Question

The reaction, \[\ce{A + B -> Products}\], follows the rate law Rate = k[A][B]².

What will be the effect on the rate if the concentration of A is doubled and that of B is halved?

Answer 1

The given rate law is

Rate = k[A][B]²

Let the initial concentrations be [A] = a, [B] = b

So,

Initial Rate = k⋅a⋅b²

Now, if:

[A] is doubled → becomes 2a

[B] is halved → becomes `b/2`

Then,

New rate = `k * (2a) * (b/2)^2`

= `k * 2a * b^2/4`

= `1/2 * (k * a * b^2)`

∴ The rate becomes half of the original rate.