Question

$A$ follows the first-order reaction.

$\left(A\right)$ $⟶$ product.

The concentration of $A$ changes from 0.1 $M$ to 0.025 $M$ in 40 minutes. Find the rate of reaction of $A$ when concentration of $A$ is 0.01 $M$?

• A. $3.47\times {10}^{-4}M$ ${min}^{-1}$
• B. $3.47\times {10}^{-5}M$ ${min}^{-1}$
• C. $1.73\times {10}^{-4}M$ ${min}^{-1}$
• D. $1.73\times {10}^{-3}M$ ${min}^{-1}$

Answer 1

The correct option is A $3.47\times {10}^{-4}M$ ${min}^{-1}$

As given, concentration changes from $0.01M$ to $0.025M$ in $40$ minutes.

As we know,

$t_{1/2}$ is the time when concentration becomes half of the original.

$\Rightarrow 2t_{1/2}=40min$

$t_{1/2}=20min$

$K=\frac{0.693}{t_{\frac{1}{2}}}$

$k=\frac{0.693}{20}$

For first order reaction,

$r=K\left[A\right]=\frac{0.693}{20}\times 0.01=3.47\times {10}^{-4}Mmi{n}^{-1}$