StudySmarter - The all-in-one study app.
4.8 • +11k Ratings
More than 3 Million Downloads
Free
Americas
Europe
Have you ever laid out a bunch of dominoes and knocked them over? As each domino topples over, it knocks over the next one, continuing until the last domino falls. That first domino technically knocked over the last one, but it took all the ones in between to get there. In chemistry, these domino-like reactions are called multistep reactions.
Explore our app and discover over 50 million learning materials for free.
Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken
Jetzt kostenlos anmeldenHave you ever laid out a bunch of dominoes and knocked them over? As each domino topples over, it knocks over the next one, continuing until the last domino falls. That first domino technically knocked over the last one, but it took all the ones in between to get there. In chemistry, these domino-like reactions are called multistep reactions.
The process of knocking down dominoes is similar to the process of multistep reactions. Pixabay.
A multistep reaction is a series of reactions that can be "summed up" by a net reaction. The reaction is made up of several steps called elementary reactions and may involve a catalyst.
In this article, we will look into these multistep reactions and see how they differ from the simpler, elementary reactions.
The formula below is an example of a net chemical equation: $$2H_2+2NO\rightarrow 2H_2O + N_2$$
There are three total steps in this reaction. Our reactants are that "first domino" that knocks over all the others until we get to the "final domino", our products. It makes sense that this reaction takes many steps. You need to break several bonds (H-H and N-O) and you also need to make several bonds (H-O-H and N-N). Reactions are caused by collisions between particles, and it takes a few collisions to make/break those bonds, hence the multiple steps.
When we write chemical reactions, we typically write the net equation (i.e. only the reactants and the final products), but if we want to get into the nuts and bolts of a reaction, we have to look at all the steps. Let's go back to that reaction above. Here is the full multistep reaction:
$$NO + NO \underset {k_{-1}} {\stackrel{k_1}{\rightleftharpoons}} N_2O_2\,\text{(fast)}$$$$N_2O_2 + H_2 \xrightarrow {k_2} H_2O + N_2O\,\text{(slow)}$$$$N_2O + H_2 \xrightarrow {k_3} N_2 + H_2O\,\text{(fast)}$$
Net Reaction: $$2H_2+2NO\rightarrow 2H_2O + N_2$$
There are a few things you'll notice about these reactions. The first is that N2O2 and N2O aren't present in our net reaction. That is because they are intermediates. An intermediate is both formed and used up before the end of the reaction. Since they are consumed during the reaction, they aren't written in the net equation.
The second thing you'll notice is the fast/slow markers. All reactions have an energy barrier that they must overcome for the reaction to proceed. This "barrier" is called the activation energy (EA), which is the minimum energy required for a reaction to happen. Reactions that have high activation energy barriers are slower than those with lower ones.
For a multistep reaction, we focus on whichever "hurdle" is the hardest to clear. In this case, the second step has the largest activation energy. Since it proceeds the slowest (has the highest activation energy barrier), it is the rate-determining step.
The rate-determining step is the step that is the slowest. The other reactions will be "halted" by this reaction and then proceed forward at a slower rate. It is "rate-determining" since the overall forward rate is based on that reaction alone.
Like our dominoes, multistep reactions proceed in a single-file line and therefore all go at the same speed. This speed is determined by our slowest step, the rate-determining step.
There is another type of multistep reaction called multistep synthesis reactions. This is a process in organic chemistry where a simple molecule (typically an alkene) is built upon through a series of reactions to create a much more complex molecule. Multistep synthesis follows the same multistep reaction mechanism, except the products are always more complex than their reactants.
We use the Bronsted-Evans-Polanyi relation of multistep reactions to calculate the activation energy of a multistep reaction. It hinges on the fact that the difference in activation energy between two reactions is proportional to the difference in their enthalpy of reaction. The formula is:
$$ E_a=E_{total}+\alpha\Delta H$$
where Ea is a reference activation energy for a similar reaction, α is the position of the transition state, and ΔH is enthalpy. We use this total activation energy to calculate the rate of reaction. This equation is way above the AP level, so don't worry about it for now!
Note, that we begin our discussion below with a derivation of a theoretical rate law from an examination of the balanced reaction equation. However, often the actual experimentally determined rate for a given reaction will not be accurately described by the theoretical rate law. If, however, it is the case that the experimentally determined rate law is described by the theoretical rate law then the following discussion concerning reaction rates will apply.
When we write a rate equation, we are focusing on the rate-determining step. Let's look at an example.
Write the rate equation for the following multistep reaction:
$$A + B \xrightarrow {k_1} X\,\text{(slow)} $$$$X + A \xrightarrow {k_2} Y\,\text{(fast)} $$$$Y + B \xrightarrow {k_3} C\,\text{(fast)}$$$$\text{Net reaction}: A + B \rightarrow C$$
Our first step is slow, so it is the rate-determining step. So, our rate equation is just: \(\text{rate}=k_1[A][B]\)
When the rate-determining step is the first in the series, writing the rate equation is pretty straightforward. But what about when it's not?
Write the rate equation for the following multistep reaction:$$A + B \underset {k_{-1}} {\stackrel{k_1}{\rightleftharpoons}} X\,\text{(fast)} $$$$X + A\xrightarrow {k_2} Y\,\text{(slow)} $$$$Y+B \xrightarrow {k_3} C\,\text{(fast)} $$$$\text{Net reaction}: A + B \rightarrow C$$
Based on this, you would normally write the equation as \(\text{rate}=k_2[X][A]\)
But there is a problem here. We can't have an intermediate in our rate equation. So, what now? Well, we need to express our intermediate in terms of our reactants. The first step in our reaction is an equilibrium reaction (shown by the double arrow), so we can use that as a conversion factor:$$K_1=\frac{k_1}{k_{-1}}=\frac{[X]}{[A][B]}$$$$[X]=K_1[A][B]$$
Now we can plug this conversion factor into our rate equation:$$\text{rate}=k_2K_1[A][B][A]$$$$\text{rate}=k[A]^2[B]$$
that is our proper rate equation (notice that, k2K1 has been simplified to k)
It is not necessary that we simplify our rate constants, but it is helpful to keep the rate equation less cluttered.
Some multistep reactions happen as a natural progression, like our dominoes. Another mode of progression for a multistep reaction is due to the presence of a catalyst.
A catalyst is a species that increases the rate of a reaction. A catalyst is not consumed during the reaction.
Here's a diagram of a generic reaction:
Certain catalysts in a multistep reaction provide an alternate "pathway" for a reaction. While extra steps may be added, the overall activation energy is lowered, so the reaction is faster. Let's look at an example.
Write the rate equation of the multistep reaction below:$$H_2O_2 +I^- \rightarrow H_2O + IO^-\,\text{(slow)}$$$$H_2O_2+IO^-\rightarrow H_2O + O_2 + I^-\,\text{(fast)}$$$$\text{Net reaction}: 2H_2O_2\rightarrow 2H_2O + O_2$$
Our catalyst here is I-. It "cancels" itself out, meaning it is on both the reactant and product side, so it isn't represented in the net equation. It also isn't consumed in the process. Unlike intermediates, a catalyst can be shown in the rate equation. In this case, it is present in the slow step, so the equation is:
$$\text{rate}=[H_2O_2][I^-]$$
Even though the catalyst here is in the "slow" step, the reaction itself is faster than it would have been without it.
Lastly, there is another mode of progression for a multistep reaction involving a catalyst. Within the field of biochemistry, multistep reactions often involve catalysts at each step of a given multistep reaction pathway. In addition, these bio-catalysts often act in a concerted fashion to carry out the biosynthesis of biomolecules. Finally, we will point out that many biocatalysts have evolved the capacity to lower the activation energy, Ea, directly. A full discussion of biochemical catalysis will not be discussed here but the interested reader can find further information on this topic by consulting Biochemistry textbooks.
Just like with elementary reactions, multistep reactions have both theoretical and actual yields. A problem with multistep reactions is that, since there are more reactions happening, there is more room for inefficiency. When we use the chemical equation to calculate the yield, that is the theoretical yield, while our actual yield is whatever is obtained during the experiment. To determine the efficiency of a reaction, we calculate the percent yield.
The percent yield of a reaction is a ratio of the expected to the actual yield. The formula is:
$$\text{percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}*100\%$$
The percent yields of a given set of reactions are shown below (Ph refers to a benzene (C6H6) ring). What is the overall yield?
$$Ph \xrightarrow {Br_2,FeBr_3} PhBr\,(75\%)$$$$PhBr \xrightarrow {Mg, CO_2,HCl} PhCOOH\,(62\%) $$$$PhCOOH \xrightarrow {CH_3OH,H_2SO_4} PhCOOCH_3\,(66\%) $$
$$\text{Net equation}: Ph \rightarrow PhCOOCH_3 (?\%)$$
The first step is to convert the yields into decimals. So, we would have 0.75 for step 1, 0.62 for step 2, and 0.66 for step 3. Now we can multiply them to get our total yield.
$$0.75*0.62*0.66=0.31$$ $$31\%$$
Our percent yield is much smaller than any of the individual steps. Any inefficiencies are compounded during a multistep reaction. However, as we saw with the catalyst reactions, these multistep reactions may be the only viable option, so we accept the low yield.
The yield of one reaction is then the concentration of reactants for the following reaction.
The overall delta h for a multistep reaction is the sum of the delta h values for each reaction in the process
The overall percent yield is the product of all the percent yields for each step.
To find the concentration of a reactant, you would use the following equation:
concentration produced-concentration consumed=net concentration
The net equilibrium constant is equal to the product of each step's equilibrium constant.
Flashcards in Multistep Reaction9
Start learningWhat is a multistep reaction?
A multistep reaction is a series of reactions that can be "summed up" by a net reaction. The reaction is made up of several steps called elementary reactions and may involve a catalyst.
What is the rate-determining step?
The rate-determining step is the step that is the slowest. The other reactions will be "halted" by this reaction and proceed at that rate. It is "rate-determining" since the rate is based on that reaction alone.
True or False: The rate-determining step has the highest activation energy
True
What is a catalyst?
A catalyst is a species that increases the rate of a reaction. It cancels itself out, but is not consumed during the reaction.
How does a catalyst lower the activation energy?
It provides an alternate reaction pathway that has a lower activation energy
How do you calculate the overall yield of a multistep reaction?
You multiply the percent yields of each step
Already have an account? Log in
Open in AppThe first learning app that truly has everything you need to ace your exams in one place
Sign up to highlight and take notes. It’s 100% free.
Save explanations to your personalised space and access them anytime, anywhere!
Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of StudySmarter.
Already have an account? Log in