Question

For reactions governed by the rate laws below, the concentration of reactant A is doubled while the concentration of reactant B is halved. Match the rate law to the change in rate of reaction that would be observed. (does the rate of the reaction double, decrease by half, quadruple, or is there no change?)

Rate = k[A][B]

Rate = k[A]^2[B]0

Rate = k[A][B^]2

Rate = k[A]^3[B]^2

Answer 1

Rate = k[A][B]

When the concentration of reactant A is doubled while the concentration of reactant B is halved, then

the new rate , r' = k[2A][B/2]

                       = k[A][B]

                       = r

So rate is no changed

Rate = k[A]²[B]⁰

When the concentration of reactant A is doubled while the concentration of reactant B is halved, then

the new rate , r' = k[2A]²[B/2]⁰

                       = 4k[A]²[B]⁰

                       = 4r

So rate becomes quadriple.

Rate = k[A][B]²

When the concentration of reactant A is doubled while the concentration of reactant B is halved, then

the new rate , r' = k[2A][B/2]²

                       = (1/2) k[A][B]²

                       = (1/2)x r

So rate decreases by half

Rate = k[A]³[B]²

When the concentration of reactant A is doubled while the concentration of reactant B is halved, then

the new rate , r' = k[2A]³[B/2]²

                       = 2x k[A]³[B]²

                       = 2 x r

So rate doubles