Question

The rate constant of a first order reaction is $6.93\times {10}^{-3}mi{n}^{-1}.$ If we start with 10 mol/L, it is reduced to 1.25 mol/L in:

• A. 100 minutes
• B. 200 minutes
• C. 30 minutes
• D. 300 minutes

Answer 1

The correct option is D 300 minutes

Given $K=6.93\times {10}^{-3}mi{n}^{-1}$
$[A{]}_0=10M,(A{)}_t=1.25M$
Hence, for a $1^{st}$ order reaction we can write
$t=\frac{1}{K}ln\frac{[A{]}_0}{[A{]}_t}$
$t=\frac{1}{6.93\times {10}^{-3}}ln\left(\frac{10}{1.25}\right)$
$\therefore t=\frac{{10}^3}{6.93}\times ln8=\frac{3\times {10}^{3}ln2}{6.93}$
\(\Rightarrow t = \frac{3 \times 0.693 \times 10^3}{6.93}

= \frac{3 \times 10^3}{1000} \times \frac{693}{693} \times 100\)

$\therefore t=300$ minutes