Zero Order Reaction

You've probably been taught that reaction rates always are affected by the concentration of the reactant(s). This makes sense, right? Reactions are all about collisions between molecules/atoms, so the more you have, the more collisions that can happen. You are likelier to bump into someone at a packed concert than in a quiet, sparse library. However, that isn't always the case. That's where zero-order reactions come in.

Zero-order reactions are not dependent on concentration. While these types of reactions are on the rarer side, it is still important that we understand how they function. In this article, we will be looking at the definition of a zero-order reaction to see what's really going on.

• This article is about zero-order reactions.
• We will learn what a zero-order reaction is and see some examples.
• Next, we will derive the equation for a zero-order reaction.
• Then we will look at the graphs for the change in concentration.
• Lastly, we will cover half-life for zero-order reactions.

Zero-order Reactions Definition

Now, zero-order reactions aren't entirely independent of concentration. Obviously, a reaction cannot continue if the concentration of the reactant(s) is zero. It is the rate that is independent, not the reaction itself.

There are two conditions that can lead to a zero-order reaction:

• The first is when a small portion of the reactant molecules are in the proper location/state to react, and once that portion is used up, it is immediately replenished from the total "pool" of molecules.

Think of it like standing in line for a roller coaster. Only a small portion of patrons can ride it at one time. Once those people go on the coaster, the next group of people in line get on. Essentially, the concentration participating in the reaction is always the same, even though the total concentration is changing.

• The second is when a small concentration of reactant reacts with a heterogeneous catalyst or a reactant with a much larger concentration.

For catalytic reactions that are zero-order, the rate depends on how much the catalyst can handle (k) rather than how much reactant is added. This is a similar situation to our roller coaster example. Think of it this way, the roller coaster itself is the catalyst, and it only has so many "seats". No matter how many people are in line, the capacity isn't going to change.

Zero-order Reaction Example

Here is an example of a heterogeneous catalytic reaction:

$$ 2N_2O \xrightarrow [high pressure] {hot Pt} 2N_2 + O_2 $$

Here, the platinum (Pt) plate is acting as a catalyst for the nitrous oxide (N₂O). The plate only has a set number of binding sites, so the concentration of nitrous oxide doesn't matter. It is also important to remember that the formulas for the rate equations do not include the presence of the catalyst. A catalyst is neither a reactant nor a product, since it is present at both the beginning and end of the reaction. We note that rate equations are only dependent on reactants and are typically written to express quantities in terms of changes in the concentrations of reactants.

The experimentally determined rate law for this reaction, in which the concentration of nitrous oxide (N₂0) is low, would be: $$\text{rate}=k$$, where the rate constant (k) is characteristic of the platinum plate (Pt) catalyst.

Another example of a zero-order reaction is a reaction catalyzed by an enzyme.

The equation for this process looks like this: $$ E + S \rightarrow ES \rightarrow P$$

where, E, symbolizes the enzyme, S, is the symbol substrate and, P, symbolizes the product. Like with the platinum plate, the enzyme has a limited number of sites. In addition, the total number of enzyme molecules may be limited in comparison to the number of substrate molecules. Since enzyme availability is what limits the rate, the concentration of the substrate doesn't matter if kept at a relatively low concentration. Again, the rate law for this type of reaction would be: $$\text{rate}=k$$, where the rate constant (k) is characteristic of the enzymatic catalyst.

Lastly, "zero-order" can refer to the whole reaction or just one of the reactants. In the case of a reactant, another reactant would be in such an excess, that its change in concentration is negligible. Here is an example: $$ CH_3I_{(aq)} + H_2O_{(l)} \rightarrow CH_3OH_{(aq)} + H^+_{(aq)} + I^-_{(aq)}$$

This reaction is taking place in water (aqueous solution), so the concentration of H₂O is much higher than the concentration of CH₃I. This means that as the reaction proceeds, the change in [H₂O] is negligable, making it zero-order in H₂O (but first-order overall).

Zero-order Reaction Formula

Since the rate is dependent on the rate constant, we need to have a formula we can use to calculate it. This formula is: $$[A]=-kt+[A]_0$$

where, [A] is the concentration of a given reactant and [A]₀ is the initial concentration of the reactant, A, in the reaction. Even though with zero-order reactions, the rate is independent of the concentration of reactants, nevertheless the concentration of the reactant, A, will still be changing over time. If the concentration of the reactant, A, meets either of the conditions for zero-order reaction we can use this change in the concentration of the reactant, A, to calculate the rate constant (k). Before we get to look into this formula further, let's see where this equation came from.

Zero-order Reaction Derivation

Let's walk through the derivation:

We first set up our equation so that the rate is equal to the change in [A] over time. To do this, we use the expression \(\frac{d}{dt}\), which represents the change in a variable (here [A]) over time. For our purposes, this is the expression for a derivative that we will be using. $$\text{rate}=-\frac{d[A]}{dt}=k$$

The rate is negative, since the concentration of a reactant decreases over time as it forms the product(s); note that the rate constant is actually a negative number, but we just use the symbol, k, instead of, -k, for convenience.

Now we multiply each side by -dt, so that both sides take a differential form. $$d[A]=-kdt$$

the differential forms in the above equation are: d[A] and dt. The reason we want differential forms is because this makes the next step, integration, much easier. We note that for any differential, d(something), integration gives: $$ \int d \left( x \right) =x$$

Next, we need to set up the integral, so we can evaluate the differential. Integration is the opposite of derivation, just like how division is the opposite of multiplication.

Often, when we integrate something, we want to integrate for a starting point and an ending point. These starting and ending points are called the bounds of the integral. These "bounds" are the minimum and maximum values of the variable we want to integrate. The variables we integrate are whatever is immediately after d, which is [A] and t, in our case. For [A], our bounds are [A]₀ (initial concentration) and [A] (concentration at a given time). For t, our bounds are 0 and t (current time).

So when we integrate both sides, it looks like this. $$ \int_{[A]_0}^{[A]} d[A]_0=-\int_{0}^{t} kdt$$

Now for solving an integral. When take the integral, we integrate whatever is to the left-hand side of the differential, d(something). So that would be 1 and k respectively (when there is no number in front, we assume it as 1). These are both constants, so we follow this formula $$\int_{x}^{y} cdt=yc-xc$$

Using this formula, our solved integration looks like this:

$$ \int_{[A]_0}^{[A]} d[A]_0=-\int_{0}^{t} kdt$$

[A] - [A]₀ = -(kt - 0•k)

[A] - [A]₀ = -kt

[A] = [A]₀ - kt

We rearranged the equation to match the equation of a line (y = mx + b). This means that when we graph the equation, the slope is equal to, -k.

Zero-order Reaction Units

For each type of ordered reaction (zero, first, second), the rate constant has different units. No matter what, the rate will always be in units of M/s (molarity/second or moles/second*liters). In this case, the rate constant will also be in units of M/s since it is the only variable present.

In the equation we just derived, the units are the same. The time, T, is in units of seconds, and the change in the concentration of a reactant, [A], and the initial concentration of the reactant, [A]₀, is in units of molarity, M, so it makes sense that the rate constant, k, would be in units of M/s.

Zero-order Reaction Graphs

To get a better understanding of these reactions, let's graph one. Let's say we had 0.6 M of HI gas. The decomposition of hydrogen iodide follows this equation: $$2HI_{(g)} \xrightarrow [high pressure] {Au} H_{2\,(g)} + I_{2\,(g)} $$

When we graph the concentration over time, we get this graph:

As time increases, the concentration decreases linearly. StudySmarter Original

As our equation suggests, there is a linear relationship between concentration and time. Only zero-order reactions will have this type of relationship. Using this relationship, we can calculate the rate constant and the concentration at a given time.

Due to the linear relationship, we can also calculate, k, using just two data points:

When we graph a zero-order catalytic reaction, the graph is slightly different.

At a certain point, the graph changes from a line to a curve, then hits a point where the concentration doesn't change. StudySmarter Original

The graph is the same until the concentration gets very low, then it trails off and becomes stable. That is because this is the point where the catalyst has accepted as much of the reactant as it can. No more of the reactant can be bound at this point, so we see a flat line. At this point, the reaction is no longer zero-order. It is only zero-order until that point.

Half-life for zero-order reactions

Something interesting happens when we look at the half-life for a zero-order reaction.

$$\begin {align}[A]&=-kt+[A]_0 \\ \frac{1}{2}[A]_0&=-kt+[A]_0\,\text{(substituted half-life eq.)} \\ -\frac{1}{2}[A]_0&=-kt \\ \frac{[A]_0}{2k}&=t_{\frac{1}{2}}\end {align} $$

Notice that the half-life equation for a zero-order reaction is dependent on concentration. How can this be? Well, as we saw in our graphs, the concentration is changing over time. The key thing to remember is that time and rate are not the same.

Back again to our roller coaster example: The size of the crowd will determine how long it takes for the line to get smaller. However, the rate that it takes for the ride to finish is fixed and independent of the length of the line. A line of 50 people will be cut down much faster than that of 500 people, but the ride is always the same speed.

As the rate of reaction (technically it's the rate constant, but there is not really a difference here) increases, the time will get shorter. However, increasing the initial concentration makes the time longer.

Let's work on an example:

Be careful when plugging in your values. Make sure you are using the initial concentration.