Question

The kinetic data for the given reaction \[A\left( g \right) + 2B\left( g \right)\xrightarrow{k}C\left( g \right)\] is provided in the following table for three experiments at 300 K

Ex. No.	[A/M]	[B/M]	Initial rate (M sec-1)
1.	0.001	0.001	\[\begin{array}{*{20}{l}} {6.930 \times {{10}^{ - 6}}} \end{array}\]
2.	0.02	0.01	\[\begin{array}{*{20}{l}} {1.386 \times {{10}^{ - 5}}} \end{array}\]

In another experiment starting with initial concentration of 0.5 and 1M respectively for A and B at 300 K. find the rate of reaction after 50 minutes from start of experiment (in M/sec):
A. \[6.93 \times {10^{ - 4}}\]
B. \[0.25 \times {10^{^{ - 7}}}\]
C. \[4.33 \times {10^{ - 5}}\]
D. \[3.46 \times {10^{ - 4}}\]

Answer 1

To solve this question, at first determine the order of the reaction. Then use the integrated rate equation of that particular order reaction to solve the second part of the question. For a reaction A \[ \to \] products. This is a first order reaction. The rate equation of first order reaction is \[r = k\left[ A \right]\] . Where, rate is r, rate constant is k and [A] is concentration of reactant A at a time t. The unit of rate depends upon the concentration of reactant and rate constant.
Formula used:
 1. \[r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }\] where, \[\alpha \] and \[\beta \] is the order with respect to A and B respectively.
2. \[k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}\]

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, according to the given data when A/M] and [B/M] are 0.001 and 0.001 respectively then rate of the reaction is, \[\begin{array}{*{20}{l}}
  {6.930 \times {{10}^{ - 6}}}
\end{array}\] . Now put these values in rate equation.
 \[
  r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }, \\
  or,\begin{array}{*{20}{l}}
  {6.930 \times {{10}^{ - 6}}}
\end{array} = k{\left[ {0.001} \right]^\alpha }{\left[ {0.001} \right]^\beta }.....(1) \\
 \]
when A/M] and [B/M] are 0.02 and 0.01 respectively then rate of the reaction is, \[\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 5}}}
\end{array}\] . Now put these values in rate equation.
 \[
  r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }, \\
  or,\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 4}}}
\end{array} = k{\left[ {0.02} \right]^\alpha }{\left[ {0.02} \right]^\beta }.....(2) \\
 \]
when A/M] and [B/M] are 0.02 and 0.02 respectively then rate of the reaction is, \[\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 5}}}
\end{array}\] . Now put these values in rate equation.
 \[
  r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }, \\
  or,\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 4}}}
\end{array} = k{\left[ {0.02} \right]^\alpha }{\left[ {0.02} \right]^\beta }.....(3) \\
 \]
Now, divide equation (2) by (3),
  \[
  \dfrac{{\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 4}}}
\end{array}}}{{\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 4}}}
\end{array}}} = \dfrac{{k{{\left[ {0.02} \right]}^\alpha }{{\left[ {0.01} \right]}^\beta }}}{{k{{\left[ {0.02} \right]}^\alpha }{{\left[ {0.02} \right]}^\beta }}} \\
  or,1 = \dfrac{{{{\left[ {0.01} \right]}^\beta }}}{{{{\left[ {0.02} \right]}^\beta }}} \\
  or,1 = {\left[ {\dfrac{1}{2}} \right]^\beta } \\
  or,{\left[ {\dfrac{1}{2}} \right]^0} = {\left[ {\dfrac{1}{2}} \right]^\beta } \\
  or,\beta = 0 \\
 \]
Put the value of \[\beta = 0\] in the equations and divide (2) by (1),
   \[
  \dfrac{{\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 4}}}
\end{array}}}{{\begin{array}{*{20}{l}}
  {\begin{array}{*{20}{l}}
  {6.930 \times {{10}^{ - 6}}}
\end{array}}
\end{array}}} = \dfrac{{k{{\left[ {0.02} \right]}^\alpha }{{\left[ {0.01} \right]}^0}}}{{k{{\left[ {0.001} \right]}^\alpha }{{\left[ {0.001} \right]}^0}}} \\
  or,20 = \dfrac{{{{\left[ {0.02} \right]}^\alpha }}}{{{{\left[ {0.001} \right]}^\alpha }}} \\
  or,20 = {20^\alpha } \\
  or,\alpha = 1 \\
 \]
So, the order and rate equation of this reaction is, \[r = k{\left[ A \right]^1}{\left[ B \right]^0}\] where, \[\alpha \] and \[\beta \] is the order with respect to A and B respectively. The order of this reaction is \[\alpha + \beta = 1\]
Now put the value of \[\beta = 0,\alpha = 1\] in equation (2) and find out the value of k,
 \[
  r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta } \\
  or,\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 5}}}
\end{array} = k{\left[ {0.02} \right]^1}{\left[ {0.01} \right]^0} \\
  or,\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 5}}}
\end{array} = k{\left[ {0.02} \right]^1} \times 1 \\
  or,\dfrac{{\begin{array}{*{20}{l}}
  {1.386 \times {{10}^{ - 5}}}
\end{array}}}{{\left[ {0.02} \right]}} = k \\
  or,k = 6.930 \times {10^{ - 4}} \\
 \]
 the integrated rate equation for 1st order reaction is
 \[k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}.......(4)\] , 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.
Now, Put the values and find out the concentration of A at time t.
 \[
  k = \dfrac{{2.303}}{t}\log \dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}} \\
  or,6.930 \times {10^{ - 4}} = \dfrac{{2.303}}{{50 \times 60}}\log \dfrac{{0.5}}{{{{\left[ A \right]}_t}}} \\
  or,\dfrac{{{\text{ }}6.930 \times {{10}^{ - 4}} \times 50 \times 60}}{{2.303}} = \log \dfrac{{0.5}}{{{{\left[ A \right]}_t}}} \\
  or,0.90273 = \log \dfrac{{0.5}}{{{{\left[ A \right]}_t}}} \\
  or,{10^{0.90273}} = \dfrac{{0.5}}{{{{\left[ A \right]}_t}}} \\
 \]
 \[
  or,{10^{0.90273}} = \dfrac{{0.5}}{{{{\left[ A \right]}_t}}} \\
  or,{\left[ A \right]_t} = \dfrac{{0.5}}{{{{10}^{0.90273}}}} \\
  or,{\left[ A \right]_t} = 0.06255 \\
 \]
Now after 50 mins the rate of the reaction would be,
 \[
  r = k{\left[ A \right]^\alpha }{\left[ B \right]^\beta }, \\
  or,\begin{array}{*{20}{l}}
  r
\end{array} = 6.930 \times {10^{ - 4}} \times {\left[ {0.06} \right]^1}{\left[ B \right]_t}^0 \\
  or,\begin{array}{*{20}{l}}
  r
\end{array} = 6.930 \times {10^{ - 4}} \times {\left[ {0.06} \right]^1} \\
  or,r = 4.33 \times {10^{ - 5}} \\
 \]

So, the rate of the reaction would be \[4.33 \times {10^5}M/\operatorname{Sec} .\]

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 become constant. For a reaction \[A + 2B \rightleftharpoons 2C\] therefore, the rates of forward 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,
  \[
  {R_f} = {\text{ }}{R_b} \\
  or,{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\
  or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
  {k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
 \]