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## Relationship between reaction concentrations and time

# Zero-order reactions

AP.Chem:

TRA‑3 (EU)

, TRA‑3.C (LO)

, TRA‑3.C.1 (EK)

, TRA‑3.C.4 (EK)

## 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.