Question

A first-order reaction is \[75\% \] completed in 100 minutes. How long will it take for its \[87.5\% \] completion?
A. 125 min
B. 150 min
C. 175 min
D. 200 min

Answer 1

Hint: 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 \[75\% \] completed in 100 minutes the remain concentration of A should be,
 \[{\left[ A \right]_t} = \dfrac{{25}}{{100}}{\left[ A \right]_0}\] .
Therefore, the value of rate constant is,
 \[
  k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}} \\
  k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{\dfrac{{25}}{{100}}{{\left[ A \right]}_0}}} \\
  k = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{25}} \\
  k = \dfrac{{2.303}}{{100}}\log 4 \\
  k = \dfrac{{2.303}}{{100}}2\log 2 \\
  k = \dfrac{{2.303}}{{100}}2 \times 0.301 \\
  k = 0.0138 \\
 \]
Now when \[87.5\% \] completed the remaining reactant would be,
 \[{\left[ A \right]_t} = \dfrac{{12.5}}{{100}}{\left[ A \right]_0}\]
So the time required for this completion is,
\[
  k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}} \\
  k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{\dfrac{{25}}{{100}}{{\left[ A \right]}_0}}} \\
  k = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{25}} \\
  k = \dfrac{{2.303}}{{100}}\log 4 \\
  k = \dfrac{{2.303}}{{100}}2\log 2 \\
  k = \dfrac{{2.303}}{{100}}2 \times 0.301 \\
  k = 0.0138 \\
 \]
Therefore, the time required is 150 mm.

The correct answer is B.

Note:
For a reversible reaction at a situation when the amount of product is formed is 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 \[{R_b} = {k_b}{\left[ C \right]^2}\] respectively. Now, at equilibrium, the forward and backward reaction rates become the same. As a result,
  \[
  {R_f} = {R_b} \\
  or,{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\
  or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
  {k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
 \]