Question

In a reaction, \[\ce{A + B -> Product}\], rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as ______.

Options

• Rate = k[A][B]²

• Rate = k[A]²[B]²

• Rate = k[A][B]

• Rate = k[A]²[B]

Answer 1

In a reaction, \[\ce{A + B -> Product}\], rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as Rate = k[A]²[B].

Explanation:

The given reaction is 

\[\ce{A + B -> Product}\]

When [B] is doubled, rate is doubled

⇒ Order w.r.t B = 1

When [A] and [B] are both doubled, rate increases 8 times.

We already know doubling [B] alone doubles the rate, so the additional increase must be due to [A]:

Total rate increase = 8

From B alone = 2

From A alone = `8/2` = 4

So, 2n = 4

n = `4/2`

n = 2

⇒ Order w.r.t. A = 2