Question

The half-life of a 1st order reaction is 1 min 40 seconds. Calculate its rate constant.
A. $6.9\times {{10}^{-3}}{{\min }^{-1}}$
B. $6.9\times {{10}^{-3}}{{s}^{-1}}$
C. $6.9\times {{10}^{-3}}s$
D. 100s

Answer 1

Hint: First-order reaction is a chemical reaction in which the rate of the reaction depends on only concentration of the only one reactant.
The formula to calculate the rate of the reaction is as follows.
\[k=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}\]
where,k =Rate constant of the 1st order reaction
             ${{t}_{{}^{1}/{}_{2}}}$ = half-life of the reactant in 1st order reaction

Complete answer:
- In the question it is given that the half-life of a 1st order reaction is 1 min 40 seconds and we have to calculate the rate of the 1st order reaction.
- The formula to calculate the rate of the reaction is as follows.
\[k=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}}\]
where, k =Rate constant of the 1st order reaction
             ${{t}_{{}^{1}/{}_{2}}}$ = half-life of the reactant in 1st order reaction
- The given half-life of the reactant is 1 min 40 seconds.
- Convert the half-life into seconds then 1 min 40 seconds = 100 sec.
- Now substitute all the known values in the above formula to get a rate constant of the 1st order reaction and it is as follows.
\[\begin{align}
  & k=\dfrac{0.693}{{{t}_{{}^{1}/{}_{2}}}} \\
  & \Rightarrow k=\dfrac{0.693}{100} \\
 & \Rightarrow k=6.93\times {{10}^{-3}}{{s}^{-1}} \\
\end{align}\]
- Therefore the rate constant of the1st order chemical reaction $k=6.93\times {{10}^{-3}}{{s}^{-1}}$ .

So, the correct option is B.

Note:
Half-life of the reactant means the time required by the chemical to reduce to half of its initial concentration. Half-life of all the radioactive elements is different and it is going to depend on the stability of the radioactive chemical.
At the times of calculation of the rate constant we have to take the half-life of the reactant in seconds.