Answer

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**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:

Note:

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

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