Answer

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**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.

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