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# First-order reactions

AP.Chem:
TRA‑3 (EU)
,
TRA‑3.C (LO)
,
TRA‑3.C.1 (EK)
,
TRA‑3.C.2 (EK)
,
TRA‑3.C.4 (EK)
The integrated rate law for the first-order reaction A → products is ln[A]_t = -kt + ln[A]_0. Because this equation has the form y = mx + b, a plot of the natural log 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?

• • Could anyone please show the derivation of the function for the first-order reactions? • This is grade-12/college-level but if you're curious I will show you below.

So for a first order reaction -- we have the reaction equals the rate constant times the concentration of the (only) reactant --> R = k[A]

1. Then we choose to re-write R as -Δ[A]/Δt
and we get -Δ[A]/Δt = k[A]
2. Then we bring -Δt to the right side
Δ[A] = -k[A]Δt
3. Then we bring [A] to the left side
Δ[A]/[A] = -kΔt
4. Then we integrate (the left side with respect to A and the right side with respect to t)
∫Δ[A] 1/[A] = -k∫Δt
Ln[A] = -kt
5. Then we evaluate both integrals from 0 to t
Ln[A]ₜ - Ln[A]₀ = -kt-(-k0)
6. Then we bring Ln[A]₀ to the right
Ln[A]ₜ = -kt -0 + Ln[A]₀

7. Finally, we have our answer:
Ln[A]ₜ = -kt + Ln[A]₀

8.Notes:

i. For the connection to y=mx+b
The natural log of the concentration of A at a given time t --> Ln[A]ₜ is basically Y, and is equal to the natural log of the initial concentration of A --> Ln[A]₀ which is basically b, minus the rate constant -->k (basically m, aka the slope of the line) multiplied by time (basically x). So we get a linear graph of the form Y=mx+b

ii. The reason the it is negative at the beginning -Δ[A]/Δt and at -kt is because the rate is positive, but the change in reactant is negative because it is decreasing, so we build in a negative sign to cancel it and make the rate positive.
• What does he mean by the "natural log" at ? • hi,

at , Jay said if the coefficient of A is 2, then -kt will become -2kt, so why don't we generalize it as -akt, where a is the coefficient of reactant A?

then according to this new convention, won't the half life equation be [At] = [Ao]e^(-akt) and t1/2 = ln2÷ak?

thank you! • I’m having trouble finding how you got -2.08 x 10 ^-4. When I add up the y and x numbers to do m= change in y divided by change in x, I get -16.189 divided by - 39800 = 4.067x10^-4
(1 vote) • The slope of a line is defined as the change in the y-direction divided by the change in the x-direction (rise over run). As a formula looks like m = Δy/Δx; where m is the slope, Δy is the change in y, and Δx is the change in x. Change here being a difference (subtraction) between two point's x and y coordinates. So we can also write the slope formula as m = (y2-y1)/(x2-x1); where x1 and y1 are the coordinates for the first point and x2 and y2 are the coordinates for the second point.

Now assuming it is a perfectly straight line, the slope should be constant at all points on the line and so we can pick any two points to calculate the slope. I'll choose the first and last points; (0,6.219) and (15000,3.109). So x1 is 0, y1 is 6.219, x2 is 15000, and y2 is 3.109. Substituting these into the previous formula yields: m = (3.109 - 6.219)/(15000 - 0) = -2.07 x 10^(-4). Which is reasonably close to what Jay got in the video. The discrepancy between my answer and Jay's is due to him using a graphing computer of some sort which takes into account all the coordinate points and also shows that the line is not perfectly straight. At any case even doing it by hand our answers should agree for the most part.

With your calculation I'm not sure why or what you added together, but having a change in y of -16.189 and a change in x of -39800 is wildly wrong. Additionally if you divide those two numbers you get a positive slope (dividing a negative by a negative) and judging solely off the graph of the line it should have a negative slope.

Hope that helps.
(1 vote)

## Video transcript

- [Instructor] Let's say we have a hypothetical reaction where reactant A turns into products and that the reaction is first-order with respect to A. If the reaction is first-order with respect to reactant A, for the rate law we can write the rate of the reaction is equal to the rate constant K times the concentration of A to the first power. We can also write that the rate of the reaction is equal to the negative of the change in the concentration of A over the change in time. By setting both of these equal to each other, and by doing some calculus, including the concept of integration, we arrive at the integrated rate law for a first-order reaction, which says that the natural log of the concentration of A at some time T, is equal to negative KT, where K is the rate constant plus the natural log of the initial concentration of A. Notice how the integrated rate law has the form of Y is equal to mx plus b, which is the equation for a straight line. So if we were to graph the natural log of the concentration of A on the Y axis, so let's go ahead and put that in here, the natural log of the concentration of A, and on the X axis we put the time, we would get a straight line and the slope of that straight line would be equal to negative K. So the slope of this line, the slope would be equal to the negative of the rate constant K, and the Y intercept would be equal to the natural log of the initial concentration of A. So right where this line meets the Y axis, that point is equal to the natural log of the initial concentration of A. The conversion of methyl isonitrile to acetonitrile is a first-order reaction. And these two molecules are isomers of each other. Let's use the data that's provided to us in this data table to show that this conversion is a first-order reaction. Since the coefficient in front of methyl isonitrile is a one, we can use this form of the integrated rate law where the slope is equal to the negative of the rate constant K. If our balanced equation had a two as a coefficient in front of our reactant, we would have had to include 1/2 as a stoichiometric coefficient. And when we set our two rates equal to each other now and go through the calculus, instead of getting negative KT, we have gotten negative two KT. However for our reaction we don't have a coefficient of two. We have a coefficient of one and therefore we can use this form of the integrated rate law. Also notice that this form of the integrated rate law is in terms of the concentration of A but we don't have the concentration of methyl isonitrile in our data table, we have the pressure of methyl isonitrile. But pressure is related to concentration from the ideal gas law, so PV is equal to nRT. If we divide both sides by V, then we can see that pressure is equal to, n is moles and V is volumes, so moles divided by volume would be molarity, so molarity times R times T. And therefore pressure is directly proportional to concentration, and for a gas it's easier to measure the pressure than to get the concentration. And so you'll often see data for gases in terms of the pressure. Therefore, we can imagine this form of the integrated rate law as the natural log of the pressure of our gas at time T is equal to negative KT plus the natural log of the initial pressure of the gas. Therefore, to show that this reaction is a first-order reaction we need to graph the natural log of the pressure of methyl isonitrile on the Y axis and time on the X axis. So we need a new column in our data table. We need to put in the natural log of the pressure of methyl isonitrile. So for example, when time is equal to zero the pressure of methyl isonitrile is 502 torrs. So we need to take the natural log of 502. And the natural log of 502 is equal to 6.219. To save time, I've gone ahead and filled in this last column here, the natural log of the pressure methyl isonitrile. Notice what happens as time increases, right, as time increases the pressure of methyl isonitrile decreases since it's being turned into acetonitrile. So for our graph, we're gonna have the natural log of the pressure of methyl isonitrile on the y-axis. And we're gonna have time on the X axis. So notice our first point here when time is equal to zero seconds, the natural log of the pressure as equal to 6.219. So let's go down and let's look at the graph. All right, so I've already graphed it here. And we just saw when time is equal to zero seconds, the first point is equal to 6.219. And here I have the other data points already on the graph. Here's the integrated rate law for a first-order reaction and I put pressures in there instead of concentrations. And so we have the natural log of the pressure of methyl isonitrile on the y-axis and we have time on the X axis, and the slope of this line should be equal to the negative of the rate constant K. So there are many ways to find the slope of this line, one way would be to use a graphing calculator. So I used a graphing calculator and I put in the data from the data table and I found that the slope of this line is equal to negative 2.08 times 10 to the negative fourth. And since if I go ahead and write y is equal to mx plus b, I need to remember to take the negative of that slope to find the rate constant K. Therefore K is equal to positive 2.08 times 10 to the negative fourth. To get the units for the rate constant, we can remember that slope is equal to change in Y over change in X. So change in Y would be the natural log of the pressure, which has no unit, and X the unit is in seconds. So we would have one over seconds for the units for K. And finally, since we got a straight line when we graphed the natural log of the pressure versus time, we know that this data is for a first-order reaction. And therefore we've proved that the transformation of methyl isonitrile to acetonitrile is a first-order reaction.