Consider the concentration-time dependence graph for two first-order reactions.

enter image description here

Which reaction has a larger rate constant?

  1. Unable to determine
  2. The reaction represented by the upper curve
  3. The reaction represented by the lower curve correct

The above question is confusing me. It gives two concentrations vs. time curves, and asks which reaction has the larger rate constant. The answer is (3) the reaction represented by the lower curve.

I know that the reaction is first-order, hence $$\mathrm{rate} = k[A].$$ Thus, algebra reveals that $$[A] = \frac{\mathrm{rate}}{k}.$$ Thus, it makes sense that increasing the value of $k$ means dividing rate by a larger number, yielding a smaller concentration at any given time. This makes sense why the lower curve is the correct answer.

However, I also understand that the value of $k$, the rate constant, does not depend/is not correlated with the concentration of the reactant(s). Thus, is my logic not applicable? How else can you reason your way to the correct answer?

  • 1
    $\begingroup$ The rate constant, k, does not depend on concentration. The reaction rate itself, $k\ce{[A]}$, does depend on concentration. For graph $k_{lower} > k_{upper}$ since the starting concentration of A is the same. $\endgroup$
    – MaxW
    Oct 28, 2018 at 16:57
  • 1
    $\begingroup$ @MaxW That is a very clear explanation. Thank you so much! Because we're comparing two different reactions, and we know that the initial concentrations are the same BUT the rates are different, k must be a different value in the two reactions. Thank you! $\endgroup$ Oct 28, 2018 at 20:36

1 Answer 1


...I also understand that the value of k, the rate constant, does not depend/is not correlated with the concentration of the reactant(s).

It would be easier to represent this in calculus, but I assume you have not gotten there yet.

Anyway, you are correct, $k$ is a constant.

$$\mathrm{rate} = k [A]^x \implies \mathrm{rate} \propto k$$

thus as $k$ increases, the rate will increase for a given concentration of $\ce A$. The question did provide that $x = 1$ but this is not needed to solve the problem. Think of rate as: $$\mathrm{rate} = -\frac{\Delta [A]}{\Delta t}$$ (the reaction is removing $\ce A$) which is in the form of a slope. thus the larger the rate value, the lower the slope of the curve. Rate is dependend on concentration, but for the same concentration, a reaction with a higher rate constant will proceed faster, thus remove $\ce A$ fast and have a more negative slope.

Lets compare for the two $k$ values, $k_{upper}$ and $k_{lower}$

$$\mathrm{rate}_{upper} = k_{upper}[\ce A]= -\frac{\Delta [A]}{\Delta t} \\ \text{and }\\ \mathrm{rate}_{lower} = k_{lower}[\ce A]= -\frac{\Delta [A]}{\Delta t} $$

Lets say arbitrarily that two reaction rates are: $k_{1} = \pu{2 s-1}$ and $k_{2} = \pu{4 s-1}$ and the instantaneous concentration is $[\ce A] = \pu{0.6M}$ then the reaction rates would be as follows:

$$\mathrm{rate}_{1} = \pu{2 s-1}\times \pu{0.6M} = \pu{1.2 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{1.2 M s-1}\\ \text{and }\\ \mathrm{rate}_{2} = \pu{4 s-1}\times \pu{0.6M} = \pu{2.4 M s-1} = -\frac{\Delta [A]}{\Delta t} \implies \frac{\Delta [A]}{\Delta t} = -\pu{2.4 M s-1}$$

Since a reaction of $\mathrm{rate}_2$ has a more negative slope, it must correspond to a curve that decreases faster than a reaction of $\mathrm{rate}_1$ and the concentration would be lower sooner, thus $\mathrm{rate}_2$ is the lower curve and $\mathrm{rate}_1$ is the upper curve.


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