Question

For a first order process $A \to B$ , rate constant ${K_1} = 0.693{\text{ }}mi{n^{ - 1}}$ and another first order process $C \to D$ , ${K_2} = x{\text{ }}mi{n^{ - 1}}$ If $99.9\%$ of $C \to D$ requires time same as $50\%$ of reaction $A \to B$ , value of x? (in min-1)

$$A.\;\;\;\;\;0.0693 \\\ B.\;\;\;\;\;6.93 \\\ C.\;\;\;\;\;23.03 \\\ D.\;\;\;\;\;13.86 \\\$$

Answer 1

First calculate the half-life of the first process because it says $50\%$ of reaction $A \to B$ has occurred. Using that, calculate the value of ‘x’ which is the rate constant for the second process. Read the question properly to understand. Important to note that the value of ’t’ for $99.9\%$ of $C \to D$ and $50\%$ of reaction $A \to B$ is the same.

Complete answer:
In the above question we are given that $A \to B$ and $C \to D$ are the two first order reactions having rate constants ${K_1} = 0.693{\text{ }}mi{n^{ - 1}}$ for the process $A \to B$ and ${K_2} = x{\text{ }}mi{n^{ - 1}}$ for the process $C \to D$ we have to find ‘x’ that is rate constant for the second reaction $C \to D$ when $99.9\%$ of reaction has taken place. Also said that the time required for $99.9\%$ of $C \to D$ is the same as that of $50\%$ of $A \to B$ . We will find ${t_{\dfrac{1}{2}}}$ value with ${K_1} = 0.693{\text{ }}mi{n^{ - 1}}$
So time taken required to complete $50\%$ completion for reaction $1$ is:
${t_{\dfrac{1}{2}}} = \dfrac{{0.692}}{{{K_1}}}$ Given that ${K_1} = 0.693{\text{ }}mi{n^{ - 1}}$
So $\Rightarrow {t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{0.693}} = {\text{ }}1{\text{ }}mi{n^{ - 1}}$
And so we get the value of ${t_{\dfrac{1}{2}}}$ as $1{\text{ }}mi{n^{ - 1}}$ . This means the time required for $50\%$ of the reaction $A \to B$ to occur is $1{\text{ }}mi{n^{ - 1}}$ . We are given the statement that $99.9\%$ of the process $C \to D$ takes the same time as that of $50\%$ of $A \to B$ . And $50\%$ of the process $A \to B$ is $1{\text{ }}mi{n^{ - 1}}$ . This means that the time required for $99.9\%$ of the process $C \to D$ is $1{\text{ }}mi{n^{ - 1}}$ . Hence with this value of ‘t’ we can find the value of $K_2$.
We are going to use the first order rate constant equation. The equation is given as:
$K = \dfrac{1}{t}\ln \dfrac{a}{{a - x}}$
Where ‘K’ is the rate constant,‘t’ is the time taken, ‘a’ is the initial concentration,‘a-x’ is the concentration at time t.
In our case this equation is:
${K_2} = \dfrac{1}{t}\ln \dfrac{a}{{a - x}}$
Where K2 value we have to find out, ‘t’ we have calculate as $1{\text{ }}mi{n^{ - 1}}$ , ‘a’ here will be initial concentration in percentage which is $100\%$ as initially $100\%$ of the reactant is reacting, ‘a-x’ will be $100 - 99.9$(after time ‘t’$100 - 99.9\%$ of the reaction has occurred)
So , ${K_2} = \dfrac{1}{t}\ln \dfrac{a}{{a - x}}$
${K_2} = \dfrac{1}{t}{\text{ }}ln{\text{ }}\dfrac{{100}}{{100 - 99.9}}$
$\Rightarrow {K_2} = \dfrac{1}{1}{\text{ }}ln{\text{ }}\dfrac{{100}}{{0.1}}$
$\Rightarrow x = 1{\text{ }}ln{\text{ }}1000$ (since K2 is ‘x’)
$\Rightarrow x = {\text{ }}ln{\text{ }}{10^3}$
$\Rightarrow x = {\text{ }}3{\text{ }}ln{\text{ }}10$
$\Rightarrow x = {\text{ }}3 \times 2.303$
$\Rightarrow x = 6.909$
Therefore rate constant $K_2$ for the process $C \to D$ is ${K_2} = 6.909{\min ^{ - 1}}$
So the correct option is $B.\;\;\;\;\;{6.93_{}}{\min ^{ - 1}}$

Note:
Different orders of reactions have different rate constant equations and different half-life equations. Sometimes in the question order of the reaction might not be given then to find out which reaction it is just check the unit of K because different orders have different units of rate constant ‘K’.