The Changes of Concentration with Time Integrated Rate Laws - 2nd Order Reactions

Introduction A second order reaction is one whose rate depends on the square of the concentration of a single reactant. A → products Rate = k [A]2

Introduction A second order reaction is one whose rate depends on the square of the concentration of a single reactant. A → products Rate = k [A]2 We also know that the rate of the reaction is equal to the rate of change in [A] with respect to time. Rate = − ∆[A] ∆t

Introduction Putting these two rate equations together, we get the differential rate law. Rate = k [A]2 = − ∆[A] ∆t

Introduction Putting these two rate equations together, we get the differential rate law. k [A]2 = − ∆[A] ∆t

Integrated Rate Law Putting these two rate equations together, we get the differential rate law. k [A]2 = − By integrating the equation, we get the relationship between [A] and time, the integrated rate law. − = kt ∆[A] ∆t 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. − = kt ➔ = kt + 1 1 1 1 [A]t [A]0 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 y = mx + b This puts it into the familiar form of the slope intercept equation of a line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. If we plot [A] vs. time, we get a curved line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. If we plot [A] vs. time, we get a curved line. If we plot 1/[A] vs. time, we get a straight line. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. If we plot [A] vs. time, we get a curved line. If we plot 1/[A] vs. time, we get a straight line. and the slope of the line is k. 1 1 [A]t [A]0

Integrated Rate Law We can rearrange the integrated rate law. = kt + 1 This puts it into the familiar form of the slope intercept equation of a line. If we plot [A] vs. time, we get a curved line. If we plot 1/[A] vs. time, we get a straight line. and the slope of the line is k. All we need is one experiment with [A] and time data. 1 1 [A]t [A]0

Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: t (s) [A] (M) 1/[A] 0.0100 100 50 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: The 1/[A] vs. time is linear. t (s) 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: We calculate k from the slope. t (s) 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: k = ∆(1/[A])/∆t = ([190] - [100])/(250 s - 0 s) t (s) [A] (M) 1/[A] 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: k = (90)/(250 s) t (s) [A] (M) 1/[A] 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Sample Problem Such as this one: k = 0.360 t (s) [A] (M) 1/[A] 0.0100 0.0100 100 50 0.0085 118 0.0074 136 150 0.0065 154 200 0.0058 172 250 0.0053 190

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration.

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0.

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0.

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. 1/[A]t½ = kt + 1/[A]0

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. 1/[A]t½ - 1/[A]0 = kt

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. 1/(½[A]0) - 1/[A]0 = kt

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. 2/[A]0 - 1/[A]0 = kt

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. (2 - 1)/[A]0 = kt

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. 1/[A]0 = kt

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. t½ = 1/(k[A]0)

Half-Life The half-life of the reaction, t½, is the time it takes for the concentration of the reactant to become ½ the value of the initial concentration. This means that [A]t½ = ½[A]0. Substituting this into our equation will give us the half- life with respect to k and [A]0. t½ = 1/(k[A]0) This means that the half-life of a second order reaction is dependent of the concentration of the reactant.

Second Order Reactions Summary: The integrated rate law for a first order reaction: 1/[A]t = kt + 1/[A]0 Plotting 1/[A] vs time produces a straight line. The slope of the line is equal to k t½ = 1 k[A]0