Rate of Reaction

Thermodynamics predicts the favored direction at specified temperature, pressure and composition. Kinetics asks how quickly that change happens. A favorable reaction can still proceed slowly because of its activation barrier.

Reaction rate measures concentration or amount changing per time. The rate constant k is a factor in a rate law; it is not itself the reaction rate. Holding temperature and the reaction conditions fixed keeps k fixed in the model, while changing concentrations can change the rate.

For an empirical power-law rate law r=k[A]^x[B]^y, x and y are experimentally determined orders. k depends on temperature and the reaction conditions. Declare whether r means a particular species disappearance rate or the stoichiometrically normalized reaction rate before calculating.

$r = k[A]^x[B]^y$

There, $r$ is the reaction rate, $k$ is the rate constant for the reaction, $[A]$ and $[B]$ are the concentrations of the reactants, while the powers they are raised to, $x$ and $y$ represent their relative impact on the reaction rate. This impact is known as order. If the equation looked like this:

$r = k[A]^2[B]$

We would say that the reaction is 2nd order in A and 1st order in B, for an overall order of 3.

It is important to note that these orders are not the same as the stoichiometric coefficients for each reactant. Sometimes they coincide, but do not assume they will always be equal. Many times you will have a reaction rate that is 2nd order in a component whose stoichiometric coefficient is 1, or perhaps you will have a reaction that is not even dependent on one of the components (0th order).

For the single-reactant model in this section, zero order in A means its disappearance rate is independent of[A] over the stated range at fixed conditions: −d[A]/dt=k. This is a declared regime, not a guarantee that the rate remains constant through every possible concentration or surface condition.

$r = k$

The units of both r and k in this equation are $\frac{mol}{L\cdot s}$. The units of r are always going to be the same, but k will change depending on the order of the reaction. For example, a first-order reaction will always have k in units of $s^{-1}$.

So okay, we have this simple rate equation. What can we do with it? To find out what this lets us do, let’s look at the units for r.

$\frac{mol}{L\cdot s}$

Do you see anything that looks familiar?
Perhaps the $\frac{mol}L$, which is the units of Molarity, a measure of concentration. Okay. Let’s replace that with M, for Molarity units. This gives us:

$\frac Ms$

The disappearance rate of a reactant A is −Δ[A]/Δt, so it is positive when [A] falls. For aA+bB→cC, the normalized reaction rate is r=−(1/a)Δ[A]/Δt=−(1/b)Δ[B]/Δt=(1/c)Δ[C]/Δt at constant volume. The coefficients connect species rates; they do not generally supply empirical rate-law exponents.

Species disappearance: −d[A]/dt. Normalized reaction rate for aA: r=−(1/a)d[A]/dt. In the following integrated-law examples, k is explicitly the constant in the stated A-disappearance law.

This lets us know how we can use the reaction rate to find out the concentration of our reaction components as a function of time. Perhaps a quick example problem could help with this point.

Reviewed replacement, Example3.1.1: In a hypothetical fixed-volume model, −d[A]/dt=k=212 mol L⁻¹ s⁻¹ over the stated range. [A]₀=30.0 M. Find the mathematical exhaustion time. These formal concentration data illustrate zero-order integration; they are not a measured automobile catalytic-converter model.

Now we have two equations that both give a bit of information about a Zeroth-Order reaction, but there’s one more major equation that you should know.

This other equation is the half-life of reaction. Half-life is defined as the time it takes for one half of a species to be either used up, or decompose, or in any way get depleted. Perhaps you have heard of half-life in reference to radioactive elements, particularly Uranium.

Radioactive decay commonly follows first-order kinetics, with a concentration-independent half-life. Chemical zero-order disappearance has t½=[A]₀/(2k), so successive halving intervals become shorter. The word half-life alone does not identify an order.

Mathematically, half-life is the time at which $[A] = \frac12 [A]_0$. Using this equation, let’s determine the half-life equation for a Zeroth-Order reaction.

We start with the integrated Zeroth-Order rate equation:

$[A] = [A]_0 - kt$

Now set $[A] = \frac12 [A]_0$:

$\frac12 [A]_0 = [A]_0-kt$

Finally, solve for $t$.

$kt = \frac12[A]_0$

$t = \frac{[A]_0}{2k}$

In summation, we now have 3 useful equations for a Zeroth-Order reaction:

The rate equation: $r =k$

The integrated rate equation: $[A] = [A]_0-kt$

The half-life equation: $t_{\frac12}=\frac{[A]_0}{2k}$

It is important to remember that these are only applicable to zero-order reactions.

If you wish to determine the reaction rate constant for a reaction that you believe is Zeroth-Order, you would need to track the concentration of a reactant over time. If you graph the reaction, you should expect it to look like a straight line.

Once you have the data points, you would need to determine a “best-fit” line. This type of line is the straight line which is as close to the data points as possible. There are many ways to determine this, but if you try to do it by hand, it is quite difficult. Instead, we’ll be giving you the best fit lines when you need them.

If you graph [A] vs. t, and it doesn’t come out like a straight line, it might not be a Zeroth-Order reaction. We will cover how to figure our higher order reactions in the upcoming sections. For now, let’s give you a bit more practice working with Zeroth-Order reactions.

Temperature often increases a reaction rate at fixed concentrations, but dissolution speed and equilibrium solubility are different quantities. Solubility can increase or decrease with temperature depending on the dissolution enthalpy. Do not infer the amount that can dissolve merely from how fast it dissolves.

Quantitative enrichment: Arrhenius relates the rate constant, not the rate alone, to temperature: k=A exp(−Eₐ/RT). AP instruction uses activation-energy and collision reasoning qualitatively; this chapter’s numerical Arrhenius calculations are additional practice.

$k = A\exp{\frac{-E_a}{RT}}$

This equation introduces a few new variables that you’ve yet to see. Let’s start with the familiar: $R$ is the gas constant, the one you will typically be using is $8.314 \frac J{mol\cdot K}$. You also have $k$, the rate constant, and $T$, the temperature (in K).

A is the pre-exponential factor and has the same units as k. Eₐ is molar activation energy. With R=8.314 J mol⁻¹ K⁻¹, convert Eₐ to J mol⁻¹ and T to kelvin. Treat A and Eₐ as constant only over the specified model range; real mechanisms can change with conditions.