Question

After how many seconds will the concentration of the reactant in a first order reaction be halved, if the rate constant is $1.155 \times {10^{ - 3}}{{\text{s}}^{ - 1}}$?
A. $600$
B. $100$
C. $60$
D. $10$

Answer 1

Hint: Chemical kinetics is based on rate of different chemical reactions. First order reaction is the reaction in which the rate of the reaction is based on only one reactant concentration. The rate is directly proportional to the concentration of the reactant.

Complete step by step answer:
It is given that the rate constant, ${\text{k = 1}}{\text{.155}} \times {\text{1}}{{\text{0}}^{ - 3}}{{\text{s}}^{ - 1}}$
Rate of the reaction is based on the reactant concentration. Reaction in which the rate depends on only one reactant is first order reaction.
Consider a chemical reaction \[{\text{A}} \to {\text{B}}\]
A is the reactant, and B is the product.
Here rate of the reaction, \[{\text{rate}} = \dfrac{{ - {\text{d}}\left[ {\text{A}} \right]}}{{{\text{dt}}}}\], where \[\left[ {\text{A}} \right]\] is the change in concentration of reactant in a specific period of time. A negative symbol is used because $\left[ {\text{A}} \right]$ decreases within the time period.
\[{\text{rate}} = \dfrac{{{\text{d}}\left[ {\text{B}} \right]}}{{{\text{dt}}}}\], where \[\left[ {\text{B}} \right]\] is the change in concentration of product B. It is positive since it increases over time period.
We know that rate of the reaction, ${\text{r}}\alpha \left[ {\text{A}} \right]$.
So the rate can be expressed as \[{\text{r}} = {\text{k}}{\left[ {\text{A}} \right]^{\text{m}}}\], where m determines the order, and k is known as the rate constant.
Given that the concentration of reactant is halved. So the time required for the concentration of reactant to be halved tells us that we have to calculate the half-life of the reactant.
Half-life can be calculated by ${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{\text{k}}}$
Substituting the value of rate constant, we get
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{1.155 \times {{10}^{ - 3}}{{\text{s}}^{ - 1}}}}$
On simplification, we get
${{\text{t}}_{\dfrac{1}{2}}} = \dfrac{{0.693}}{{1.155 \times {{10}^{ - 3}}{{\text{s}}^{ - 1}}}} = 600{\text{s}}$


So, the correct answer is Option A .

Note:
During each half-life, half of the substance decay into atoms or molecules. Each radioactive nuclide has its own half-life. Half-lives can be as short as a fraction of a second or as long as billions of years. Half-life is the time required to reach steady state.