Question

The value of rate of reaction for a first order reaction is $2.303 \times {10^{ - 2}}\;{{se}}{{{c}}^{ - 1}}$. What will be the time required to reduce the concentration to $\dfrac{1}{{10}}th$ of its initial concentration.
A. 100 s
B. 10 s
C. 2303 s
D. 230.2 s

Answer 1

Rate of reaction is directly proportional to the product of concentration of reactants, each concentration term raised to the power equal to stoichiometric coefficients.
$aA + bB \to Products$
Then,
$Rate \propto {\left[ A \right]^a}{\left[ B \right]^b}$

Complete answer:
Here, we have given that the reaction is of first order. Also, we know that in first order reaction, the rate of reaction only depends upon one concentration term.
For the reaction,
$A \to Products$
Then $Rate = k\left[ A \right]$
Where $k$ is a constant, known as rate constant or velocity constant.
Now, the integrated rate expression for the above reaction can be written as
$k = \dfrac{{2.303}}{t}log\dfrac{a}{{a - x}}$
Where $k$ is rate constant
$t$ is reaction time
$a$ is initial concentration of reactant
And $x$ is concentration of reactant decomposed in time $t$
In question, we have to find the time taken to reduce the concentration to $\dfrac{1}{{10}}th$ of its initial concentration. That means suppose $a$ is initial concentration taken here.
Then, the final concentration ($a - x)$ will be $\dfrac{1}{{10}}th$ of $a$
i.e., $a - x = \dfrac{a}{{10}}$
So we have the integrated rate equation, where rate constant is given as $2.303 \times {10^{ - 2}}\;{\text{se}}{{\text{c}}^{ - 1}}$. And to find time, t we rewrite the equation as
$\Rightarrow t = \dfrac{{2.303}}{k}{\text{log}}\left( {\dfrac{a}{{a - x}}} \right)$
Substituting the value of $k$
$\Rightarrow t = \dfrac{{2.303}}{{2.302 \times {{10}^{ - 2}}}}{\text{log}}\left( {\dfrac{{\text{a}}}{{{\text{a}}/10}}} \right)$
Now, after cancellation,
$t = 100\,\log\left( {10} \right)$
We know that value of $\log\left( {10} \right) = 1$
Therefore, time ($t$) = 100s

**The time required to reduce the concentration to $\dfrac{1}{{10}}th$ of its initial concentration is 100 seconds.

Note:**
We should not be confused with reducing the concentration to $\dfrac{1}{{10}}th$ of its initial concentration, this only means the final concentration. Most of the time we end up reducing $\dfrac{1}{{10}}$ from the initial concentration. We should read this as only the final concentration.