Question

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

S. No	$[A]M$	$[B]M$	Rate of reaction$(M\,sec^{-1})$
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 $( R )= k [ A ][ B ]$