Question

The rate constant for a first order reaction is $60s^{-1}$. How much time it will take to reduce the initial concentration of the reactant to its 1 / 10 value?

• A. $4.2\times {10}^{-2}s$
• B. $3.38\times {10}^{-2}s$
• C. $2.5\times {10}^{-2}s$
• D. None of these

Answer 1

The correct option is B $3.38\times {10}^{-2}s$

For the first-order reaction we have:-

$K=\frac{1}{t}ln\left(\frac{a}{a-x}\right)$

where, $K$= rate constant

$t$= time

$a$= initial concentration of the reactant

$x$= amount of reactant reacted in time $t$

Now, as given in question

$K=60s^{-1}$, $t=?$, $a-x=\frac{1}{10}\times a$

$\Rightarrow$ $t=\frac{1}{60}ln10=3.38\times {10}^{-2}s$