Question

The half-life of first order reaction is $1.5$ hours. How much time is needed for 94% of the reactant to change to product?

Answer 1

In the above question, the half life of first order reaction is given and we are asked about the time needed for 94% of the reactant to change to product. We can find the rate constant from the first order half life equation and put this value of rate constant in concentration after time t equation to get the desired time.

Formula Used-
${{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
where ${{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ = }}$ half-life time.
k= rate constant
${{\text{[A]}}_{\text{t}}}{\text{ = [A}}{{\text{]}}_{\text{0}}}{\text{.}}{{\text{e}}^{{\text{ - kt}}}}$
Where ${{\text{[A]}}_{\text{t}}}$= concentration of reactant after time t.
${{\text{[A]}}_{\text{0}}}$=initial concentration
k= rate constant.

Complete step-by-step answer:
We know that half life is the time required for a substance to reduce to half of its concentration. In a first order reaction, the half life of the element is independent of the concentration of the element and is given by:
${{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.693}}}}{{\text{k}}}$
Rearranging it, we get:
${\text{k = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}}}$
Substituting the value, we get:
${\text{k = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}}}{\text{ = }}\dfrac{{{\text{0}}{\text{.693}}}}{{{\text{1}}{\text{.50}}}}{\text{ = 0}}{\text{.462}}{{\text{h}}^{{\text{ - 1}}}}$
The second question asked is when the reactant will reduce it to 94%.
So, let at time t, reactant concentration (${{\text{[A]}}_{\text{t}}}$) be $\left( {{\text{1 - }}\dfrac{{{\text{94}}}}{{{\text{100}}}}} \right){{\text{[A]}}_{\text{0}}}{\text{ = }}\dfrac{{{\text{6[A}}{{\text{]}}_{\text{0}}}}}{{{\text{100}}}}$
We know that:
${{\text{[A]}}_{\text{t}}}{\text{ = [A}}{{\text{]}}_{\text{0}}}{\text{.}}{{\text{e}}^{{\text{ - kt}}}}$
Substituting the value of ${{\text{[A]}}_{\text{t}}}$we get:
$\dfrac{{{\text{6[A}}{{\text{]}}_{\text{0}}}}}{{{\text{100}}}}{\text{ = [A}}{{\text{]}}_{\text{0}}}{\text{.}}{{\text{e}}^{{\text{ - kt}}}}$
Or, $\dfrac{{\text{6}}}{{{\text{100}}}}{\text{ = }}{{\text{e}}^{{\text{ - kt}}}}$
Taking ln on both the side of the equation, we get:
${\text{ln}}\left( {\dfrac{{\text{6}}}{{{\text{100}}}}} \right){\text{ = ln(}}{{\text{e}}^{{\text{ - kt}}}}{\text{)}}$
Simplifying:
${\text{ - kt = ln(6) - ln(100)}}$
Substituting the value of k, we get:
$- 0.462t = - 2.813$
${\text{t = }}\dfrac{{{\text{ - 2}}{\text{.813}}}}{{{\text{ - 0}}{\text{.462}}}}{\text{ = 6}}{\text{.08}}$ hr
Hence,${\text{6}}{\text{.08}}$ hour is needed for 94% of the reactant to change to product.

Note: The term half life is commonly used in nuclear physics to describe how quickly an unstable atom undergoes radioactive decay or how long stable atoms survive.