For a reaction R → P, the plot shown in the figure is a straight line.
![Fig. 3.4 Plot of ln[R] against t](https://assets.qrepo.in/img/Uploads/admin/cuet-chem-s3-fm06-q14-full.png)
The slope of the line is −0.0231 min−1. What is the half-life of the reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 30 min
The figure plots ln[R] against t and gives a straight line, which is the plot for a first order reaction. It marks k = −slope, so the rate constant is read from the slope. For a first order reaction the half-life is t1/2 = 0.693/k.
\( \displaystyle k = -\text{slope} = -(-0.0231\ \mathrm{min^{-1}}) = 0.0231\ \mathrm{min^{-1}} \)
\( \displaystyle t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231\ \mathrm{min^{-1}}} = 30\ \mathrm{min} \)
NCERT (2026-27 edition):
3.3.2 First Order Reactions – “Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]0”
3.3.3 Half-Life of a Reaction
\( \displaystyle t_{1/2} = \frac{0.693}{k} \qquad (3.17) \)
“The half-life of a first order equation is readily calculated from the rate constant and vice versa.”
Fig. 3.4: A plot between ln[R] and t for a first order reaction