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First-order reaction (with calculus)

Deriving the integrated rate law for first-order reactions using calculus. How you can graph first-order rate data to see a linear relationship.

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• At , I read below that "d" stands for derivative, but what exactly is a derivative and why do we need it in this equation? I would suggest that the person making this video explain for us non-calculus viewers what a derivative is and it's function or reference a video where we can find a simple explanation. Additionally, at , what is integration? Why do we need to "integrate"?
• In this case we' re saying that the rate of consumption of the reagent d{A}/dt is directly proportional to the concentration {A}. Therefore when the concentration {A} decreases, the rate of consumption also decreases, it' s a differential equation and you have to integrate in order to come up with a function of {A} respect to time.
• , How did you know it was a first order reaction?
• He is saying, IF we know that the reaction is first order, this is how to do the math.
• how is the integration of d(A)/(A) = ln(A) at
• Let's just say that A is equal to x for now.
d/dx (ln x) = 1/x
To derive the rate law, I'm taking the antiderivative of dx/x, or 1/x dx.
So then if I take the antiderivative of both sides:
In x = antiderivative of 1/x
• We have been taught first order formula as 0.693/k...can't we use that one??
(1 vote)
• yes, it is exactly the same thing because the natural log of 2 is 0.693
• Why does taking the indefinate integral of
(d (A))/[A] = -kt
only leaves the left side of the equation with a constant and not also the right?
Int[ (d (A))/[A] = -kt ] =
ln[A]t + X = -kt + Y, where X & Y are both constants.
Is it that
Y - X = ln[A]o
after setting the general boundaries?
• One of the constants is zero.
dA/A = -kdt
∫dA/A = -k∫dt
lnA] = -kt]
A runs from A₀ to A, and t runs from 0 to t.
lnA – lnA₀ = -kt + 0
lnA = -kt + lnA₀
• At 0408, he says this formula is on the AP chemistry equation sheet. Are these videos really made with the MCAT content in mind? Any thoughts?
• I dont think they were made for mcat review but they were chosen because they go over mcat content. The mcat doesnt use calculus so I'm assuming this is just for deeper understanding
• Don't we get an exponential graph if we solve the equation this way or am I making a mistake in my calculations?

ln(At) - ln(a0) = -kt

ln(At/A0) = -kt

e^-kt = At/A0

At = A0*e^-kt
• At , why can't you take the derivative of both sides of the equation?
(1 vote)
• You could take the derivative of both sides, but it wouldn't achieve a meaningful result. The goal here is to get a function that allows us to calculate [A] at any given time, t. We start out with the rate of how [A] changes with time, and this rate is changing, so we need to integrate in order to be able to calculate [A] for a particular time, t. If we took a derivative, as you suggested, then we'd be able to determine that rate at which the rate is changing. This, however, is not what we're interested in here.
• What is the integrated rate law? I don't see any videos on that.
• Any place you see the concentration of a reactant as a function of time, it is an integrated rate law. Just to mention, the other type, the differential rate law, gives the rate of a reaction as a function of concentration of a reactant, assuming the effects of the reverse reaction are negligible.
(1 vote)
• In this video are we assuming that the reaction is a first order reaction because there is only one reactant.
(1 vote)
• Nope, we are assuming it is a first order reaction because this reaction only has one step. And this step only has one reactant with a coefficient of 1.

If this reaction has multiple steps, even if it has one reactant, the order of reaction may not necessarily be equal to 1.

Hence, the order of reaction depends on whether the reaction is one-step or multi-step.
If it is one-step, you can take the coefficient of reactant as the order
If it is multi-step, there is a video later on that will tell you how to derive the order