Question

The reaction A$+$B$\to$products is first order in each of the reactants. What is the change in the rate of reaction if the concentration of A is halved and concentration of B is doubled?

Answer 1

Since, the reaction is first order in each reactant, the rate law expression is rate $=k\left[A\right]\left[B\right]$.....( 1)

The concentration of A is halved and concentration of B is doubled.

$[A{]}^{\prime }=0.5[A]$, $[B{]}^{\prime }=2[B]$.....( 2)

rate' $=k\left[A{]}^{\prime }\right[B{]}^{\prime }$.....( 3)

Substitute equation (2) in equation (3)

rate' $=k\left(0.5\right[A\left]\right)\left(2\right[B\left]\right)$

rate' $=k\left[A\right]\left[B\right]$.....( 4)

Divide equation (4) with equation (1)

$\frac{{\text{rate}}^{\prime }}{\text{rate}}=\frac{k\left[A\right]\left[B\right]}{k\left[A\right]\left[B\right]}=1$
Hence, the rate will remain unchanged.