Question

A reaction is first order in $A$ and second order in $B$:
(i) Write differential rate equation
(ii) How is the rate affected when the concentration of $B$ is tripled?
(ii) How is the rate affected when the concentration of both $A$ and $B$ is doubled?

Answer 1

Reaction rate, in chemistry the speed at which chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of reactant that is consumed in a unit of time.

Complete step by step answer:
It is given that a reaction is first order in $A$ and second order in $B$.
(i) The differential equation is as follows:
${\left( {{\text{Rate}}} \right)_I} = K\left[ A \right]{\left[ B \right]^2}$
${\text{rate = }}\dfrac{{ - d\left[ A \right]}}{{dt}} = \dfrac{{ - 1d\left[ B \right]}}{{2dt}}$

(ii) On increasing concentration of $B$three times:
${\left( {{\text{Rate}}} \right)_{II}} = K\left[ A \right]{\left[ {3B} \right]^2}$
$= K\left[ A \right]{3^2}{\left[ B \right]^2}$
${\left( {{\text{Rate}}} \right)_{II}} = 9K\left[ A \right]{\left[ B \right]^2} = 9{\left( {{\text{Rate}}} \right)_I}$
The rate becomes $9$times.

(iii) When the rate of both $A$ and $B$ are doubled
${\left( {{\text{Rate}}} \right)_{III}} = K\left[ {2A} \right]{\left[ {2B} \right]^2}$
$= K\left[ A \right]8{\left[ B \right]^2}$
$= 8K\left[ A \right]{\left[ B \right]^2}$
${\left( {{\text{Rate}}} \right)_{III}} = 8{\left( {{\text{Rate}}} \right)_I}$

Additional information:
Factors affecting rate of reaction:
(i) Reactant concentration increasing concentration of one or more reactants will increase the rate of reaction.
(ii) Physical state of reactants and surface area
(iii) Temperature
(iv) Presence of a catalyst

Note:
The units of $K$ can be calculated by the formula:
${M^{1 - N}}{S^{ - 1}}$
$N \to$ overall order of reaction.