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### Course: AP®︎/College Chemistry>Unit 5

Lesson 3: Concentration changes over time

# First-order reactions

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 ?
• why does the natural log of pressure have no unit?
• It still has the original unit of pressure because all we've done to it by taking the ln(P) is a mathematical operation. We haven't added a new unit or eliminated the original unit. So technically the unit of ln(P) should still be Torr and they should have indicated the unit on graph.

Hope that helps.
• what does the t subscript mean in ln[A]_t
• In the same way that ln[A]_0 is concentration at time 0, ln[A]_t is concentration at time t where t can be any time desired
• please what does calculus and log mean
(1 vote)
• Calculus is a more advanced branch of math that people usually learn after algebra and geometry. The reason we use calculus here is because the rate law defining the disappearance of the reactant, R = -Δ[A]/Δt, can be written with derivatives. Derivatives are one of the main concepts in calculus which essentially are a more precise way of imagining slope or rate of change. So, R = -Δ[A]/Δt is rate of change of a reaction using algebra, but R = -d[A]/dt, is rate of change using calculus where the ‘d’ stands for derivative. And the derivative form is technically what is being used here (even though it’s not being shown).

The equation at the bottom of the screen at is known as the integrated rate law which uses the two equations above to derive. This requires using another concept in calculus known as the integral.

Without getting into the details, all you have to know is that the integrated rate law for a certain order is the result of calculus related math. They omit the actual calculus here so as not to confuse people with the math so they can focus on the chemistry.

Log, short for logarithm, is a way to express exponential problems and solve them. An exponential problem is one where the unknown to be solved for, the variable, is an exponent of another number called the base. So a problem like: 10^(x) = 100, is an exponential problem; we’re trying to find which power 10 must be raised to to get 100. This can be rewritten with logs as: log10 (100) = x. Logs have a base written as a subscript in the lower right side which we want to raise by a certain power to get the number in parentheses. Here we know x is 2 because 10^(2) = 100, but logs allow us to solve any exponential problem no matter how difficult they get.

The most common bases for logarithms are 10 and e. Log10 is so common that it is the default base and most calculators will express it as simply log. e is a mathematical constant known as Euler’s number equal to approximately 2.718, so a logarithm using base e would look like loge. Again it’s so commonly used that it gets its own symbol ‘ln’ which stands for natural logarithm from the Latin logarithmus naturalis.

Hope that helps.
• 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!
• Yeah, the calculus works out to that.
(1 vote)
• I don't get why you need to use the integrated rate law to find k.
If you have the table shown in this example, can't you just use that to find R, and then use R to find k?
(1 vote)
• So essentially, why use the integrated rate law when you can just use the regular rate law to find the rate constant, k? You can use either to find the rate constant, however I would say using the integrated rate law is less calculation intensive. Assuming we have the same data, time and the concentrations of a reactant, and that we know that the reaction is a first order reaction.

Using the integrated rate law, we simply need to know the original concentration, a concentration at time t and that time to find the rate constant. We’ll have all the variables of the first order integrated rate law except k, and then it’s simply some algebra to solve for k.

Using the regular rate law becomes more challenging because we need to know the rate of the reaction at some time. Given the data, we’d have to calculate that first. We can use: rate = -Δ[A]/Δt, to find the rate where we find the difference in concentration of reactant A at two different times and divide by the change in time. However, this assumes that the graph of [A] vs. time for a first order reaction is linear, which it isn’t (the graph of ln([A]) vs. time is). So to really find the rate at any time of the reaction, we need to find the slope of a graph of [A] vs. time using a derivative. Only once we’ve done that can we find the reaction rate at some time, then solve for the rate constant given the reactant concentration at the corresponding time.

So you can do either method, however I feel that the integrated rate law is the easier of the two methods given the data shown in the video.

Hope that helps.
• Do I have to memorize the integrated rate law?
(1 vote)
• Only if you want to do kinetic problems. The alternative is that you need to derive it using calculus every time you need to use an integrated rate law.
• does ap chem require calculus?
(1 vote)
• Not really. It helps for actually deriving the integrated rate laws. Other than that you can get through general chemistry with only algebra knowledge.
• 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.