Question

Given,
\[{{\rm{T}}_{{\rm{50\% }}}}\] of the first-order reaction is 10 min. starting with \[{\rm{10mol}}{{\rm{L}}^{{\rm{ - 1}}}}\], rate after 20 min is:
(A) \[{\rm{0}}{\rm{.0693mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]
(B) \[{\rm{0}}{\rm{.0693}} \times {\rm{2}}{\rm{.5mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]
(C) \[{\rm{0}}{\rm{.0693}} \times {\rm{5mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]
(D) \[{\rm{0}}{\rm{.0693}} \times 10{\rm{mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]

Answer 1

Hint: If the rate of the reaction is determined by the change of one concentration term only, then the reaction is said to be a first-order reaction. \[{{\rm{T}}_{{\rm{50\% }}}}\]of the first-order reaction means the half-life period or \[{T_{\dfrac{1}{2}}}\]of the first order reaction.

Complete answer:
We can say that the rate varies as the first power of the concentration of the reactant that is rate increases as the number of times the concentration of the reactant is increased.
-The time required to reduce the concentration of the reactant to half of its initial value is known as the half-time of a reaction. Half-time of the reaction is denoted as \[{{\rm{T}}_{{\rm{50\% }}}}\]or \[{T_{\dfrac{1}{2}}}\].
-We know that the first-order reaction is given as-
\[k = \dfrac{1}{t}\ln \dfrac{{{{[A]}_o}}}{{[A]}}\]
For deriving half-time formula, we will substitute
\[t = {t_{\dfrac{1}{2}}}\]and \[[A] = \dfrac{{{{[A]}_o}}}{2}\]
We will get,
\[k = \dfrac{1}{t}\ln \dfrac{{{{[A]}_o}}}{{\dfrac{{{{[A]}_o}}}{2}}}\]
\[k = \dfrac{1}{{{t_{\dfrac{1}{2}}}}}\ln 2\]
\[{t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{k}\]
Here ‘k’ is a constant for a given reaction at a given temperature. Since the half-time equation lacks any concentration term, it is proved that half-time of the first-order reaction is a constant independent of the initial concentration of the reactant.
-According to the question, we are given
\[{t_{50\% }} = 10\min \],
Initial concentration= \[ = 10mo{l^{ - 1}}\]
First, we will find the concentration at \[{t_{\dfrac{1}{2}}}\]
\[k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{0.693}}{{10}} = 0.0693\]
-The concentration of the reaction after 10 min will first become half (half-time). With the passage of
another 10 min more, the reaction will again become half of the half-time, that is it is now one-fourth of the initial concentration.

Remaining concentration \[ = \dfrac{{10}}{4} = 2.5mo{l^{ - 1}}\]
Now finding the rate of the reaction after 20 min,
\[{\rm{Rate = k \times [Reactant] = 0}}{\rm{.0693}} \times {\rm{2}}{\rm{.5mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]

Hence, the correct answer is option (B) \[{\rm{0}}{\rm{.0693}} \times {\rm{2}}{\rm{.5mol}}{{\rm{L}}^{{\rm{ - 1}}}}{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\]

Note:
Following are the characteristics of the first-order reaction,
- When the concentration is increased by ‘n’ times, we will observe the increase in the rate of reaction with ‘n’ times.
- The unit of the rate of reaction for the first order is \[{\sec ^{ - 1}}\]or \[{\rm{mi}}{{\rm{n}}^{{\rm{ - 1}}}}\].
- The time required to complete a definite fraction of reaction in the first order, the reaction is independent of the initial concentration. Let \[t\dfrac{1}{u}\] be the time of one ‘u’ the fraction of reaction to take place, then
\[{k_1} = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}\]