Courses
Courses for Kids
Free study material
Offline Centres
More
Store

For a first order reaction, $A \to P$, ${t_{1/2}}$ (half-life) is $10\;days$. The time required for $\dfrac{1}{4}th$ conversion of $A$ (in days) is: $\left( {\ln \;2 = 0.693,\;\ln \;3 = 1.1} \right)$ A. $3.2$ B. $2.5$ C. $4.1$ D. $5$ Given:- The chemical equation for the reaction is : $A \to P$- The order of the reaction : $n = 1$- The half-life for the reaction is: ${t_{1/2}} = 10\;days$- Fraction of converted reactant: $\dfrac{1}{4}th$- Some useful values: $\left( {\ln \;2 = 0.693,\;\ln \;3 = 1.1} \right)$

Last updated date: 12th Aug 2024
Total views: 420.6k
Views today: 12.20k
Verified
420.6k+ views
Hint:
We can use the relationship between the half-life period and the rate constant for first order reactions to determine the latter which can be further used along with rate law expression to determine the required time.

Complete step by step solution:
We know that the half-life period can be written in terms of rate constant of a reaction if we know the order of the reaction. Here, our reaction is of first order. So, we can write the expression relating the half-life period, $\left( {{t_{1/2}}} \right)$ to the rate constant as shown below:
$k = {\rm{ }}\dfrac{{{\rm{ln 2}}}}{{{t_{1/2}}}}$

We can find out the rate constant for the given reaction by substituting the given value of half life period and the constant into the above expression as follows:
$k = {\rm{ }}\dfrac{{{\rm{0}}{\rm{.693}}}}{{{\rm{10}}\;{\rm{days}}}}\\ = 0.0693\;{\rm{da}}{{\rm{y}}^{ - 1}}$

Now we will write the rate law for the given reaction as follows:
$- \dfrac{{d\left[ {\rm{A}} \right]}}{{dt}} = k\left[ {\rm{A}} \right]\\ \dfrac{{d\left[ {\rm{A}} \right]}}{{\left[ {\rm{A}} \right]}} = - kdt$

We will integrate both the sides as follows:
$\int_{{{\left[ {\rm{A}} \right]}_0}}^{{{\left[ {\rm{A}} \right]}_t}} {\dfrac{{d\left[ {\rm{A}} \right]}}{{\left[ {\rm{A}} \right]}}} = - \int_0^t {kdt}$
We can write the integrated rate law for the given reaction as follows:
$\ln \left( {\dfrac{{{{\left[ {\rm{A}} \right]}_t}}}{{{{\left[ {\rm{A}} \right]}_0}}}} \right) = - kt$
We can rearrange this for time as follows:
$t = \dfrac{{\ln \left( {\dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{{{\left[ {\rm{A}} \right]}_t}}}} \right)}}{k}$
Now, as we know that $\dfrac{1}{4}th$ of the reactant has to be converted leaving $\dfrac{3}{4}th$ of the reactant. So we can write:
$\dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{{{\left[ {\rm{A}} \right]}_t}}} = \dfrac{{{{\left[ {\rm{A}} \right]}_0}}}{{\left( {\dfrac{3}{4}} \right){{\left[ {\rm{A}} \right]}_0}}}\\ = \dfrac{4}{3}$

Finally, we can calculate the required time by using the above ratio and values as follows:
$t = \dfrac{{\ln \left( {\dfrac{4}{3}} \right)}}{{0.0693\;{\rm{da}}{{\rm{y}}^{ - 1}}}}\\ = \dfrac{{\left( {2 \times \ln \;2} \right) - \ln \;3}}{{0.0693\;{\rm{da}}{{\rm{y}}^{ - 1}}}}\\ = \dfrac{{\left( {2 \times 0.693} \right) - 1.1}}{{0.0693\;{\rm{da}}{{\rm{y}}^{ - 1}}}}\\ = {\rm{4}}{\rm{.1}}\;{\rm{days}}$

Hence, the time required is ${\rm{4}}{\rm{.1}}\;{\rm{days}}$ which makes option C to be the correct option.

Note:

Rate law expression and the relationship between the half-life period and rate constant is different for different order of reactions.