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The reaction \[cis - Cr{\left( {en} \right)_2}\left( {OH} \right)_2^ + \rightleftharpoons {\text{ }}trans - Cr{\left( {en} \right)_2}{\left( {OH} \right)_2}^ + \] is first order in both directions. At \[25^\circ C\] the equilibrium constant is 0.16 and the rate constant \[{k_1}\] is \[3.3{\text{ }}x{\text{ }}{10^{ - 4}}{\text{ }}{s^{ - 1}}\]. In an experiment starting with the pure cis form, how long would it take for half the equilibrium amount of the trans isomer to be formed?
A. 4.83 min
B. 5.66 min
C. 7.2 min.
D. 9 min

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Last updated date: 17th Jun 2024
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Answer
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Hint: 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 the first-order reaction is \[r = {k_1}\left[ A \right]\]. Where the rate is r, the rate constant is ${k_1}$ and \[\left[ A \right]\] 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:
 \[[B] = \dfrac{{{{[B]}_{eq}}}}{2}. = \dfrac{{{k_1}{{[A]}_0}}}{{2({k_{1}} + {k_2})}}\] , \[{[B]_{eq}} = \dfrac{{{k_1}{{[A]}_{eq}}}}{{{k_2}}} = \dfrac{{{k_1}{{[A]}_0}}}{{{k_{1}} + {k_2}}}\] \[[B] = \dfrac{{{k_1}{{[A]}_0}}}{{{k_{1}} + {k_2}}}\;[1 - {e^{ - ({k_{1}} + {k_2})t}}]\] where \[{k_{1}}\] and \[{k_2}\] are the rate constant of forwarding and backward reaction.

Complete step by step answer:
At equilibrium, the forward and backward reaction rates become the same. As a result, the equilibrium constant can be written as the ratio of product side concentration to reactant side concentration.
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.
According to this the forward rate equation for the reaction, \[cis - Cr{\left( {en} \right)_2}\left( {OH} \right)_2^ + \rightleftharpoons {\text{ }}trans - Cr{\left( {en} \right)_2}{\left( {OH} \right)_2}^ + \]
 \[{r_f} = {k_1}\left[ {cis - Cr{{\left( {en} \right)}_2}\left( {OH} \right)_2^ + } \right]\] and the backward rate is \[{r_b} = {k_2}\left[ {trans - Cr{{\left( {en} \right)}_2}\left( {OH} \right)_2^ + } \right]\]
Now, at equilibrium,
  \[
  \dfrac{{{{[B]}_{eq}}}}{{{{[A]}_{eq}}}} = \dfrac{{{k_1}}}{{{k_2}}} \\
  {[B]_{eq}} = \dfrac{{{k_1}{{[A]}_{eq}}}}{{{k_2}}} = \dfrac{{{k_1}{{[A]}_0}}}{{{k_{1}} + {k_2}}} \\
 \]
Where A and B are, \[cis - Cr{\left( {en} \right)_2}\left( {OH} \right)_2^ + \] and \[trans - Cr{\left( {en} \right)_2}{\left( {OH} \right)_2}^ + \] respectively.
Now, according to the question when the concentration of B is half of the equilibrium concentration, then
  \[[B] = \dfrac{{{{[B]}_{eq}}}}{2}.\]
Now, put the value of \[{[B]_{eq}} = \dfrac{{{k_1}{{[A]}_0}}}{{({k_{1}} + {k_2})}}\] in this equation we get,
 \[[B] = \dfrac{{{{[B]}_{eq}}}}{2}. = \dfrac{{{k_1}{{[A]}_0}}}{{2({k_{1}} + {k_2})}}\].
Now, for reversible reaction the amount of B is
 \[[B] = \dfrac{{{k_1}{{[A]}_0}}}{{{k_{1}} + {k_2}}}\;[1 - {e^{ - ({k_{1}} + {k_2})t}}]\]
Rearrange this equation as follows,
 \[
  \dfrac{{[B]}}{{\dfrac{{{k_1}{{[A]}_0}}}{{{k_{1}} + {k_2}}}}} = \;[1 - {e^{ - ({k_{1}} + {k_2})t}}] \\
  \dfrac{1}{2} = \;[1 - {e^{ - ({k_{1}} + {k_2})t}}] \\
  \dfrac{1}{2} - 1 = \;[ - {e^{ - ({k_{1}} + {k_2})t}}] \\
   - \dfrac{1}{2} = \;[ - {e^{ - ({k_{1}} + {k_2})t}}] \\
  2 = \;[{e^{({k_{1}} + {k_2})t}}] \\
  \dfrac{{\ln 2}}{{({k_{1}} + {k_2}}} = t \\
 \]
Now the equilibrium constant \[k = \dfrac{{{k_1}}}{{{k_2}}}\]
Now, from the given values, the value of k2 is,
 \[
  0.16 = \dfrac{{3.3 \times {{10}^{ - 4}}{\text{ }}{s^{ - 1}}}}{{{k_2}}} \\
  {k_2} = \dfrac{{3.3 \times {{10}^{ - 4}}{\text{ }}{s^{ - 1}}}}{{0.16}} \\
  {k_2} = 2.06{\text{ }}x{\text{ }}{10^{ - 3}}{\text{ }}{s^{ - 1}} \\
 \]
So, it takes for half the equilibrium amount of the trans isomer to be formed is,
 \[
  \dfrac{{\ln 2}}{{({k_{1}} + {k_2})}} = t \\
  \dfrac{{\ln 2}}{{(3.3{\text{ }}x{\text{ }}{{10}^{ - 4}} + 2.06{\text{ }}x{\text{ }}{{10}^{ - 3}})}} = t \\
  \dfrac{{\ln 2 \times {{10}^3}}}{{(0.33 + 2.06{\text{ }})}} = t \\
  0.289 \times {10^3} = t \\
  289\sec = t \\
  4.83\min = t \\
 \]

So, the correct answer is, A.

Note:
The rate of a reaction can be expressed in terms of concentration of reactants or products. The magnitude of the rate value does not depend upon the way of expression ( by reactant or product). For reactants, the rate shows as negative as throughout the reaction the concentration of the reactant decreases. On the other hand, the rate for the product shows as positive as the concentration of the product increases throughout the reaction.