Question

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

• A. 3
• B. 1
• C. 2
• D. 0

Answer 1

Answer: (A)

$A-⟶B$

Rate $=k[A{]}^n$

${\text{ (Rate) }}_1=k(0.05{)}^n=2\times 10^{-3}...(i)$

${\text{ (Rate) }}_2=k(0.1{)}^n=1.6\times 10^{-2}...(ii)$

Dividing the Eq. (ii) by Eq. (i).

$\frac{{\text{ (Rate) }}_2}{{\text{ (Rate) }}_1}=\frac{k(0.1{)}^n}{k(0.05{)}^n}=\frac{1.6\times 10^{-2}}{2\times 10^{-3}}$

$(2{)}^n=8\text{ or }(2{)}^n=2^3$

$\therefore n=3$