Question

The time taken for $90\%$ of a first-order reaction to complete is approximate:
A. 1.1 times that of half-life
B. 2.2 times that of half-life
C. 3.3 times that of half-life
D. 4.4 times that of half-life

Answer 1

For a reaction $A \to$ products. This is a first-order reaction. The rate equation of the first-order reaction is $r = k\left[ A \right]$ . Where, the rate is r, the rate constant is k, and [A] is the concentration of reactant A at a time t. the unit of the rate depends upon the concentration of reactant and rate constant.
Formula used: $k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$

Complete step by step answer:
Let, the rate law of this equation, $A\left( g \right) + 2B\left( g \right)\xrightarrow{k}C\left( g \right)$ is $r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }$ where, $\alpha$ and $\beta$ is the order with respect to A and B respectively.
Now for the first-order reaction, the rate law is $r = k\left[ A \right]$ .
The integrated rate equation for a 1st order reaction is
$k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$ , where k is rate constant, t is time, and ${\left[ A \right]_0}$ , ${\left[ A \right]_t}$ are concentrations of A at initial and at time t respectively.
When is $90\%$ completed in 100 minutes the remain concentration of A should be,

{\left[ A \right]_t} = \dfrac{{100 - 90}}{{100}}{\left[ A \right]_0} \\\ {\left[ A \right]_t} = \dfrac{{10}}{{100}}{\left[ A \right]_0} \\\ {\left[ A \right]_t} = \dfrac{1}{{10}}{\left[ A \right]_0} \\\ $$ . The equation of half-life of the first-order reaction is $k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$ Now this value in the rate equation to find out the time required for $$90\% $$ completion of the first-order reaction as follows,

t = \dfrac{{2.303}}{k}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}t}}} \\
t = \dfrac{{2.303}}{{\dfrac{{0.693}}{{{t{\dfrac{1}{2}}}}}}}\log \dfrac{{{{\left[ A \right]}0}}}{{\dfrac{1}{{10}}{{\left[ A \right]}0}}} \\
t = 3.323 \times {t{\dfrac{1}{2}}}\log 10 \\
t = 3.323 \times {t{\dfrac{1}{2}}} \\

So, the time taken for $$90\% $$ of a first-order reaction to complete is approximately 3.32. **Therefore, the correct option is, C** **Note:** For a reversible reaction at a situation when the amount of product is formed equal to the amount of reactant is formed then it is called equilibrium. At equilibrium the amount of product and reactant becomes constant. For a reaction $$A + 2B \rightleftharpoons 2C$$ , therefore, the rates of forwarding and backward reactions are, $${R_f} = {k_f}\left[ A \right]{\left[ B \right]^2}$$ and $${\text{ }}{R_b} = {k_b}{\left[ C \right]^2}$$ respectively. Now, at equilibrium, the forward and backward reaction rates become the same. As a result,

{R_f} = {R_b} \\
or,{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\
or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
{k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\

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