Slope of the graph given is \(-8.4 \text{ sec}^{-1}\), the rate constant for the reaction is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(19.34 \text{ sec}^{-1}\)
The graph of log [A] against time is a falling straight line, which is the mark of a first order reaction. Its integrated rate equation, in base-10 logarithms, is
\( \displaystyle \log [A] = -\frac{k}{2.303}\,t + \log [A]_0 \)
Comparing with y = mx + c, the slope is
\( \displaystyle \text{slope} = -\frac{k}{2.303} \)
With slope = −8.4 s−1:
\( \displaystyle k = -2.303 \times \text{slope} = 2.303 \times 8.4 = 19.34\ \text{s}^{-1} \)
NCERT (2026-27 edition):
First Order Reactions – “… The first order rate equation (3.10) can also be written in the form
\( k = \frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
\( \log\frac{[R]_0}{[R]} = \frac{kt}{2.303} \)
If we plot a graph between log [R]0/[R] vs t, (Fig. 3.5), the slope = k/2.303”