WBJEE 2024 Physics & Chemistry Paper

To understand the slope of the plot of $\log$ (reactant concentration) against time for a first order reaction, let's start by examining the fundamental relationship governing first order reactions:A first order reaction can be described by the differential rate law:$\; \frac{d[A]}{dt} = -k[A]$where $[A]$ is the concentration of the reactant and $k$ is the rate constant.To solve this differential equation, we can separate variables and integrate:$\int\; \frac{1}{[A]} d[A] = -k \int dt$This gives us the integrated form of the first order rate law: $\ln [A] = -k t + C$where $C$ is the integration constant that can be determined from the initial conditions. If at $t=0$, $[A]=[A]_0$ (initial concentration of $A)$, then:$\ln [A]_0 = C$Thus, we can rewrite the equation as:$\ln [A] = \ln [A]_0 - k t$The next step is to convert the natural logarithm to a common logarithm (base 10) using the relationship: $\log [A] = \; \frac{\ln [A]}{\ln 10}$This results in: $\square$$\log [A] = \log [A]_0 - \frac{k t}{2.303}$We can clearly see that this equation is in the form $y = m x + c$ where $y$ is $\log [A], c$ is $\log [A]_0$, and the slope $m$ is $-\frac{k}{2.303}$.Therefore, the slope of the plot of $\log$ (reactant concentration) against time for a first order reaction is:Option C: $-\frac{k}{2.303}$