Question

For a first order reaction with half – life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

1. 1500s
2. 500s
3. 900s
4. 600s

Answer 1

Half – life of any reaction is defined as the time in which the concentration of the reactants gets reduced to half of their initial concentrations. Half life of a first order reaction is inversely proportional to the rate constant, but is independent of initial concentration of reactants.
Formula used:
Half life for first order reaction, ${{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{K}$ , where K is rate constant.
Integrated form of first order reaction, $-K=\dfrac{2.303}{t}\log \dfrac{[{{R}_{0}}]}{[R]}$ , where $[{{R}_{0}}]$ is initial concentration of reactants, and $[R]$is final concentration.

Complete answer:
We have been given a first order reaction with half life, 150 seconds, from this data we will first find the rate constant value of the reaction.
As we know, half life of first order reaction, ${{t}_{{}^{1}/{}_{2}}}=\dfrac{0.693}{K}$ , where K is rate constant. So, rearranging we have, $K=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}$, where ${{t}_{{}^{1}/{}_{2}}}=150s$, so rate constant will be,
$K=\dfrac{0.693}{150}$
Now, we know for the first order reaction, the integrated form of the rate law is, $-K=\dfrac{2.303}{t}\log \dfrac{[{{R}_{0}}]}{[R]}$ , where $[{{R}_{0}}]$ is initial concentration of reactants, and $[R]$is final concentration. We have been given that the concentration fall from M/10 to M/100, so these are the initial and the final concentrations respectively. Rearranging this equation for rate law, for finding time, and putting the given values we have,
$t=\dfrac{2.303}{K}\log \dfrac{[{}^{M}/{}_{10}]}{[{}^{M}/{}_{100}]}$, putting the value of rate constant K as, $K=\dfrac{0.693}{150}$, we have
$t=\dfrac{2.303}{\dfrac{0.693}{150}}\log (10)$
$t=\dfrac{2.303\times 150\times 1}{0.693}$
t = 493.4 s
Therefore, by rounding off, it will be $t\simeq 500s$.
Hence, time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately 500 sec, so option 2 is correct.

Note:
For a first order reaction the concentration of reacting species is raised to the power of one. The rate law of the first order reaction is integrated to give a relationship between rate constant, time and initial and final concentrations. It also tells the values for graphical form through the slope- intercept form which is y = mx + b.