Animated Solution for Chemistry - Chemical Kinetics: Which of the following plots is(are) correct for the given reaction? ($[P{]}_0$ is the initial concentration of P)

Identifying the Mechanism

Before we can dive into the graphs, we must first understand the chemistry happening in the beaker. We are given the reaction of tert-butyl bromide with sodium hydroxide.

Tert-butyl bromide is a tertiary alkyl halide. Because the carbon attached to the bromine is bonded to three other bulky methyl groups, it is highly sterically hindered. This makes a direct backside attack by the hydroxide ion (an $S_{N}2$ mechanism) virtually impossible. Instead, the molecule prefers to first lose the bromide ion, forming a highly stable tertiary carbocation.

This two-step process is the hallmark of an $S_{N}1$ mechanism. The rate-determining step is the slow formation of the carbocation, which depends only on the concentration of the alkyl halide. Therefore, this reaction strictly follows first-order kinetics.

Now that we know the reaction is first-order, let's evaluate Plot A, which graphs the half-life ($t_{1/2}$) against the initial concentration ($[P{]}_0$).

For any first-order reaction, the half-life is given by the beautifully simple equation:

Notice what is missing from this equation? There is absolutely no concentration term! This means that no matter how much reactant you start with, it will always take the exact same amount of time for half of it to react. Because $t_{1/2}$ is a constant, its graph against $[P{]}_0$ must be a perfectly horizontal line. Plot A shows exactly this, making it the correct answer.

Plot B attempts to trick us by showing the initial rate as a horizontal line against $[P{]}_0$. Let's consult the rate law:

If we look at the very beginning of the reaction, the initial rate ($R_0$) is simply $k[P{]}_0$. This is a linear equation of the form $y=mx$, which describes a straight line passing through the origin.

A horizontal line would imply that the rate does not change even if you add more reactant. That behavior is exclusive to zero-order reactions. Since our reaction is first-order, Plot B is fundamentally incorrect.

Plot C graphs the ratio of the product formed ($[Q]$) to the initial reactant ($[P{]}_0$) over time. Let's derive this mathematically.

We know the reactant decays exponentially:

By the law of conservation of mass, the amount of product formed is whatever reactant has disappeared:

Dividing both sides by $[P{]}_0$, we get:

This function starts at 0 (when $t=0$) and asymptotically approaches 1 as time goes to infinity. Visually, this creates a curve that is concave downwards. However, Plot C shows a curve that is concave upwards (like a parabola or $e^{kt}$). Thus, Plot C is incorrect.

Finally, let's examine Plot D, which graphs $\ln \left(\frac{[P]}{[P{]}_0}\right)$ against time.

Starting from the integrated rate law:

Rearranging this gives:

This is a linear equation ($y=mx$) where the y-variable is the logarithmic ratio, the x-variable is time $t$, and the slope is $-k$. Because the rate constant $k$ is always positive, the slope must be negative, meaning the line should slope downwards.

Plot D, however, shows a straight line with a positive slope, heading upwards. This contradicts the math entirely, making Plot D incorrect.

By trusting the mathematical equations of first-order kinetics, we have systematically proven that only Plot A is correct!