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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.

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  • blobby green style avatar for user Michael YANG
    Could anyone please show the derivation of the function for the first-order reactions?
    (3 votes)
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    • blobby green style avatar for user BilalKawsara
      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.
      (16 votes)
  • blobby green style avatar for user Kokes, Joshua
    What does he mean by the "natural log" at ?
    (3 votes)
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  • blobby green style avatar for user Minsu Park
    why does the natural log of pressure have no unit?
    (4 votes)
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    • leaf red style avatar for user Richard
      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.
      (2 votes)
  • blobby green style avatar for user imtheturtleguy
    what does the t subscript mean in ln[A]_t
    (2 votes)
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  • starky tree style avatar for user McJay
    please what does calculus and log mean
    (1 vote)
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    • leaf red style avatar for user Richard
      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.
      (4 votes)
  • blobby green style avatar for user bob ross
    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!
    (2 votes)
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  • sneak peak green style avatar for user yonsuissa
    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)
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    • leaf red style avatar for user Richard
      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.
      (2 votes)
  • male robot johnny style avatar for user John
    Do I have to memorize the integrated rate law?
    (1 vote)
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  • blobby green style avatar for user ssilberstein
    does ap chem require calculus?
    (1 vote)
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  • blobby green style avatar for user Ria Rose
    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)
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    • leaf red style avatar for user Richard
      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.
      (2 votes)

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.