Question

A reagent undergoes $90\% $ decomposition in 366 minutes. According to the first-order reaction, its half-life is:
A. $366 \times 100\left( {\dfrac{{\ln 2}}{{90}}} \right)$
B. $366\left( {\dfrac{{\ln 2}}{{\ln 10}}} \right)$
C. $\dfrac{1}{{366}}$
D. 183

Answer 1

Hint: A first-order reaction is a reaction whose rate is determined by the change in concentration one concentration term only. Or in other words, a reaction whose rate varies as the first power of the concentration of a reactant. Let us consider the following reaction:
$A \to B$, where A is the reactant and B is the product.
For the above reaction to being a first-order reaction,
Rate $ = \dfrac{{ - d[A]}}{{dt}} = - k[A]$, where k is the rate constant.

Complete step by step answer:
Half-life of a substance is the time required for a particular quantity of that substance to reduce to half of its initial value. It is represented by ${t_{1/2}}$.
For a first-order reaction, the half-life is given by:
${t_{1/2}} = \dfrac{1}{k}\ln 2$
Here k, the rate constant is given by: $k = \dfrac{1}{t}\ln \left( {\dfrac{{{{[A]}_i}}}{{{{[A]}_t}}}} \right)$ , where t is time,${[A]_i}$ is the initial concentration of the reactant, and ${[A]_t}$ is the concentration of reaction at time t.
According to the question, $90\% $ of the decomposition takes 366 minutes. That means if the initial concentration is 100, the concentration left at t $ = $ 366 minutes is 10. Putting these values in the equation for k we have:
$k = \dfrac{1}{{366}}\ln \left( {\dfrac{{100}}{{10}}} \right)$
$k = \dfrac{1}{{366}}\ln 10$
Putting the value of k in the equation for ${t_{1/2}}$ we have:
${t_{1/2}} = \dfrac{1}{{\dfrac{1}{{366}}\ln 10}}\ln 2$
${t_{1/2}} = 366\left( {\dfrac{{\ln 2}}{{\ln 10}}} \right)$

So, the correct answer is Option B.

Note:
The order of a reaction is defined as the sum of powers of the reactants of a chemical reaction in its rate law equation. While the rate of a reaction is defined as the speed at which the reaction proceeds it is rate constant multiplied by the concentration of each reactant each raised to some power which may or may not be equal to the stoichiometric coefficients of the reactants.
${\text{aA + bB}} \to c{\text{C + dD}}$
The rate of a reaction is :
Rate = ${\text{k[A}}{{\text{]}}^{\text{x}}}{{\text{[B]}}^{\text{y}}}$
And $\therefore$ the order the reaction is x+y.