Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first order reaction, the rate constant $k$ is given by the expression:

$k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$

Given:
Initial amount, $a=20$ millimole
Amount after time $t$, $a-x=10$ millimole
Time, $t=0.3010$ minute

Substituting the given values into the equation:

$k=\frac{2.303}{0.3010}\log \left(\frac{20}{10}\right)$

$k=\frac{2.303}{0.3010}\log 2$

Using the standard value $\log 2=0.3010$:

$k=\frac{2.303}{0.3010}\times 0.3010$

$k=2.303{\text{minute}}^{-1}$

Answer: $2.303{\text{minute}}^{-1}$