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### Course: MCAT > Unit 9

Lesson 18: Kinetics- Kinetics questions
- Introduction to reaction rates
- Rate law and reaction order
- Worked example: Determining a rate law using initial rates data
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reaction (with calculus)
- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Elementary reactions
- Reaction mechanism and rate law
- Catalysts
- Kinetic and thermodynamic enolates

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# Zero-order reaction (with calculus)

Deriving the integrated rate law for zeroth order reactions using calculus. How you can graph zero order rate data to see a linear relationship.

## Want to join the conversation?

- At9:17Please can you give some other examples for zero order?(3 votes)
- Reactions catalysed by enzymes are often zero order with respect to substrate. The enzymes can only process so many substrate molecules per second so increasing the substrate concentration has no effect on rate.(5 votes)

- does this system of increased half life with increased concentration apply in any way to radioactive decay. Like is there any place in nuclear chemistry this is true like in this video.(2 votes)
- Radioactive decay is a first order process so the half-life is constant and doesn't depend on the concentration.(4 votes)

- hello everyone i have a doubt...in this video we're told that the increase in concentration of A doesn't affect the rate...but won't it be false if the reaction occurs in a closed container as it would lead to a greater pressure and thus the compression of the gas and thus increase in it's density....which will eventually increase the number of molecules in contact with the metal surface...??(1 vote)
- If the reaction rate increases when you increase the pressure, then the reaction isn't zero order.(5 votes)

- In my Chemistry book, it states that sublimation is normally a zero-order reaction or process, as the concentration of molecules that have sufficient energy to sublime is limited by the solid's surface area. My book states that the concentration of these molecules does not change when the amount of subliming substance decreases. However, I am struggling to understand one thing: if the amount of subliming substance decreases and the object gets smaller, wouldn't that have an effect on the surface area of the solid as well as a consequent effect on the rate of reaction? Why exactly is sublimation zero-order when clearly the surface area is variable in different instances? The only way for sublimation to be zero-order is if the solid is held in a sealed container and is under dynamic equilibrium with its vapor pressure. Am I correct about this? I would appreciate it greatly if someone could help me out about all this.(1 vote)
- Zero order reaction simply means that the rate of reaction is independent of concentration of reactants. And if you put a substance in a box then the change in its area will be negligibly small compared to the amount of gas evolved. for example if there is 1 mole of dry ice aka solid
*CO2*and 0.1 mole of it sublimes then evolved*CO2*will have huge 2.27 litres volume compared to the volume of 0.1 moles of solid CO2 and the change in active surface area will be small so it can be said that rate of reaction is almost constant but will surely change if amount of surface area is decreased considerably but this is generally not the case under "normal" conditions.

"normal" refers to the usual temperature at which a particular reaction happens or is feasible.(2 votes)

- At0:45, how can a reaction have a zero order while the molecularity can never be zero? Because the powers raised over the concentration in an elementary reaction while using rate law is equal to the stoichiometric coefficients of reactants and these coefficients can also never be zero.(0 votes)
- I think you are gravely mistaken the powers are not the stoichiometric coefficients, they are the order of reaction with respect to the given atom or molecule, and order as we discussed is experimental and molecularity is very different from order they are two different things..i hope you got your error...(4 votes)

- If K is always the same regardless of concentration, and Rate = k, is K still the same over time as the reaction continues? Does time effect K?(1 vote)
- The reaction rate is independent of concentration of the reactant. However, what would change the rate of reaction is using a different amount of catalyst, or changing the temperature.(1 vote)

- Hi - at 0.29 why is change[a]/change of time negative?(1 vote)
- Because the concentrations of A decrease to produce the products in this particular reaction, so the change in concentration will be negative.(1 vote)

- At3:39, is the integrated rate law used always the same for zero order reactions?(1 vote)
- At8:30, we added more ammonia molecules but it did not affect the rate of reaction as the added molecules were not on ammonia surface.What if we added the extra molecules on the ammonia surface??Also, if we removed some molecules from the ammonia surface, will it decrease the rate of reaction??(1 vote)
- If more molecules of ammonia were on the platinum surface this would increase the reaction rate. This could be achieved by increasing the surface area of the platinum by using more platinum and/or by grinding it to a smaller particle size.

Conversely, decreasing the surface area of the platinum so that there were fewer ammonia molecules on the surface would decrease the reaction rate.(1 vote)

- What's the highest order reaction possible?(1 vote)

## Video transcript

- Let's see we have a zero order reaction where A turns into our products and when time is equal to zero we're starting with our
initial concentration of A and after some time period T, we would have the concentration of A at that time. So we could express the
rate of our reaction, one way to do it would be to
say the rate of the reaction is equal to the negative change in
the concentration of A over the change in time. And another way to do this
would be to right the rate law. So the rate of our reaction is equal to the rate constant K times the concentration of A and since I said this is
a zero order reaction, this would be A to the zero power. And any number to the zero power is equal to one. Therefore, the rate of the reaction would be equal to K times one or the rate is just equal
to the rate constant K. So the rate of a zero order reaction is a constant. It's independent of
the concentration of A. Next, we can set these equal. Right we can say that K is equal to the negative change in
the concentration of A over the change in time. So we have the negative change
in the concentration of A over the change in time is equal to the rate constant K. And next we could think
about our calculus, alright, instead of writing change in A over change in time, we're going to write the rate of change of
the concentration of A with respect to time. We have our negative sign in here and then we have our K on the side. So we're ready to think about our differential equation and we're gonna multiple
by both sides by DT. Let's go ahead and multiple both sides by negative DT, so then we would have DA on the left side, right, and then we would have
negative KDT on the right side. And we're ready to integrate. Right, so we're gonna
integrate on the left, K is a constant so we can pull it out of our integral on the right. And we go back up here
to refresh our memories about what would be integrating from. So we'd be going from
time is equal to zero to time is equal to T and from our initial concentration to our concentration at time T. So you plug those in we're going from time is equal to zero to time is equal to T and then for our concentration we're going from our initial concentration to our concentration at time T. So we have some easy
integrals here, right? What's the integral of DA? That would be of course A or the concentration of A. So we have the concentration of A, right. We are evaluating this from
our initial concentration to our concentration of A at time T. On the right side we have
another easy integral, integral of DT. That's just T. So we have negative KT from zero to T. Next fundamental theorem
of calculus, right, so we would get the concentration of A minus the initial concentration of A is equal to on the right side, that would be negative KT so we get negative KT here. And so this is one form
for the integrated rate law for a zero order reaction. We could rearrange this. We could move the initial
concentration to the right side so we would get the final concentration is equal to negative KT plus the initial concentration of A. And so this is just another way to write our integrated rate law. So here's the integrated rate
law for a zero order reaction. And if we look at the form of that, it's Y is equal to MX plus B. Y is equal is to MX plus B. So we would put time on the X axis. If we graph time on the X axis, on the Y axis we have
the concentration of A. We're gonna get a straight line and the slope of that line, the slope of that line M, the slope is equal to negative K. The slope is equal to the
negative of the rate constant and the Y intercept would be
the initial concentration of A. So if we just sketch out
a quick little graph here. Let me put our axis down. So we put time on the X axis, right, so down here we'd have time and on the Y axis we
have our concentration of A. We're going to get a straight line, right, we're gonna get a straight line and let me see if I can. Let me just switch to a straight line. I don't wanna draw one here. So we got a straight line like this. Okay. And the slope of this line, right, the slope of our line here. The slope would be equal to negative K. So once again we can see
that from up here, right, the slope is equal to negative K. So M is equal to negative K. And our Y intercept of course would be the initial concentration. This would be the initial
concentration of A. So this is the integrated rate law for a zero order reaction. Next let's think about the half life. So remember the half
life is the time required for the concentration of a reactant to decrease to half of
its initial concentration. So when the time is
equal to the half life, T one half, we'd plug this in for T. The concentration of A when time is equal to the half life would be half of the
initial concentration. So the initial concentration
divided by two. So that would go into here. So let's rewrite what we have. We now have the initial concentration divided by two is equal to negative K times the half life plus the
initial concentration of A. So let's solve for half life. On the left side we have the
initial concentration of A divided by two or one half of
the initial concentration of A on the right side we have the
initial concentration of A. So this would be one half minus one, which is negative one half. So negative one half of the
initial concentration of A is equal to negative K times T one half. So solve for T one half or our half life. T one half is equal to this would be the initial
concentration of A divided by two times K. Two times the rate constant so here is the half life for a zero order reaction. Notice, if you increase the
initial concentration of A, if you increase the
initial concentration of A what happens to the half life? They're directly proportional. So if you increase the
initial concentration of A, the half life should increase. So the half life should increase. Let's look at an example
of a zero order reaction and this will help us understand
this idea of half life a little bit better. So our example is the
decomposition of ammonia. So the decomposition of ammonia into nitrogen and hydrogen. And this reaction occurs on the
surface of a metal catalyst. So let's say that we're
using platinum here. So on the left let's call
this situation one on the left we have a piece of platinum. Imagine this is our platinum surface and since the reaction
occurs on the surface of the platinum, the ammonia molecules have to be in contact
with our platinum surface. So let me draw in some ammonia molecules in blue here in contact
with our platinum surface. Remember this is a zero order reaction. So the rate is equal to the rate constant times the concentration of ammonia to the zero power. So increasing the
concentration of ammonia, doesn't affect the rate. And this picture helps us understand that a little bit better. If you increase the
concentration of ammonia, right, I'm adding some more
ammonia molecules in here. You're not affecting the
overall rate of the reaction because the ammonia molecules I just added are not in contact with
our platinum surface. So only these down here are in contact and are able to react. So we increase the
concentration of ammonia but we didn't have any effect on the rate because the rate of the
reaction is limited by how much surface area of
our metal that we have. So that situation number one. Let's think about situation
number two over here. We have our platinum surface. Let's draw in a bunch
of ammonia molecules. So here we have some in contact
with our platinum surface and then here we have even more. We have even more ammonia molecules and obviously most of
them are not touching the surface of our metal catalyst. So we've increased the concentration, we've increased the
initial concentration of A, right, of course, A here is our ammonia. And so we must increase the half life. That's what we learned up here. If you increase the
initial concentration of A, right, you have a longer half life. And that makes sense because our reaction has a constant rate, the more molecules we have present, the longer it takes to
consume half of them. So we have all these
molecules up here of ammonia that aren't touching our metal surface and so we're limited by our constant rate and so it's gonna take a lot longer to consume half of the molecules in blue in situation number two. So that's comparing situation number one with situation number two, we've increased the
initial concentration of A and therefore we have a longer half life.