Question

Calculate the time taken in seconds for 40% completion of a first-order reaction, if its rate constant is \[3.33 \times {10^{ - 4}}{\sec ^{ - 1}}\] .

Answer 1

Hint: A reaction is called first-order if the rate of the reaction relies on one concentration term only. For the reaction:\[A \to B\] the rate of reaction is directly proportional to the concentration of A i.e., [A].

Formula Used:
\[k = \dfrac{{(2.303)}}{t}\log \left( {\dfrac{{{{[A]}_0}}}{{[A]}}} \right)\]

Complete Step by Step Solution:
We know that for a first-order reaction, the rate constant, k can be calculated by the following equation.
\[k = \dfrac{{(2.303)}}{t}\log \left( {\dfrac{{{{[A]}_0}}}{{[A]}}} \right)\]

where
\[{[A]_0}\] = initial concentration of the reactant,
\[[A]\] = final concentration of the reactant,
t = time taken

Here, we are given the rate constant is \[3.33 \times {10^{ - 4}}{\sec ^{ - 1}}\].
There is a 40% completion of the reaction, which means 40% of the concentration of reactant is used up to form the products.

Let us consider the initial concentration of reactants to be 100%.
We are left with \[(100 - 40)\% = 60\% \] of the concentration reactant.
Therefore, \[{[A]_0} = 100\] and \[[A] = 60\]


Establishing these values in the equation (1), we obtain
\[k = \dfrac{{(2.303)}}{t}\log \left( {\dfrac{{{{[A]}_0}}}{{[A]}}} \right)\]
\[ \Rightarrow 3.3 \times {10^{ - 4}}{\sec ^{ - 1}} = \left( {\dfrac{{2.303}}{t}} \right)\log \left( {\dfrac{{100}}{{60}}} \right)\]
\[ \Rightarrow t = \dfrac{{2.303}}{{3.33 \times {{10}^{ - 4}}}}\log \left( {\dfrac{5}{3}} \right)\sec \]
\[ \Rightarrow t = (0.6915 \times {10^4})\log (1.66)\sec \]
\[ \Rightarrow t = (6915 \times 0.22)\sec \]
\[ \Rightarrow t = 1521\sec \]

So, 1521 seconds is required for $40\%$ completion of a first-order reaction, given that its rate constant is \[3.33 \times {10^{ - 4}}{\sec ^{ - 1}}\] .

Note: While attempting the question, a student must retain in mind that the question says that some per cent of the reaction is complete. This implies that this amount of reactant is consumed. Hence, by assuming the initial concentration of the reactant to be 100%, the final concentration can be calculated. Here, the order of the reaction should also be kept in mind and the formula for the rate constant of that order of reaction needs to be applied. The time should be expressed with the correct unit.