Question

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

• A. 3
• B. 1
• C. 2
• D. 0

Answer 1

Answer: (A)

$A-\longrightarrow 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 $