Animated Solution for Chemistry - Chemical Kinetics: For a first-order reaction R $⟶$ P at a given temperature, $k$ is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time $t$ are [R] and [P], respectively. The correct graphical representation(s) for this reaction is(are)

The study of chemical kinetics is not just about crunching numbers; it is about visualizing the dynamic dance of molecules over time. In this problem, we are tasked with identifying the correct graphical representations for a classic first-order reaction, $R⟶P$.

To master this, we must translate our fundamental rate laws into geometric shapes—lines, curves, and slopes. Let's break down each graph systematically.

Before we even look at the graphs, we need our mathematical toolkit. For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant $R$:

Integrating this differential rate law gives us the concentration of the reactant as a function of time:

Since the reactant $R$ converts into product $P$, the concentration of the product at any time $t$ is the initial concentration of $R$ minus what remains:

With these equations in hand, we are ready to interrogate the graphs.

The Product Growth Curve

Graph A plots the concentration of the product, $[P]$, against time, $t$. According to our integrated equation, $[P{]}_t=[R{]}_0(1-e^{-kt})$, the product concentration should start at zero and grow exponentially, eventually plateauing at $[R{]}_0$ as $t\to \infty$.

However, look closely at the shape of the curve in Graph A. It is concave upwards. A concave upward curve implies that the rate of production is increasing over time. Physically, this is impossible for a simple first-order reaction because as the reactant gets consumed, the reaction must slow down. The true curve must be concave downwards, reflecting a decreasing rate of production. Therefore, Graph A is incorrect.

The Rate vs Concentration Line

Graph B plots the rate of disappearance of the reactant, $\frac{d[R]}{dt}$, against its concentration, $[R]$. Let's look at our differential rate law:

This is the equation of a straight line passing through the origin, analogous to $y=mx$, where the slope $m$ is $-k$. Since the rate constant $k$ is always a positive value, the slope must be negative.

The graph provided in option B shows a line with a positive slope, shooting up into the first quadrant. This would bizarrely imply that the reactant concentration is increasing! The correct line should slope downwards into the fourth quadrant. Thus, Graph B is a trap and is incorrect.

The Rate of Production Decay

Graph C plots the rate of formation of the product, $\frac{d[P]}{dt}$, against time, $t$. Let's substitute our integrated reactant equation into the rate law:

This equation describes an exponential decay. At $t=0$, the rate is at its maximum ($k[R{]}_0$), and as time progresses, the rate decays exponentially towards zero.

Looking at Graph C, we see exactly this: a curve starting from a positive y-intercept and decaying exponentially. This perfectly matches our mathematical derivation. Graph C is absolutely correct!

The Constancy of 'k'

Finally, Graph D plots the rate constant, $k$, against time, $t$. This is a test of fundamental definitions. The rate constant $k$ is a measure of the intrinsic speed of the reaction at a given temperature. It is completely independent of time or concentration.

Therefore, a plot of $k$ versus $t$ must be a perfectly horizontal line, indicating that $k$ remains unchanged as the reaction proceeds. Graph D shows exactly this horizontal line. Graph D is correct!

By rigorously applying the differential and integrated rate laws, we successfully navigated the graphical traps. The correct representations are indeed (C) and (D). Always let the math guide your visual intuition!