Question

Half-life period of a first-order reaction is $10$ minutes. Starting with $10$ ${\text{mol L}}^{-1}$, rate after $20$ minutes will be :

• A. $0.0693\text{mole L}{\text{min}}^{-1}$
• B. $0.0693\times 2.5{\text{mol L}}^{-1}{\text{min}}^{-1}$
• C. $0.693\times 5{\text{mol L}}^{-1}{\text{min}}^{-1}$
• D. $0.693\times 10{\text{mol L}}^{-1}{\text{min}}^{-1}$

Answer 1

The correct option is B $0.0693\times 2.5{\text{mol L}}^{-1}{\text{min}}^{-1}$

Given, half life of the first order reaction is $10$ minutes.
So, basically $20$ minutes is the second half life of the reaction.
So, concentration of the reactant after $20$ minutes
$={\left(\frac{1}{2}\right)}^2\times$ (Initial concentration)
$={\left(\frac{1}{2}\right)}^2\times 10=2.5M$
Also, for a first order reaction,
$t_{\frac{1}{2}}=\frac{0.693}{k}$
Where, $k=\text{Rate constant}$
$\therefore k=\frac{0.693}{t_{\frac{1}{2}}}$
Then, rate of the reaction $=k\left[a_t\right]=\frac{0.693}{10}\times 2.5{\text{mol L}}^{-1}{\text{min}}^{-1}$
$=0.0693\times 2.5{\text{mol L}}^{-1}{\text{min}}^{-1}$