Animated Solution for Chemistry - Chemical Kinetics: Rate constant, $k$ of the first order reaction when initial concentration $(C_0)$ and concentration $(C_t)$ at time $t$, is given by equation $kt=\log C_0-\log C_t$ Graph is a straight line if we plot

The beauty of chemical kinetics lies in its ability to translate abstract molecular behavior into simple, visual geometry. In this problem, we are given the integrated rate equation for a first-order reaction and asked to identify which plot yields a straight line.

We start with the given equation for a first-order reaction:

Here, $C_0$ is the initial concentration of the reactant, $C_t$ is the concentration at time $t$, and $k$ is the rate constant. To visualize this relationship, we need to map it to the standard equation of a straight line in coordinate geometry, which is:

To make our kinetic equation look like the straight-line equation, we need to isolate the variable that changes with time. Since $C_0$ is a constant, $\log C_0$ is also a constant. The variables are $C_t$ and $t$. Let's rearrange the equation to solve for $\log C_t$:

Now, let's compare our rearranged equation with $y=mx+c$: - y-axis ($y$): $\log C_t$ - x-axis ($x$): $t$ - Slope ($m$): $-k$ - y-intercept ($c$): $\log C_0$

This direct comparison reveals that if we plot $\log C_t$ on the y-axis and $t$ on the x-axis, we will obtain a perfect straight line. The negative slope ($-k$) visually represents the continuous decrease in the reactant's concentration over time, while the y-intercept ($\log C_0$) marks the starting point of our reaction.

Therefore, plotting $t$ versus $\log C_t$ gives a straight line, making option (b) the correct choice.