Rate Constant

If you're reading this, you're probably diving into reaction rates, rate laws, and rate constants right now in your chemistry studies. A key skill in chemical kinetics is the ability to calculate the rate constant for chemical reactions mathematically. So let's talk about rate constants now!

• First, we will review reaction rates and look at the definition of rate constant.
• Then, we will look at the units for the rate constant and the equation for rate constant.
• After, we will solve some problems involving rate constant calculations.

Rate Constant Definition

Before diving into the rate constant, let's review reaction rates and rate laws.

The reaction rate is directly proportional to temperature, so when temperature increases, the reaction rate becomes faster than before! This is because the more energy the reaction mixture has, the quicker the particles move around, successfully colliding with others more frequently.

Two other important factors that affect reaction rates are concentration and pressure. Similar to the effects of temperature, an increase in concentration or pressure will also lead to an increase in the rate of the reaction.

The equation for the instantaneous reaction rate can be expressed as a change in product concentration over a series of very short time intervals, for example over 10 seconds. Since the concentrations of products increase with time, the reaction rate in terms of products will be positive. On the other hand, if the instantaneous reaction rate is expressed in terms of reactants, because the concentrations of reactants decrease with time, the reaction rate will be negative.

$$ \text{aA + bB}\longrightarrow \text{cC + dD} $$

$$ \text{Reaction rate} = \text{ }\color {red}- \color {black}\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta \text{t}} = \text{ } \color {red} - \color {black}\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta \text{t}} = \text{ } \frac{1}{c}\frac{\Delta[\text{C}]}{\Delta \text{t}} = \text{ } \frac{1}{d}\frac{\Delta[\text{D}]}{\Delta \text{t}} $$

Let's look at an example. Suppose that you are dealing with the chemical reaction below. What would be the reaction rate of N₂?

$$ 2\text{ NH}_{3}(\text{g})\text{ }\rightleftharpoons \text{N}_{2} (\text{g})\text{ + 3 H}_{2}\text{(g)} $$

This is fairly simple to answer. All we need to do is look at the reaction and apply the equation for the instantaneous reaction rate! So, for N₂, the instantaneous reaction rate would be \( \frac{1}{1}\frac{\Delta[\text{N}_{2}]}{\Delta \text{t}} \), where, Δ[N₂], is the change in concentration (Final concentration - Initial concentration), and Δt is a very short time interval.

Now, what if you were given the same exact chemical reaction and were told that the instantaneous reaction rate of N₂ is equal to 0.1 M/s? Well, we could use this instantaneous reaction rate to find the instantaneous reaction rate of H₂! Since 3 moles of H₂ are produced for every 1 mole of N₂, then the reaction rate for H₂ will be three times that of N₂!

The second topic we need to review is rate law. Rate laws must also be determined experimentally, and its general equation for a power rate law is as follows:

$$ \text {Rate} = \color {#1478c8}k \color {black}[\text{A}]^{\text{X}}[\text{B}]^{\text{Y}}... $$

Where,

• A and B are reactants.

• X and Y are the reaction orders of the reactants.

• k is the rate constant

When it comes to reaction orders, the greater the value, the more that a change in the concentration of that reactant will affect the overall reaction rate.

• Reactants whose exponents (reaction orders) equal zero will not have an effect on reaction rates when their concentration is changed.

• When the reaction order is 1, doubling the concentration of the reactant will double the reaction rate.

• Now, if the reaction order is 2, if the concentration of that reactant gets doubled, the rate of reaction will be quadrupled.

For example, the experimentally determined rate law for a reaction between NO and H₂ is \( \text{Rate = }k[\text{NO}]^{2}[\text{H}_{2}]^{1} \). By adding the reaction orders, we can determine the overall reaction order of the rate law expression, which is 3 in this case! Therefore, this reaction is third-order overall.

$$ 2\text{ NO (g) + 2 H}_{2}\text{ (g)}\longrightarrow\text{ N}_{2}\text{ (g) + 2 H}_{2}\text{O (g)} $$

Now, take another look at the rate law equation above. Notice that there is a rate constant (k) present in its formula! But what exactly does it mean? Let's take a look at the definition of rate constant.

Rate Constant Units

Rate constant units vary based on the order of reactions. In zero-order reactions, the rate law equation is Rate = k and the unit of rate constant in this case is, \( \text{mol L}^{-1}\text{s}^{-1} \).

For first-order reactions, Rate = k[A]. The constant rate unit, in this case, is \( \text {s}^{-1} \). On the other hand, second-order reactions have a rate law of, Rate = k[A][B], and rate constant unit of. \( \text{mol}^{-1}\text{L}\text{ s}^{-1} \).

Rate Constant Equation

Depending on the reaction order we are dealing with, the equation to calculate the rate constant differs. Zero-order reactions are by far the easiest ones to solve for the rate constant because k is equal to the rate of the reaction (r).

$$ k = r $$

In the case of a first-order reaction, k will be equal to the rate of the reaction divided by the reactant concentration.

$$ k = \frac{r}{[A]} $$

Now, for second and third-order reactions, we would have the rate constant equations \( k = \frac{r}{[A][B]} \) and \( k = \frac{r}{[A]^{2}[B]} \), respectively.

First Order Rate Constant

To better understand the rate constant, let's talk about the first-order reactions and first-order rate constant.

When a kinetic plot is done for a first order reaction, the kinetic graph of ln[A]_t versus t yields a straight line with a slope of negative k.

Figure 2. ln [A] vs. time graph for a first-order reaction, Isadora Santos - StudySmarter Originals.

Rate Constant Calculations

Lastly, let's walk through how to do calculations involving rate constant, similar to what you will most likely encounter during the AP chemistry exam.

Solving a Multi-Step Problem

Sometimes analyzing a chemical equation doesn't tell the full story. As you should be aware, final chemical equations are usually the overall chemical equations. This means that there may be more than one step that produces the overall equation. For example, take the following overall chemical equation, where each step is fully written out, including how fast each step relatively occurs.

$$ 1. \text{ NO}_{2}\text{ + NO}_{2}\longrightarrow \text{NO}_{3}\text{ + NO } (slow) $$

$$ 2. \text{ NO}_{3}\text{ + CO}\longrightarrow \text{NO}_{2}\text{ + CO}_{2}\text{ } (fast)$$

$$ \rule{8cm}{0.4pt} $$

$$ \text{ NO}_{2}\text{ + CO}_{2}\longrightarrow \text{NO}\text{ + CO}_{2}\text{ } $$

As you can see, the overall chemical equation is found by canceling the common reactants and products. This applies to the entire system of chemical equations. (For example, the NO₂ in the reactants of step 1 cancels the NO₂ in the products of step 2, which is why NO₂ doesn't appear in the products of the overall reaction.) But how would you figure out what the rate law is for a problem like this? Take a second to think about what determines how fast this reaction occurs.

Intuitively, the overall reaction is only as fast as its slowest step. This means that the overall rate law for this reaction would be its slowest step, which would be Step 1. This also means that Step 1 would be the rate-Determining step. As for solving the rate constant, we now just follow the same process we have before. We need to set up a rate law equation using the rate-determining step, and then solve for k.

$$ \text{Rate = }k[\text{NO}_{2}][\text{CO}_{2}] $$

$$ k = \frac{\text{Rate}}{[\text{NO}_{2}][\text{CO}_{2}]} $$

Solving an Experimental Problem

As mentioned earlier in this lesson, chemists have to experimentally determine a chemical equation's unique rate law. But how do they do this? As it turns out, the AP test has problems that are just like this.

For example, let's say that we have chlorine gas reacting with nitric oxide, and we want to determine the rate law and rate constant from the following experimental data. How would we do this? Let's take a look!

$$ 2 \text{ NO (g) + Cl}_{2}\text{ (g)} \rightleftharpoons \text{2 NOCl (g)} $$

In this type of calculation, the first step is to find the rate law. The basic rate law expression, in this case, can be written as:

$$ \text{Rate = }k [\text{NO}]^{X}[\text{Cl}_{2}]^{Y} $$

However, we don't know the reaction orders of the reactions, so we need to use the experimental data collected from three different experimental trials to find out what type of reaction order we are dealing with!

First, choose two trials where only one concentration changes. In this case, let's compare experiments 2 and 3. Experiment 2 used 0.10 M of NO and 0.20 M of Cl₂, whereas experiment 3 used 0.20 M of NO and 0.20 M of Cl₂. When comparing them, notice that doubling the NO concentration (from 0.10 M to 0.20 M) and keeping the concentration of Cl₂ constant causes an increase in the initial rate from 0.36 M/s to 1.44 M/s.

So, if you divided 1.44 by 0.36, you will get 4, which means that doubling the concentration of NO, quadrupled the initial rate from experiment 1. So, the rate law equation, in this case, will be:

$$ \text{Rate = }k [\text{NO}]^{2}[\text{Cl}_{2}]^{1} $$

Now that we know the rate law expression, we can re-arrange it to solve for the rate constant, \( k \)!

$$ k = \frac{\text{Rate}}{[\text{NO}]^{2}[\text{Cl}_{2}]} $$

$$ k = \frac{\text{1.44 M/s}}{[\text{0.20 M}]^{2}[\text{0.20 M}]} = \textbf {180} \textbf{ M}^{-2}\textbf{s}^{-1} $$

Hopefully, you now feel more confident when approaching problems involving rate constant. Remember: take your time with these sorts of calculations, and always double-check your work!