Question

The decomposition of a compound A follows a first order rate law. If it takes $15min$ for $20\%$ of the inital compound to decompose, then calculate the value of rate constant:

• A. $3.34mi{n}^{-1}$
• B. $2.12mi{n}^{-1}$
• C. $0.22mi{n}^{-1}$
• D. $0.014mi{n}^{-1}$

Answer 1

The correct option is D $0.014mi{n}^{-1}$

For a first order reaction:
$A\to \text{Products}$

$\left[A\right]=\left[A_0\right]e^{-kt}...\left(i\right)$

Here,
$\left[A\right]=\text{Concentration of A at time t}$
$\left[A_0\right]=\text{Concentration of A initially}$
$k=\text{Rate constant}$
Given:
$Att=15min\Rightarrow \left[A\right]=0.8\left[A_0\right]$

Putting the value in $\left(i\right)$

$0.8\left[A_0\right]=\left[A_0\right]e^{-k\times 15}$
$\ln \left(0.8\right)=-k\times 15k=\frac{0.22}{15}=0.014mi{n}^{-1}$