Question

$ln$ a first order reaction the concentration of reactant decreases from 800 $mol/d{m}^3$ to 50 $mol/d{m}^3$ is $2\times {10}^4$ sec. The rate constant of reaction in ${\sec }^{-1}$ is:

• A. $2\times {10}^4$
• B. $3.45\times {10}^{-5}$
• C. $1.386\times {10}^{-4}$
• D. $2\times {10}^{-4}$

Answer 1

The correct option is C $1.386\times {10}^{-4}$

Assuming decay is happening in the above example, we can say it will take 4 half-lives in $2\times {10}^4$ sec.

So, each half-life will take $5\times {10}^3$ sec.

Now using, $t_{1/2}=\frac{0.693}{k}$ and putting values we get

$k=\frac{0.693}{5\times {10}^3}$

$k=1.386\times {10}^{-4}$

Option C is correct.