Question

In a $1st$ order reaction $A⟶product$, the concentration of the reactant decreases to $6.25$% of its initial value in $80minutes$. What is
(i) the rate constant and
(ii) the rate of the reaction. $100minutes$ after the start, if initial concentration is $0.2mole/litre$?

• A. $2.17\times {10}^{-2}{min}^{-1},3.47\times {10}^{-4}\cdot {litre}^{-1}{min}^{-1}$
• B. $3.465\times {10}^{-2}{min}^{-1},2.166\times {10}^{-4}\cdot {litre}^{-1}{min}^{-1}$
• C. $3.465\times {10}^{-3}{min}^{-1},2.17\times {10}^{-3}\cdot {litre}^{-1}{min}^{-1}$
• D. $2.166\times {10}^{-3}{min}^{-1},2.667\times {10}^{-4}\cdot {litre}^{-1}{min}^{-1}$

Answer 1

The correct option is B $3.465\times {10}^{-2}{min}^{-1},2.166\times {10}^{-4}\cdot {litre}^{-1}{min}^{-1}$

1st Order Reaction

$k=\frac{2.303}{t}log\frac{A}{A-x}$

$k=\frac{2.303}{80}log\frac{100}{6.25}$

$k=3.465\times {10}^{-2}{min}^{-1}$

$3.465\times {10}^{-2}=\frac{2.303}{100}log\frac{0.2}{A-x}$

$A-x=6.26\times {10}^{-3}$

$Rate=k\left[A\right]$

$Rate=6.26\times {10}^{-3}\times 3.465\times {10}^{-2}$

$Rate=2.166\times {10}^{-4}{lit}^{-1}{min}^{-1}$