Question

Question: For a first order reaction $A \to B$, the reaction rate at reactant concentration of 0.01M is found to be $2.0 \times {10^{ - 5}}\;mol{L^{ - 1}}{s^{ - 1}}$. The half-life period of the reaction is:
A) 220 s
B) 30 s
C) 300 s
D) 347 s

Answer 1

The half-life period of the reaction is the time period in which the concentration of reactant becomes half of the initial concentration. The half-life period of first order reaction is independent of any concentration term.

Complete step by step answer:
Here, we have given a first order reaction $A \to B$
In the first order reaction, the rate of reaction depends upon one concentration term only.
For the reaction shown
Rate of reaction $= \;k\left[ A \right]$
Where k is a constant, known as rate constant.
It is also given that reactant concentration is 0.01M.
i.e., $\left[ A \right]\; = \;0.01M$
And the rate of reaction $= \;2.0 \times {10^{ - 5}}\;mol{L^{ - 1}}{s^{ - 1}}$
Therefore, $k = \dfrac{{Rate\;of\;reaction}}{{\left[ A \right]}}$
Substituting the values, we get
$\Rightarrow k = \dfrac{{2.0 \times {{10}^{ - 5}}}}{{0.01}}$
$\Rightarrow k = 2.0 \times {10^{ - 3}}\;{s^{ - 1}}$
Now, to find the half-life of first order reaction.
${t_{1/2}} = \dfrac{{0.693}}{k}$
Where ${t_{1/2}}$ is the half-life period
Now, substituting the value of k in to the equation, we get
$\Rightarrow {t_{1/2}} = \dfrac{{0.693}}{{2.0 \times {{10}^{ - 3}}{s^{ - 1}}}}$
$\Rightarrow {t_{1/2}} = 346.5s$
Approximating we can say that ${t_{1/2}} = 347s$

Therefore, option(d) is correct.

Note: In this question it was already given that this was a first order reaction. And we can clearly see why it is called a first order reaction. Still there are some reactions which may not look like a first order reaction but actually it is a first order reaction. Such reactions are called pseudo-unimolecular reactions or simple pseudo first order reactions.
For example, acidic hydrolysis of an ester.
$C{H_3}COO{C_2}{H_5} + {H_2}O\xrightarrow{{{H^ + }}}\;C{H_3}COOH + {C_2}{H_5}OH$
In these reactions, concentration of water (one of the reagents) is in excess and its concentration remains constant throughout the reaction.
Thus, $rate\; \propto \;\left[ {C{H_3}COO{C_2}{H_5}} \right]$