Question

A first order reaction is 40% complete after 8 min. How long will it take before it is 90% complete? What is the value of the rate constant?

Answer 1

Hint: The initial concentration is 100, 40% reaction complete in 8 mins and the time required to calculate the 10 % reaction after 90% completion is required. The formula used to calculate the first order is
\[K = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}\]

Complete step by step answer:
Given,
Time required to complete a 40% first order reaction is 8 min.
The reaction is said as a first order reaction when the reaction rate depends on the concentration of only one reactant. The unit referring the first order reaction is ${s^{ - 1}}$.
The rate constant for the first order reaction is given by the formula as shown below.
\[K = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}\]
Where,
K is the rate constant.
t is the time
$a$ is the initial concentration.
$a - x$ is the left out concentration.
Let the initial concentration be 100. So after completion of 40% reaction left concentration will be 60.
To calculate the rate constant, substitute the values in the equation.
\[\Rightarrow K = \dfrac{{2.303}}{8}\log \dfrac{{100}}{{60}}\]
\[\Rightarrow K = \dfrac{{2.303}}{8}\log \dfrac{5}{3}\]
\[\Rightarrow K = 0.06386{\min ^{ - 1}}\]
When the reaction is 90% complete, then the left out concentration is 10.
\[\Rightarrow 0.06386 = \dfrac{{2.303}}{t}\log \dfrac{{100}}{{10}}\]
\[\Rightarrow t = \dfrac{{2.303}}{{0.06886}}\]
\[\Rightarrow t = 36.01\]min.

Thus, the rate constant for the first order reaction is 0.06886 and the time required to complete 90% reaction is 36.01min.

Note: The value of log with base 10 is 1. In this question they refer to two situations of reaction taking place in a specific duration of time.The rate law is an equation which gives a relation between the rate of chemical reaction with the concentration of the reactant or the partial pressure of the reactant.