Rate Constants and Rate Law - Online Tutor, Practice Problems & Exam Prep

Rate constants, denoted as k, are crucial in determining the efficiency of chemical reactions, particularly in enzyme-catalyzed processes. Key constants include k₁, k_-1, k₂, and k_-2, with k₂ often being the rate-limiting step. During initial reaction stages, the initial velocity (v₀) can be simplified by ignoring k_-2, allowing for easier calculations. The rate law relates reaction velocity to initial reactant concentrations and the rate constant, facilitating the understanding of reaction dynamics in biochemical contexts.

Rate Constants and Rate Law

In this video, we're going to begin our discussion on rate constants and the rate law. I'm sure most of you have already covered rate constants and the rate law in your previous chemistry courses. A lot of the information in this video might sound familiar to you. In this video, we're only going to talk about the rate constants. But later, in a different video, we'll talk about the rate law. What I want you to recall from your previous chemistry courses is that every reaction has what's known as a rate constant, which can be abbreviated with the lowercase variable k. A rate constant is a constant positive value for a reaction that indicates the reaction rate efficiency or probability under very specific set conditions. The higher the value of the rate constant, the more likely it is that the reaction will be faster. It's important to note that when the rate constant is equal to 0, no reaction will occur or take place. The rate constant is a positive value, which means that the rate constant will never be negative. When we're considering a typical or a standard enzyme-catalyzed reaction, there are 4 rate constants that we will be considering: k₁, k_-1, k₂, and k_-2.

Starting with k₁, it is the free enzyme and the free substrate association rate constant that forms the enzyme-substrate complex. Notice that the free enzyme and the free substrate associating forwards to form the enzyme-substrate complex is a reaction that has a rate constant of k₁. Of course, k_-1 will be the exact opposite reaction or essentially the enzyme-substrate complex dissociation rate constant backwards to reform the free enzyme and the free substrate. Going from the enzyme-substrate complex and dissociating backwards to reform the free substrate and free enzyme, this is going to be k_-1. Now k₂ is going to be the enzyme-substrate complex dissociation rate constant forwards to form the product. Down below, you can see that here we have the enzyme-substrate complex and it can dissociate backwards, but it can also dissociate forwards to form the product and the free enzyme. This rate constant here will be k₂. Of course, k_-2 is going to be the exact opposite or essentially taking the enzyme, the free enzyme, and the free product, and associating them together to reform the enzyme-substrate complex. Taking the free product and the enzyme and going backwards to reform this enzyme-substrate complex will be k_-2.

What I want you to recall is that v is the variable that represents the reaction rate or the reaction velocity, which we already covered in our previous lesson videos. We know that typically when we're measuring the reaction velocity, we're looking at the change in the product concentration over the change in time. Both k₂ and k_-2 directly affect the change in the product concentration. k₂ will increase the product concentration, whereas k_-2 will decrease the product concentration directly. k_-1 and k₁ do not directly affect the product concentration; they only directly affect the concentrations of the enzyme-substrate complex and the free enzyme and free substrate. Because we're focusing on the change in the product concentration when it comes to reaction velocity, we're very interested in measuring these two rate constants right here. When there are 2 rate constants to consider, that makes calculating the reaction velocity a lot more complicated. If we focus on a reaction when it's at its initial stages very early on, then we're able to eliminate one of these rate constants, k_-2, and only have one rate constant that affects the product concentration. In our next lesson video, we're going to talk more about the idea of the rate constants that are going to be considered during the initial stages of an enzyme-catalyzed reaction. I'll see you guys in that video.

Rate Constants and Rate Law

So in our last lesson video, we said that there are 4 rate constants to consider at any given time during a typical enzyme catalyzed reaction. And those four rate constants are k1, k−1, k2, and k−2. Now, instead of focusing on the enzyme catalyzed reaction at any given time, if we instead focus on one specific period of time at the very, very beginning of the enzyme catalyzed reaction, essentially at initial stages of the enzyme catalyzed reaction, then there are only 3 rate constants for us to consider instead of 4. And so which of these 4 rate constants here can we ignore during the initial stages of an enzyme catalyzed reaction? Well, it turns out that it's actually the k−2 rate constant that we can ignore during the initial stages of an enzyme catalyzed reaction. And so, essentially over here we can put that k−2 can be ignored.

Free enzyme and free substrate are added to the reaction mixture, but they do not form any enzyme-substrate complex or any product. Thus, we can say that initially, at the very beginning of an enzyme catalyzed reaction, the product concentration is going to be 0, and so there's going to be no product. And so if there is no product, then that means that there's no product available to associate with the enzyme for this reverse reaction to occur. And so, that means that, at the initial stages of an enzyme catalyzed reaction, k−2 is ignored. Biochemists will focus on the initial reaction velocity or the v0 of an enzyme catalyzed reaction for this reason.

The initial reaction velocity is just the velocity at the very beginning of an enzyme catalyzed reaction during the initial stages, specifically where the rate constant k−2 is negligible and we can ignore it. Notice down below what we have are the initial enzyme catalyzed reaction conditions at the very beginning of the reaction, essentially, where the time is approximately equal to 0 since we're just a few microseconds into the reaction. And so, the product concentration just a few microseconds into the reaction is going to be equal to 0 as well.

Notice that the reverse rate constant k2 is not present here. It's notable that k−2 is the one that is negligible. In our course moving forward, we're always going to assume k2 is going to be the slowest step of the enzyme catalyzed reaction, thereby becoming the rate-limiting step. Recall from your previous chemistry courses that the slowest step, or the rate-limiting step, is always going to be the one that dictates the rate of the reaction. The overall reaction cannot occur any faster than its slowest step. Thus, k2 is what biochemists are measuring because with the initial reaction velocity, the k−2 variable is being ignored, meaning the change in the product concentration over the change in time is affected only by one rate constant instead of two like it is at any given time during the reaction.

Moving forward in our course, we're going to talk about how biochemists tend to focus on the initial reaction velocity at the very beginning of a reaction. It's this reaction velocity that is measurable to biochemists because it's affected only by one rate constant, which is a lot easier to consider and measure compared to considering two rate constants. This is a reminder that we're going to revisit these concepts later in our course when we discuss Michaelis-Menten enzyme kinetics and the assumptions. But for now, this concludes our lesson on the rate constants, and in our next lesson video, we'll talk about the rate law. So I'll see you guys there.

Which of the following rate constants is negligible for the initial velocity (V ₀) of an enzyme-catalyzed reaction?

• A. k₁.
• B. k_-1.
• C. k₂.
• D. k_-2.

Rate Constants and Rate Law

So now that we've brushed up a little bit on rate constants, in this video we're going to refresh our memories on the rate law from our previous chemistry courses. Recall that the rate law uses rate constants to help us calculate the reaction rate v. Recall from our previous lesson videos that the reaction rate or the reaction velocity v is literally just equal to the change in the product concentration over the change in time. The change in product concentration really just means the final product concentration minus the initial product concentration. The change in time just means the final time minus the initial time.

Imagine a scenario where the final concentrations of reactants and products are unknown. In that scenario, we could not use this method here to help us calculate the reaction rate or the reaction velocity because we are missing the final product concentration. What we can do is we can still calculate the reaction rate or the reaction velocity, but we need to use the rate law to do that. The rate law is just a mathematical relationship between the reaction rate v, the rate constant k, and each initial reactant concentration. We only need initial concentrations here and we do not need final concentrations. Typically, we know the initial concentrations of reactants because when biochemists are studying reactions in a lab, they only add the reactants and so they know exactly how much of the reactants they are adding. For the rate law, all we need to do is multiply the rate constant k by all of the initial reactant concentrations raised to the order, the reaction order. We'll talk a lot more about this reaction order later in our course in a different topic.

Down below, we have an example of the rate law, and notice we have the reaction velocity or the reaction rate, which we know is symbolized with the variable v. In our previous lesson videos, we said that this reaction rate or reaction velocity is just defined as the change in the product concentration over the change in time. But again, if we do not know the final concentration of the product, then that means that we cannot use this method. But, we can still use the rate law to help us get the reaction rate or the reaction velocity. The rate velocity v is equal to the rate constant, k, times the initial concentration of the reactants raised to their reaction order. If you have 2 reactants, then you'll have the rate constant times reactant number 1 concentration raised to the reaction order, times reactant number 2 concentration raised to the reaction order. This is all multiplication right here, and really, that is the rate law.

v = k ⁢ C 1 r ⁢ C 2 r

We're going to talk a lot more about these reaction orders later in our course in different videos, but it's important to know that these reaction orders must be experimentally determined. Moving forward, we're going to see that frequently the reaction orders are just equal to the coefficients of the reactants.

If we take a look at our example and try to determine the rate law for each of the simple reactions that are shown, you can see over here on the left we have a simple reaction where we're converting reactant A into product B. And if we're trying to determine the rate law, remember the rate law is always going to be the reaction velocity, v, is going to be equal to the rate constant k times the concentration of the reactants. In this reaction, we only have one reactant, so it's going to be times the concentration of a, and then it's going to be raised to the power of the order. As we said previously, frequently we'll see that the coefficients of the reactant will equal the reaction order. So the rate law for this simple reaction assumes the reaction order will be 1.

Now, if we do the same for this reaction over here, notice that this time, if we want to get the rate law, we have 2 reactants to consider. The rate law will be the reaction velocity v is equal to the rate constant k times the concentration of the first reactant a raised to the order of 1 times the second reactant b, the concentration of the second reactant b, raised to the coefficient as well, which is going to be a 1. This is the rate law for this simple reaction.

In our next lesson video, in our next practice video, we'll be able to get some practice applying the rate law concepts that we refreshed our memories on in this video, and then in our next lesson video, we'll be able to apply the rate law to enzyme catalyzed reactions. So I'll see you guys in those videos.

Calculate the reaction rate for the following simple reaction if k = 1.3 x 10 ^-1 M^-1s^-1, initial [A] = 4.0 x 10^-3 M, and the initial [B] = 6.0 x 10 ^-3 M (for simple reactions, assume coefficients are reaction orders):

A + B → 2C

• A. 2.13 x 10^-6 M.
• B. 3.12 x 10^-6 M/s.
• C. 1.32 x 10^-6 s.
• D. 3.12 x 10^-6 M^-1s^-1.

Rate Constants and Rate Law

Alright. So now that we brushed up a bit on rate constants and the rate law, in this video, we're going to talk a little bit about how to apply rate laws to enzyme-catalyzed reactions. And so recall from our previous lesson videos that when biochemists are studying the reaction velocity or the reaction rate, which can be expressed with the variable v, they tend to express the reaction velocity or rate as the change in product concentration over the change in time. But we also know from our previous lesson videos that the reaction rate or the reaction velocity here can also be expressed and rewritten as the rate law. And so we already covered the rate law in our last lesson video. So let's take a look at our example down below at our typical enzyme-catalyzed reaction shown here. And we know that this enzyme-catalyzed reaction is specifically at the initial stages, at the very, very beginning of the reaction, and we know that because notice that the reverse reaction here, expressed by k_-2, is not being included. And so because k_-2 is negligible, we know that this is going to be the initial enzyme-catalyzed reaction. And so let's write the rate law for the product formation step. And so the product formation step is going to be expressed with the rate constant k₂ since that's the one that leads to the product formation. And so the product formation rate law, we already know that the rate law is literally just going to be the reaction velocity. And in this case, because this is the very beginning of the reaction, this is going to be the initial reaction velocity, is going to be equal to the rate constant, which is going to be k₂ here, times the reactant concentration, and the reactant of this particular reaction right here is actually the enzyme-substrate complex. And so, what we can say is that the enzyme-substrate complex is our reactant for this particular product formation step. And then of course, because we're going to assume that all of our enzyme-catalyzed reactions are simple reactions, we're going to assume that the coefficient in front of our reactant is going to be the reaction order. And so, because it has no coefficient listed here, we'll assume that the coefficient is 1. And so this here is the rate law for the product formation step. And so we can apply all of these rate laws, each of these reactions here have their own particular rate laws, and in our next practice problem we'll be able to apply the rate laws to these rate constants as well. Now, what I want you guys to realize in addition to applying the rate laws to enzyme-catalyzed reactions is that there's only one rate constant in this expression right here that directly affects the change in the product concentration, and that is k₂. And so for enzyme-catalyzed reactions, this formation of the product will only depend on k₂, and that's because k_-2 again is negligible early on at the very beginning of the reaction. And k₁ and k_-1 alone do not affect the product concentration. So if we were to look at k₁ alone, notice that it doesn't directly affect the complex concentration and the free enzyme and free substrate concentration, and the same applies for k_-1. And so these two do not affect the product concentration. And so, essentially what we can say is that when biochemists are studying and measuring and plotting the reaction velocities of an enzyme-catalyzed reaction, they're really going to be plotting the initial reaction velocities or the v₀ for this particular product formation step. So when they're plotting their velocities, they're only plotting the velocity of this step right here and these steps over here are not going to be the ones that the biochemists tend to focus on. And remember that this step right here, we're always going to assume is the slowest step. So these two steps here are going to occur much faster than this step right here. And so over here, essentially what we can see is a graph where we have the product concentration on the y axis and the time on the x axis, and we've seen this graph before in our previous lesson videos. So we know that the curve is going to look like this, where the product concentration levels off. And we know that the slope of the line that's tangent to any point on this curve is going to represent the reaction velocity. And so, if we take a point that's very, very early on in our reaction here and we take the slope of the line that's tangent to this point, we will get the reaction velocity. And so this is early early on in our reaction where the time is 0, and so that's why we, this slope here is going to represent the initial reaction velocity. And really because we're measuring product concentration and only k₂ affects the product concentration, this reaction velocity that we see here is really just the reaction velocity for this product formation step k₂ step here. So that's something important to keep in mind as we move along in our course. But again, we're going to be able to get some more practice applying rate laws to enzyme-catalyzed reactions in our next practice video. So I'll see you guys there.

Write out the rate law equations for each association/dissociation indicated below.

a) Rate law for ES dissociation into E + P: V = ______________

b) Rate law for E + S association: V = ______________

c) Rate law for ES dissociation back into E + S: V = ______________

All right. So this practice problem wants us to write out the rate law equations for each association and each dissociation indicated below in this enzyme catalyzed reaction. Now, notice again that in this enzyme catalyzed reaction, the enzyme-catalyzed reaction is occurring very early on at the very beginning of our reaction, making it essentially the initial enzyme-catalyzed reaction. Looking at our first one here, the rate law for the enzyme substrate complex dissociation into the enzyme and product is essentially the rate law that includes \(k_2\). We know that rate laws are the reaction velocities times the rate constants, and the rate constant for this particular reaction is \(k_2\) times the concentration of the reactant, which in this case is the enzyme-substrate complex. We can express this as: v = k 2 • [ enzyme-substrate complex ] Here, the reaction order is assumed to be 1. This expression represents the rate law for the product formation step, effectively matching the enzyme substrate dissociation into enzyme and product. Moving on to the next rate law, for the enzyme and substrate association, we have the forward rate constant forming the enzyme-substrate complex, denoted by \(k_1\). The rate law equation becomes: v = k 1 • [ enzyme ] • [ substrate ] This rate law describes the association of enzyme and substrate to form the enzyme-substrate complex. Lastly, we have the rate law for the enzyme-substrate complex dissociation backwards into the free enzyme and substrate. For this reverse reaction, the rate constant is \(k_{-1}\), and the rate law is: v = k − 1 • [ enzyme-substrate complex ] This concludes the expression of the reaction velocities for each of these reactions within our enzyme-catalyzed reaction. It's important to apply this knowledge moving forward. I'll see you guys in our next video.

In a typical enzyme-catalyzed reaction, when & why is the rate constant k  _-2 negligible?

• A. At the very beginning of a reaction because the [S] & [P] are at equilibrium and not yet disturbed by the enzyme.
• B. Initially towards the beginning of a reaction because enzymes are getting off to a slow start.
• C. At the start of a reaction when [S] are at their highest, [P] are at their lowest, & the reverse reaction is unlikely.
• D. At the end of a reaction when the substrate and product are at equilibrium with each other.
• E. As soon as the reaction begins when the reaction rate is at its lowest.

Here’s what students ask on this topic:

What is the significance of the rate constant (k) in a chemical reaction?

The rate constant (k) is a crucial parameter in chemical kinetics, indicating the speed at which a reaction proceeds under specific conditions. It is a positive value that reflects the efficiency or probability of a reaction occurring. A higher rate constant means a faster reaction. For enzyme-catalyzed reactions, different rate constants (k₁, k_-1, k₂, k_-2) describe various steps, such as substrate binding and product formation. The rate constant is essential for calculating reaction rates and understanding reaction dynamics.

How do you determine the rate law for a chemical reaction?

The rate law for a chemical reaction is determined by the relationship between the reaction rate (v), the rate constant (k), and the initial concentrations of the reactants. The general form of the rate law is:

$v=k\cdot {\mathrm{[A]}}_m\cdot {\mathrm{[B]}}_n$

where [A] and [B] are the initial concentrations of the reactants, and m and n are the reaction orders, which are often determined experimentally. The rate law helps predict how changes in reactant concentrations affect the reaction rate.

Why is k_-2 ignored during the initial stages of an enzyme-catalyzed reaction?

During the initial stages of an enzyme-catalyzed reaction, the product concentration is effectively zero, meaning there is no product available to reverse the reaction. As a result, the reverse rate constant k_-2 is negligible and can be ignored. This simplification allows biochemists to focus on the initial reaction velocity (v₀), which is primarily influenced by the forward rate constants (k₁, k_-1, and k₂), making calculations more straightforward.

What is the rate-limiting step in an enzyme-catalyzed reaction?

The rate-limiting step in an enzyme-catalyzed reaction is the slowest step that dictates the overall reaction rate. Typically, this is the step involving the rate constant k₂, which represents the conversion of the enzyme-substrate complex to the product and free enzyme. Since k₂ is the slowest, it controls the reaction's pace, and the overall reaction cannot proceed faster than this step.

How does the rate law apply to enzyme-catalyzed reactions?

In enzyme-catalyzed reactions, the rate law helps describe the relationship between the reaction velocity (v) and the concentrations of the enzyme-substrate complex. For the product formation step, the rate law is:

$v=k_2\cdot {\mathrm{[ES]}}_1$

where [ES] is the concentration of the enzyme-substrate complex. This rate law indicates that the initial reaction velocity depends on the rate constant k₂ and the concentration of the enzyme-substrate complex, simplifying the analysis of reaction dynamics.