Slope of the graph given is $-8.4 \text{ sec}^{-1}$, the rate constant for the reaction is :

$19.34 \text{ sec}^{-1}$

The correct answer is option (4) → $19.34 \text{ sec}^{-1}$

1. Identify the Equation

For a first-order reaction, the relationship between concentration $[A]$ and time $t$ is:

$\ln[A] = -kt + \ln[A]_0$

However, the graph uses $\log_{10} A$ (common logarithm) rather than $\ln A$ (natural logarithm). To convert, we use the factor $2.303$:

$\log[A] = -\frac{k}{2.303}t + \log[A]_0$

2. Relate to the Slope

This equation follows the straight-line format $y = mx + c$, where:

• $y$ is $\log[A]$
• $x$ is time $t$
• Slope ($m$) is $-\frac{k}{2.303}$

3. Calculate the Rate Constant ($k$)

From the problem statement, the slope is given as $-8.4 \text{ sec}^{-1}$. We can set up the equality:

$\text{Slope} = -\frac{k}{2.303}$

$-8.4 = -\frac{k}{2.303}$

Multiply both sides by $-2.303$:

$k = 8.4 \times 2.303$

$k \approx 19.3452 \text{ sec}^{-1}$