Main content

### Course: Chemistry archive > Unit 10

Lesson 2: Relationship between reaction concentrations and time- First-order reactions
- 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 reactions
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reactions
- Zero-order reaction (with calculus)
- Kinetics of radioactive decay
- 2015 AP Chemistry free response 5

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Zero-order reactions

The integrated rate law for the zero-order reaction A →

*products*is [A]_*t*= -*kt*+ [A]_0. Because this equation has the form*y*=*mx*+*b*, a plot of the concentration of A as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to -*k*. Created by Jay.## Want to join the conversation?

- Is there a half-life for zero-order reaction?(2 votes)
- Yes, zero-order reactions have a half-life equation as well. We can derive it the same way we derive the half-life equations for the first and second-order reactions.

The given integrated rate law of a zero-order reaction is: [A]t = -kt +[A]0. At half-life the concentration is half of its original amount, so [A]t = [A]0/2.

[A]0/2 = -kt + [A]0, after the substitution.

-[A]0/2 = -kt, subtract [A]0 from both sides of the equation.

[A]0/2 = kt, divide by -1 on both sides.

[A]0/2k = t, divide by k on both sides.

So half-life, t1/2, for a first-order reaction is the original reactant concentration divided by two times the rate constant: [A]0/2k

Hope that helps.(7 votes)

- If we increase the concentration of ammonia present above the platinum surface then it will have no effect on the rate of reaction but what if we increase the concentration of ammonia present on the platinum surface? Will the rate of reaction be affected?

According to me, I guess that the rate of reaction will be affected and it will increase. Correct me if am wrong. :)(1 vote)- Essentially the platinum metal surface is saturated with ammonia molecules when this reaction is happening. We can’t force more ammonia molecules onto the existing surface because it’s at maximum capacity. The one way we can increase the reaction rate though is by increasing the surface area of the platinum since more surface area can accommodate more ammonia molecules than before. We can do this by either having a larger single sheet of platinum or have more finely divided platinum particles.

Hope that helps.(3 votes)

- So on the AP Chem exam, when the coefficient of A is 2, the slope of the line is 2k instead of k?(1 vote)
- Well technically if the coefficient of A is 2, then the slope would be -2k. If you follow the steps of deriving an integrated rate law, and the coefficient is some integer n, the slope will be –nk.

Hope that helps.(1 vote)

- why did we take the stoichiometric coefficient to be half when calculating the rate of the reaction? shouldn't we double it because the number o moles also doubles?(1 vote)

## Video transcript

- [Instructor] Let's say we
have a hypothetical reaction where reactant A turns into products. And let's say the reaction is
zero order with respect to A. If it's zero order with
respect to A, we can write that the rate of the reaction is equal to the rate constant k, times
the concentration of A to the zero power. And since any number to the
zero power is equal to one, then the rate of the reaction would just be equal to
the rate constant k. We can also write that
the rate of the reaction is equal to the negative and the change in the concentration of A,
over the change in time. If we set these two ways of
writing the rate of reaction equal to each other, and
we use some calculus, including the concept of
integration, we will arrive at the integrated rate law
for a zero-order reaction, which says that the concentration
of A at time t is equal to the negative of the rate
constant k times the time, plus the initial concentration of A. Notice that the integrated rate law is in the form of Y is equal to mx plus b, which is the equation for a straight line. So if we graph the concentration
of A on the Y axis, and the time on the X axis, we will get a straight line
if the reaction is zero order. So if we write the concentration
of A on the Y axis, and time on the X axis, the graph will be a straight line, and the slope of that line
is equal to the negative of the rate constant k, so the slope is equal to -k, and the Y intercept of that line, so right where the line
intersects with the y-axis, this point is the initial
concentration of A. So everything we've talked
about assumes that there's a coefficient of 1 in front
of the concentration of A. However, let's say we
have a coefficient of 2 in front of A in our balanced equation. That means we need a
stoichiometric coefficient of 1/2, which changes the math. Now, instead of getting
-kt, we would get -2kt after we integrate, which means
that the slope of the line, when we graph the
concentration of A versus time, the slope of the line
would be equal to -2k. It's important to note
that textbooks often just assume the coefficient
in front of A as a 1, which would give the slope as equal to -k. However, if the coefficient
in front of A is a 2, then technically the slope of the line should be equal to -2k As an example of a zero-order
reaction, let's look at the decomposition of ammonia
on a hot platinum surface to form nitrogen gas and hydrogen gas. In our diagram, we have
four ammonia molecules on the surface of our platinum catalyst, and then we have another four that are above the
surface of the catalyst. Only the ammonia molecules on
the surface of the catalyst can react and turn into
nitrogen and hydrogen, the ammonia molecules above
the surface can't react. And even if we were to add in
some more ammonia molecules, so let's just add in some more here, those molecules still can't
react, and therefore the rate of the reaction doesn't
change as we increase the concentration of ammonia. So we can write that the
rate of the reaction is equal to the rate constant k times
the concentration of ammonia, but since increasing the
concentration of ammonia has no effect on rate, that's why this is
raised to the zero power. And therefore we get
the rate of the reaction is just equal to the rate constant k. Normally, increasing the
concentration of a reactant increases the rate of the reaction. However, for this reaction,
since we're limited by the surface area of the catalyst, if the catalyst is covered
with ammonia molecules, increasing the concentration
of ammonia molecules will have no effect on
the rate of the reaction. And therefore this reaction, the decomposition of ammonia
on a hot platinum surface, is an example of a zero-order reaction.