A reaction follows first order kinetics, its rate constant is found to be 4.794 × 10−3 min−1 at 25° C. The half-life of the reaction is-
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 144.5 min
For a first order reaction, the integrated rate equation is:
\( \displaystyle k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
At the half-life, [R] = [R]0/2, so the ratio [R]0/[R] = 2:
\( \displaystyle t_{1/2}=\frac{2.303}{k}\log 2=\frac{2.303\times0.3010}{k} \)
\( \displaystyle t_{1/2}=\frac{0.693}{k} \)
The half-life of a first order reaction is independent of the initial concentration. Here k = 4.794 × 10−3 min−1:
\( \displaystyle t_{1/2}=\frac{0.693}{4.794\times10^{-3}\ \mathrm{min^{-1}}} \)
\( \displaystyle t_{1/2}=\frac{693}{4.794}\ \mathrm{min} \)
\( \displaystyle t_{1/2}=144.56\ \mathrm{min} \)
So the half-life of the reaction is about 144.5 min (the last digit differs only because of rounding). The temperature (25° C) is not needed, because k is already given at that temperature.
Shortcut: 0.693 is a little less than 0.7, and 0.7/(4.8 × 10−3) ≈ 146 min. Only Option (3) is close to this value; Options (1) and (4) are 100 and 10 times too small.