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Main content
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.