deriving the integrated rate law

[Vuong_2F]

how do you derive the integrated rate law?

[Chem_Mod]

See the detailed step-by-step process I did in class.

[Scot Widjaja Dis 1J]

For first order reaction, we know that rate = -(1/a) * (d[A]/dt) = k[A]. We also assume that our a = 1 so 1/a just becomes 1.

We then have to separate the variables by putting all the [A] variables in one side and everything else on the other. Rearranging that gives us:

(d[A]/[A}) = -kdt. Now we have to integrate both sides. We know that the integral of 1/X is ln(X), and since the side on the left matches this exact same form, the integral of (d[A]/[A}) is ln[A]. We also have to integrate the right side but since k is just a constant, our integral of -kdt is -kt + C. Our constant, C, represents the initial concentration of reactants, ln[A]0

Now we have something in the form ln[A] = -kt + ln[A]0. This is our integrated rate law for a first order reaction.

[Scot Widjaja Dis 1J]

For second order reaction, we know that rate = -(1/a) * (d[A]/dt) = k[A]^2. We also assume that our a = 1 so 1/a just becomes 1.

We then have to separate the variables by putting all the [A] variables in one side and everything else on the other. Rearranging that gives us:

(d[A]/[A}^2) = -kdt. Now we have to integrate both sides. We know that the integral of 1/X^2 is -1/X, and since the side on the left matches this exact same form, the integral of (d[A]/[A}^2) is 1/[A]^2. (Both our negatives cancel out). We also have to integrate the right side but since k is just a constant, our integral of kdt is kt + C. Our constant, C, represents the initial concentration of reactants, 1/[A]^2 (initial).

Now we have something in the form 1/[A]^2 = kt + 1/[A]^2 (initial). This is our integrated rate law for a second order reaction.

[Scot Widjaja Dis 1J]

For zero order reaction, we know that rate = -(1/a) * (d[A]/dt) = k. We also assume that our a = 1 so 1/a just becomes 1.

We then have to separate the variables by putting all the [A] variables in one side and everything else on the other. Rearranging that gives us:

d[A] = -kdt. Now we have to integrate both sides. We know that the integral of a constant is just that times the variable, so our integral of d[A] is equal to [A]. We also have to integrate the right side but since k is just a constant, our integral of -kdt is -kt + C. Our constant, C, represents the initial concentration of reactants, [A]0.

Now we have something in the form [A] = -kt + [A]0. This is the integrated rate law for a 0 order reaction.