Question

Calculate the time taken in seconds for 40% completion of a first order reaction, if its rate constant is $3.3\times {10}^{-4}{s}^{-1}.$
(Rounded off to the nearest integer)
Take:
$log\frac{5}{3}=0.22$

Answer 1

For a first order reaction,
$k=\frac{1}{t}ln\frac{\left[A_o\right]}{\left[A\right]}$

$t=\frac{1}{k}ln\frac{\left[A_o\right]}{\left[A\right]}=\frac{1\times 2.303}{3.33\times {10}^{-4}}log\frac{\left[100\right]}{\left[60\right]}=\frac{1\times 2.303}{3.33\times {10}^{-4}}log\frac{\left[5\right]}{\left[3\right]}=\frac{1\times 2.303}{3.33\times {10}^{-4}}\times 0.22$

$=30\times {10}^2\times 0.506=1518s$