Question

The reaction A+2B →products , was found to have the rate law , rate= k[A][B]². Predict by what factor the rate of reaction will increase when the concentration of A and B is doubled?

Answer 1

By a factor of `8`.

Explanation:

For your reaction

`"A" + 2"B" -> "products"`

the rate law takes the form

`"rate" = k * ["A"] * ["B"]^2`

As you can see, the rate is proportional to the concentration of `"A"` and the square of the concentration of `"B"`.

Let's say that initially, you have `["A"] = x` and `["B"] = y`. The rate law in this case is equal to

`"rate 1" = k * x * y^2`

Now you double the concentrations of `"A"` and `"B"` to new values of `2x` and `2y`. The new rate law will be

`"rate 2" = k * (2x) * (2y)^2`

`"rate 2" = k * 2x * 4y^2 = 8 * x * y^2`

As you can see, you have

`"rate 2"/"rate 1" = (color(red)(cancel(color(black)(k))) * 8 * color(red)(cancel(color(black)(x * y^2))))/(color(red)(cancel(color(black)(k))) * color(red)(cancel(color(black)(x * y^2)))) = color(green)(8)`

Therefore, the rate of the reaction has increased by a factor of `8`.