Reaction Orders - Online Tutor, Practice Problems & Exam Prep

Reaction orders describe the relationship between reaction rates and substrate concentrations. Zero order reactions show no dependence on substrate concentration, often occurring when enzymes are saturated. First order reactions are directly proportional to one substrate concentration, while second order reactions depend on two substrate concentrations or the square of one. Pseudo first order reactions appear first order due to one substrate being in excess. Understanding these concepts is crucial for analyzing enzyme kinetics and reaction mechanisms.

Reaction Orders

In this video, we're going to begin our discussion on reaction orders. Recall in our previous lesson videos we talked about the rate law and we said that the rate law is just another way to express the reaction velocity or \( v \) here. The rate law states that the reaction velocity is equal to the rate constant \( k \) times each initial reactant concentration raised to the power of the order. This reaction order is exactly what we're going to be focusing on in this video. The reaction order is a proportionality relationship between the reaction rate \( v \) and each initial substrate concentration or each initial reactant concentration. Remember that substrates are the reactants in enzyme-catalyzed reactions. Recall from our previous lesson videos we said that the reaction orders are frequently going to be the coefficients. However, the coefficients of the substrates do not always equal the reaction orders; thus, the reaction orders need to be experimentally determined. It turns out that the substrate coefficients will only equal the reaction order for elementary reactions. Elementary reactions are just those reactions with only one transition state. If the reaction only has one transition state, then it's considered an elementary reaction, and we know that the substrate coefficients will equal the reaction order. However, if it's a non-elementary reaction, meaning that it's not elementary, then we know this will not be true, and the substrate coefficients will not equal the reaction order. It's important to note that the overall reaction order, regardless of whether the reaction is non-elementary or elementary, will always be equal to the sum of the individual reaction orders for all substrates, and we'll see how that works better down below in our example.

In this example, we're going to determine the overall reaction order for the following reactions. In the reaction on the left, notice that we have ozone gas plus oxygen gas being converted into 2 moles of oxygen dioxide gas. This is an elementary reaction, so we know that this reaction here only has one transition state, represented by the single arrow here in this reaction. If we were to draw the rate law for this particular reaction, notice that the rate law is expressed here as the reaction velocity being equal to the rate constant \( k \) times the concentration of the substrates; here, ozone gas and oxygen. Since this is an elementary reaction, we know that the substrate coefficients will equal the reaction orders. We notice that the reaction orders placed here match the substrate coefficients of 1. Since we have the first reaction order as 1 and the second reaction order as 1, we know that \( 1 + 1 \) will give us the overall reaction order for this reaction.

In the reaction on the right, notice that we have nitrogen dioxide gas plus carbon monoxide gas giving us nitrogen monoxide gas and carbon dioxide gas. This is a non-elementary reaction, which means that it has more than one transition state. In this reaction, we can see this because there is more than one reaction arrow, indicating some kind of intermediate that forms this overall reaction here. Just because there are two reaction arrows here doesn't mean that this reaction will always be expressed with two reaction arrows. Sometimes we'll see that this will be expressed with one reaction arrow, making it look like it's an elementary reaction when, in reality, it's actually not. It's broken up into two smaller reactions. Thus, we can't rely on the arrows to let us know if this is a non-elementary reaction or not. However, moving forward in our course, we know that all of the enzyme-catalyzed reactions we're going to assume are elementary reactions with only one transition state. Looking at this non-elementary reaction, if we were to draw out the rate law for this reaction, notice that the rate law is equal to the reaction velocity, which is equal to the rate constant \( k \) times the initial concentration of the reactants; here, nitrogen dioxide and carbon monoxide. Because this is a non-elementary reaction, the substrate coefficients do not necessarily equal the substrate reaction order, as we can see here. The reaction order for this nitrogen dioxide gas is 2, but the coefficient here is not 2. The same applies for the carbon monoxide gas; its reaction order is 0, even though its coefficient is 1. Since they don't match, we know that's acceptable for a non-elementary reaction. However, even though this is a non-elementary reaction, the overall reaction order is always going to be the sum of the individual reaction orders for all the substrates. Thus, we have a reaction order of 2 plus a reaction order of 0, so \( 2 + 0 \) is still 2, which means that the overall reaction order is still second order. Notice that the overall reaction order for this elementary reaction is second order, as is the overall reaction order for this non-elementary reaction. The overall reaction order does not necessarily give us information about the individual reaction orders for each reactant, which is why individual reaction orders must be experimentally determined.

Professors are not going to expect us to use data to determine the reaction orders for particular enzyme-catalyzed reactions that we'll be looking at. Next, we're going to talk about the three common overall reaction orders: 0 order reactions, 1st order reactions, and second order reactions. Moving forward in our course, we're going to discuss each of these common overall reaction orders in their separate videos. We'll start off by talking about 0 order reactions. I'll see you guys in that video.

The rate law for an elementary and/or nonelementary reaction is _________________________:

• A. always determined from the balanced chemical reaction.
• B. the rate law for the fastest step in the reaction mechanism.
• C. experimentally determined.
• D. always 2nd order overall if there are 2 reactants.

Reaction Orders

Alright, so now that we know a little bit about reaction orders, in this video we're going to talk about reactions that have an overall reaction order of 0 or zero-order reactions. Zero-order reactions are reactions where the substrate concentration has zero effect on the reaction rate or the reaction velocity. If the substrate concentration has zero effect on the reaction rate, that means that even when we change the substrate concentration, the reaction rate will not change. Enzyme-catalyzed reactions will actually exhibit zero-order kinetics when an enzyme is completely saturated with substrate, and we'll be able to see an example of that down below in our image.

The zero-order rate constant units are specifically molarity times inverse seconds, which is expressed right here. But I really don't want you guys to focus on the units of the rate constant just yet because later in our course, in a different video, we're going to focus specifically on all of the rate constant units. If these units are a bit confusing to you right now, that's okay because we're going to talk about the units later in a different video.

Notice here we have a specific reaction where we have reactant A being converted into product B. We're told that this is a zero-order reaction, which means that the overall reaction order is 0. The rate law for this reaction, which is another way to express the reaction rate or the reaction velocity, is equal to k times the concentration of the reactant or the substrate, raised to the reaction order. Since we're told that this is a zero-order reaction, the reaction rate or the reaction velocity will be equal to the rate constant k.

Notice that we have the reaction rate on the y-axis and the substrate concentration on the x-axis, specifically a zero-order reaction. The value of this reaction rate is exactly equal to the rate constant value.

In this graph, what we have is the initial reaction rate on the y-axis, and the substrate on the x-axis when we plot this data here. There comes a point in the reaction where it displays zero-order kinetics, as seen by the horizontal line. By changing the substrate concentrations in this area where it's displaying zero-order kinetics, the reaction velocity does not change.

This means that within this region where zero-order kinetics are reached, enzymes, when they're saturated with substrate, display zero-order kinetics. That concludes our lesson here on zero-order reactions, and we'll be able to take a look at an example of this in our next video. So I'll see you guys there.

Reaction Orders Example 1

Alright. So, in our last lesson video, we said that zero-order reactions are reactions where the substrate concentration has zero effect on the reaction rate or the reaction velocity. And so here, we're showing you guys an example of zero-order kinetics. Notice on the left over here, we have a bunch of cars that are lined up, and these cars represent our substrate. And on the right, what we have here is a one-lane bridge that represents our enzyme. And so, the cars crossing over to the other side represent the reaction. And so up above, we have a question that's asking, does increasing the number of cars, or does increasing the substrate concentration also increase the rate that the cars cross the one-lane bridge? And so, recall that here we have so many cars that are lined up to cross this one-lane bridge. Essentially what we're saying is that this one-lane bridge, this enzyme here, is pretty much already saturated with substrate. And so, when it's saturated with substrate, we know that it's going to display zero-order kinetics and that means that increasing the substrate concentration will not increase the rate at which the cars will cross this bridge. And so even if we were to add 10 more lanes of cars over here, that would not increase the rate that all of these cars would cross this one-lane bridge because remember, this one-lane bridge is already operating at its maximum rate, and adding more cars will not increase the rate that the cars will cross the bridge. So essentially, what we're saying is that the answer to this problem is no. That increasing the substrate concentration or the number of cars will not increase the rate that the cars will cross the one-lane bridge and that is because the enzyme or the bridge is already saturated with substrate. Now, one of the only ways that we can actually increase the rate of a reaction when the enzyme is saturated with substrate is by adding more enzyme. And so, if we were to add more one-lane bridges, then that would increase the rate that the cars would cross the bridge, because there are now more lanes, and there are more cars that can be crossed within a smaller amount of time. And so, this concludes our example of zero-order kinetics, and I'll see you guys in our next video.

Reaction Orders

Alright, so now that we've covered 0 order reactions, in this video we're gonna talk about first order reactions. First order reactions are reactions where the rates of the reaction are directly proportional to only one substrate concentration, and this includes unimolecular reactions or reactions that only have 1 molecule, "uni" meaning 1. These reactions involve only a single substrate. For example, converting substrate a into product b. Now the first order rate constant units are specifically inverse seconds. But again, I don't want you guys to focus on the units of the rate constant just yet because later in our course, we're going to have a specific video dedicated to helping you guys understand rate constant units. So if this is confusing to you guys right now, that's okay because later we'll talk more about it to clear it up. So down below in our example, what we have is an example of first order kinetics. Notice that we're looking at a unimolecular reaction that involves only 1 molecule and we're converting substrate a into product b, and we're told that this is a first order reaction. When we consider the rate law, which is another way to define the reaction rate or the reaction velocity, we know that it's going to be equal to the rate constant k times the concentration of the only substrate here, which is substrate a. And, of course, because we're told this is a first order reaction, we know that the reaction order must be 1 for this substrate. If we take a look at this graph over here where we have the reaction rate on the y-axis and the substrate concentration on the x-axis, notice that instead of getting a horizontal line like what we saw for 0 order reactions, this time what we're seeing is a diagonal line that's going up like this and that's because the rate of the reaction is directly proportional to the substrate concentration. So we can say here that the rate of the reaction is directly proportional to the substrate concentration. This symbol means directly proportional. All that means is that as the substrate concentration increases, so does the reaction rate. If we take a look at this other graph over here of a typical enzyme-catalyzed reaction, where we have the initial reaction rate on the y-axis and the substrate concentration on the x-axis. Notice again from our last lesson video we already know that if we increase the concentration to a point where it's saturating the enzyme, then the enzyme will be displaying 0 order kinetics where the rate of the reaction does not depend on the substrate concentration. And where we have a lower substrate concentration in the reaction notice that, the reaction, the enzyme-catalyzed reaction is displaying first order kinetics where the rate of the reaction is directly proportional and depends on the substrate concentration. So you can kinda see that it's going up in this, diagonally in this fashion similar to how we can see it's going diagonally over here in this fashion. And so, we'll be able to apply these concepts here on first order reactions as we move along through our course. But for now this concludes our lesson here on First Order Reactions, and I'll see you guys in our next video.

Reaction Orders

So now that we've refreshed our memories on first-order reactions in our last lesson video, in this video, we're going to focus on second-order reactions. Second-order reactions are reactions where the rates of those reactions are directly proportional to 2 substrate concentrations, instead of just one substrate concentration like we saw with first-order reactions. What's also important to note is that second-order reaction rates could also be proportional to the square of only one substrate concentration. Essentially, there are 2 common ways to express the rate law for a second-order reaction. It could be expressed as being directly proportional to 2 substrate concentrations or directly proportional to the square of only one substrate concentration, and we'll be able to see how that works better down below when we talk about the 2 common rate laws for a second-order reaction.

Second-order reactions include bimolecular reactions that involve 2 substrates. For example, this reaction right here where we have substrate A interacting with substrate B to form product C. The second-order rate constant units for k are actually in units of inverse molarity and inverse seconds.

With second-order reaction rates, there are really 2 different ways to express the rate laws for a second-order reaction. The rate law says that the reaction velocity v is equal to the rate constant k times the concentration of the substrates or the reactants, and we have 2 reactants or substrates, A and B. So, it's going to be multiplied by those substrate concentrations and raised to the reaction order. This is one way that the rate of a second-order reaction can be expressed. Another way is where the rate is proportional to the square of only one substrate concentration. In this case, the rate law would say that the reaction rate v is equal to the rate constant k times the concentration of substrate A squared, and the concentration of B is zero order with respect to the reaction rate.

Notice that we have the graph specifically for second-order reactions. Instead of getting a horizontal line like we got for zero-order reactions, and instead of getting a straight line with a positive slope like we got for first-order reactions, in a second-order reaction, what we get is this exponential curve when we plot the reaction rate v on the y-axis and the substrate concentration on the x-axis.

Also, there is a reaction within our cells showing the breakdown of glycogen. Glycogen, recall, is a carbohydrate made up of a bunch of individual glucose molecules linked together. The glycogen molecule interacts with an inorganic phosphate molecule to produce glucose 1 phosphate and glycogen with one less glucose molecule. This reaction separates the green structure from the blue structure in the visual representation provided.

If we were to write the rate law for this reaction, we know that the rate law is another way to express the reaction rate or the reaction substrates. We have substrates glycogen and inorganic phosphate. The concentrations are going to be raised to their reaction orders. We always assume with the reactions that we're going to cover moving forward in this course, that these are simple reactions, so the coefficients are the reaction orders. Both coefficients are just a one, raising both to the power of 1. Adding these individual reaction orders, 1+1, shows us that this is a second-order reaction. This concludes our lesson on second-order reactions, and we'll be able to get more practice utilizing these concepts as we move forward in our course. I'll see you guys in our next lesson video.

What is the overall reaction order for the following rate law? v = k [A] ¹ [B]¹ [C]⁰

• A. Zero order.
• B. 1st order.
• C. 2nd order.
• D. 3rd order.

Reaction Orders

Alright. So now that we've covered both first and second order reactions, in this video, we're going to briefly talk about pseudo first order reactions. And so pseudo is just a word that pretty much just means fake. And so pseudo first order reactions are just fake first order reactions, and really all they are are reactions that appear to be first order reactions. But in reality, they're actually second order reactions and really other than that, there's nothing special about them. And so, pseudo first order reactions can occur when the concentration of one substrate is much greater than the concentration of the other. And so for instance, if the concentration of B is much, much, much greater than the concentration of A, then we could have ourselves a pseudo first order reaction. And so remember that when the concentrations of a substrate are really, really, really high, like the concentration of B here, then they're going to be at saturating concentrations. And when the enzyme is saturated with substrate, it's going to have 0 order kinetics and it's going to appear that the concentration of B does not affect the reaction rate when indeed it might actually do that in a second order reaction. And when the concentration of A is so small in comparison to B, this makes the substrate A the limiting reagent. And as the limiting reagent, it's going to appear that only the concentration of A alone will dictate the rate of the reaction when in reality, it's not just the concentration of A, it's also the concentration of B. And so, if we take a look up above at the graph from our second order kinetics reaction that we talked about in our previous lesson video, what we're saying is that a pseudo first order reaction is really just it's actually just a second order reaction, and there's nothing special about that. However, when we graph a pseudo first order reaction, it appears to look just like a first order reaction. And that's because, again, the concentration of one substrate is much greater than the other. And so even though with a pseudo first order reaction the reaction appears to be first order, we know that it's actually a second order reaction and so this can sometimes cause a little bit of complications and challenges for a biochemist when they're trying to study a reaction. So they have to vary the concentrations to make sure that these conditions here don't exist because otherwise, it could fake out a biochemist. And so that concludes our brief lesson on pseudo first order reactions and we'll be able to get some practice as we move along with our practice video. So, I'll see you guys there.

Which of the following options is true for a reaction with the provided rate law: v = k [NO]² [O₂]

• A. The reaction has an overall order of 3.
• B. The reaction is first order with respect to the reactant O₂.
• C. The reaction is endothermic.
• D. The reaction has an overall order of 2.
• E. a & b are true.
• F. b & c are true.

Consider the nonenzymatic elementary reaction from A → B. When the initial [A] = 20 mM, the reaction velocity is measured as 5 μM/min. Determine the reaction order and calculate the rate constant for the reaction.

• A. 1st order reaction; k= 100 min^-1.
• B. 1st order reaction; k= 0.25 min^-1.
• C. 1st order reaction; k= 2.5 x 10^-4 min^-1.
• D. 1st order reaction; k= 4.0 x 10³ min^-1.
• E. 2nd order reaction; k= 100 min^-1.
• F. 2nd order reaction; k= 0.25 min^-1.
• G. 2nd order reaction; k=2.5 x 10^-4 min^-1.
• H. 2nd order reaction; k= 4.0 x 10³ min^-1.

Consider the nonenzymatic elementary reaction A  B.  When the [A] = 20 mM, the reaction velocity is measured as 5 µM of 'B' produced per minute.  Calculate the rate constant for the reaction. Hint: Consider the rate law.

• A. 6.0 x 10²min^-1
• B. 2.5 x 10^-4min^-1
• C. 3.7 x 10^-5min^-1
• D. 9.1 x 10^-2min^-1

The hypothetical elementary reaction 2A --> B + C has a rate constant of 10-6 M-1s-1.  What is the reaction velocity when the concentration of A is 10 mM?

• A. 8 x 10^-3Ms^-1
• B. 2 x 10^-7Ms^-1
• C. 4 x 10^-5Ms^-1
• D. 1 x 10^-10Ms^-1

Here’s what students ask on this topic:

What is the difference between zero order and first order reactions?

Zero order reactions are characterized by a reaction rate that is independent of the substrate concentration. This means that even if you change the concentration of the substrate, the reaction rate remains constant. This often occurs when an enzyme is saturated with substrate. The rate law for a zero order reaction is given by:

$v=k$

First order reactions, on the other hand, have a reaction rate that is directly proportional to the concentration of one substrate. This means that if you double the concentration of the substrate, the reaction rate also doubles. The rate law for a first order reaction is:

$v=k{\mathrm{[A]}}_1$

How do you determine the overall reaction order?

The overall reaction order is determined by summing the individual reaction orders of all the reactants in the rate law. For an elementary reaction, the reaction orders are equal to the stoichiometric coefficients of the reactants. For non-elementary reactions, the reaction orders must be determined experimentally. For example, in a reaction where the rate law is:

$v=k{\mathrm{[A]}}_1{\mathrm{[B]}}_2$

The overall reaction order is 1 + 2 = 3.

What are pseudo first order reactions?

Pseudo first order reactions are reactions that appear to be first order but are actually second order. This occurs when the concentration of one reactant is much greater than the other, making the reaction rate primarily dependent on the concentration of the limiting reactant. For example, if the rate law for a second order reaction is:

$v=k{\mathrm{[A]}}_1{\mathrm{[B]}}_1$

and [B] is in large excess, the reaction rate will appear to depend only on [A], making it pseudo first order.

What are the units of the rate constant for zero, first, and second order reactions?

The units of the rate constant (k) vary depending on the reaction order:

• Zero order: $k=M{s}^{-1}$
• First order: $k=s^{-1}$
• Second order: $k=M^{-1}{s}^{-1}$

How do you experimentally determine reaction orders?

Reaction orders are experimentally determined using methods such as the method of initial rates or the integrated rate law method. In the method of initial rates, the initial reaction rate is measured for different initial concentrations of reactants. By analyzing how the rate changes with concentration, the reaction order can be deduced. For example, if doubling the concentration of a reactant doubles the rate, the reaction is first order with respect to that reactant. The integrated rate law method involves measuring the concentration of reactants over time and fitting the data to integrated rate equations to determine the reaction order.