Question

For a chemical reaction $A\to B,$ the rate of the reaction is $2\times {10}^{-3}$ mol $d{m}^{-3}{s}^{-1},$ when the initial concentration is 0.05 mol $d{m}^{-3}$ . The rate of the same reaction is $1.6\times {10}^{-2}mold{m}^{-3}{s}^{-1}$ when the initial concentration is $0.1mold{m}^{-3}$ . The order of the reaction is

• A. 2
• B. 0
• C. 3
• D. 1

Answer 1

Answer: (C)

$A\to B$ Rate $=k{[A]}^n$ ${(Rate)}_1=k{(0.05)}^n=2\times {10}^{-3}$ ${(Rate)}_2=k{(0.1)}^n=1.6\times {10}^{-2}$ Dividing the Eq. (ii) by Eq. (i) $\frac{{(Rate)}_2}{{(Rate)}_1}=\frac{k{(0.1)}^n}{k{(0.05)}^n}=\frac{1.6\times {10}^{-2}}{2\times {10}^{-3}}$ ${(2)}^n=8$ or ${(2)}^n=2^3$ $\therefore$ $n=3$