are elementary reactions, for which the stoichiometric coefficients give the order with respect to the reactants involved.

If such a mechanistic step is slow, then it gives you the rate law directly. This rate law has already incorporated the behavior of `B`.

So, the rate law can be written as:

`r(t) = k[A]`

`a)`

Thus, if `[A]` doubles, `r(t)` doubles...

`color(red)(2) xx r(t) = k xx color(red)(2) xx [A]`

`color(blue)(r(t)' = 2r(t) = "0.0060 M/s")`

`b)`

If `[B]` is doubled, we know the rate law has no `B` in it, i.e. `B` is in some mechanistic step that is fast, and is the overall reaction is effectively zero order with respect to `B`.

Hence, doubling `[B]` does not do anything, and...

`color(blue)(r(t)' ~~ r(t) = "0.0030 M/s")`

given the original concentrations.

`c)`

Again, as in `(b)`, changing `[B]` does nothing, so focus on `[A]` and rehash what you did in `(a)`.

`color(red)(5) xx r(t) = k xx color(red)(5) xx [A]cancel([B]^(0))^(1)`

Evidently the reaction rate then multiplies by `bb5` to maintain mathematical equivalence. Thus:

`color(blue)(r(t)' = 5r(t) = "0.0150 M/s")`