Question

At 298 K, for a first order reaction (A → P) the following graph is obtained. The rate constant (in s$^{-1}$) and initial concentration (in mol L$^{-1}$) of ‘A’ are respectively: 

• A. $2.303; 10^{-1}$
• B. $10^{-2}; 2.303$
• C. $10^{-1}; 10^{-2}$
• D. $10^{-2}; 10^{-1}$

Hint:

For first-order reactions, the slope of the $\ln[A]$ vs. time graph gives the rate constant $k$.

Answer 1

The Correct Option is D

Step 1: Identifying the Rate Constant.
From the integrated rate equation for a first-order reaction:
$$\ln[A] = -kt + \ln[A]_0$$
- The slope of the graph is given as $-10^{-2}$, which corresponds to the rate constant: $$k = 10^{-2} \text{ s}^{-1}$$ Step 2: Identifying the Initial Concentration.
- The y-intercept of the graph corresponds to $\ln[A]_0$. Given $\ln[A]_0 = -2.303$, $$[A]_0 = e^{-2.303} = 10^{-1} \text{ mol L}^{-1}$$ Thus, the correct answer is $k = 10^{-2} \text{ s}^{-1}$ and $[A]_0 = 10^{-1} \text{ mol L}^{-1}$.