Question

A and B react as per the reaction
$A\left(g\right)+2B\left(g\right)\to C\left(g\right)$
From the data given below identify the rate law.

S. No	$[A]M$	$[B]M$	Rate of reaction$(Mse{c}^{-1})$
1	$0.3$	$0.4$	$2\times (10{)}^{-3}$
2	$0.6$	$0.8$	$8\times (10{)}^{-3}$
3	$0.6$	$0.4$	$4\times (10{)}^{-3}$

• A. $R=k{\left[A\right]}^2{\left[B\right]}^1$
• B. $R=k\left[A\right]\left[B\right]$
• C. $R=k{\left[A\right]}^1{\left[B\right]}^0$
• D. $R=k\left[A\right]{\left[B\right]}^{-1}$

Answer 1

Answer: (B)

Rate $R=k[A{]}^m[B{]}^n$
From Exp. (i) $R_1=2\times 10^{-3}=k[0.3{]}^m[0.4{]}^n$
From Exp. (ii) $R_2=8\times 10^{-3}=k[0.6{]}^m[0.8{]}^n$
From Exp. (iii) $R_3=4\times 10^{-3}=k[0.6{]}^m[0.4{]}^n$
$\frac{R_3}{R_1}\Rightarrow \frac{4\times 10^{-3}}{2\times 10^{-3}}=\frac{k[0.6{]}^m[0.4{]}^n}{k[0.3{]}^m[0.4{]}^n}$
$2^{\prime }=2^m\therefore m=1$
$\frac{R_2}{R_3}\Rightarrow \frac{8\times 10^{-3}}{4\times 10^{-3}}=\frac{k[0.6{]}^m[0.8{]}^n}{k[0.6{]}^m[0.4{]}^n}$
$2=2^n\therefore n=1$
$\therefore$ Rate law $(R)=k[A][B]$