Question

What is the order of a chemical reaction $A+2B\stackrel{k}{\to }C$, if the rate of formation of $C$ increases by a factor of 2.82 on doubling the concentration of $A$ and increases by a factor of 9 on tripling the concentration of $B$?

• A. 7/2
• B. 7/4
• C. 5/2
• D. 5/4

Answer 1

The correct option is A 7/2

The rate law expression is $R=K\left[A{]}^m\right[B{]}^n$ ......(1)

The rate of formation of $C$ increases by a factor of 2.82 on doubling the concentration of $A$.

The rate law expression becomes $R^{\prime }=2.82R=K{2}^m\left[A{]}^m\right[B{]}^n$ ......(2)

Divide equation (2) with equation (1)

$\frac{2.82R}{R}=\frac{K{2}^m\left[A{]}^m\right[B{]}^n}{K\left[A{]}^m\right[B{]}^n}$

Hence,

$2.82=2^m$ or $m=\frac{3}{2}$

The rate of formation of $C$ increases by a factor of 9 on tripling the concentration of $B$

The rate law expression becomes $R"=9R=K\left[A{]}^m{3}^n\right[B{]}^n$ ......(3)

Divide equation (3) with equation (1)

$\frac{9R}{R}=\frac{K\left[A{]}^m{3}^n\right[B{]}^n}{K\left[A{]}^m\right[B{]}^n}$

Hence,

$9=3^n$ or $n=2$

The overall order of the reaction is $m+n=\frac{3}{2}+2=\frac{7}{2}$