Question

What fraction of a reactant showing first order remains after 40 minute if $t_{1/2}$ is 20 minute?

• A. 1/4
• B. 1/2
• C. 1/8
• D. 1/6

Answer 1

The correct option is A 1/4

Its a 1st Order Reaction,

Taking the Initial Concentration as 100 (For Convenience)

$t_{1/2}=\frac{0.693}{k}$

$k=0.03465{min}^{-1}$

Now, $k=\frac{2.303}{t}log\frac{a}{a-x}$

$0.03465=\frac{2.303}{40}log\frac{a}{a-x}$

$\frac{a}{a-x}=\frac{1}{4}$