AP Chemical Kinetics by justinsociety.com

The study of chemical kinetics studies the rates of chemical reactions. In chemistry, we focus on how fast the concentration changes. i.e. change in molarity over time.  (Remember molarity is moles per liter.)  In a balanced equation, the molar coefficients will reflect in the relative rates.  Consider the generic chemical equation 2A + 4B à 4C + 6D.  The concentration of compound B is consumed (drops) twice as fast as compound A.  Likewise, compound D will form three times as fast as compound A is used.  All of this is from molar ratios. As a chemical reaction proceeds, the reaction rate decreases, after all, there’s less stuff.  Doesn’t the same thing happen when you put popcorn in a microwave?  A tangent slope gives the instantaneous rate of change.  The reaction is initially fast (steep slope).  You could get the average reaction rate between any two reaction times by getting the slope of a line connecting the two points on the curve.  (FYI, This is called a secant line.)

In our reaction 2A + 4B à 4C + 6D, the rate of the reaction will be proportional concentration A (noted [A]) raised to some power, let’s say m.  ( is the symbol proportional symbol)

Rate [A] m

The same could be said for [B] raised to some other power.

Rate [B] n

The combination equation is Rate [A] m [B] n.

Throw in proportionality constant and you get:  Rate [A] m [B] n

The exponents m and n are NOT necessarily obtained from the balanced equation, though sometimes they match.  We get them from experimenting in the lab.  Different concentrations are tried in the lab and the reactions are timed.  Just like in math, we can solve for unknowns, of course assuming that only one thing is unknown. m and n are numbers that represent what is called the “order of the reaction” for their associated reactant.  Overall order of the (entire) reaction here is the sum m+n.

AP favorite test question.  They’ll give you a table and you have to figure out what the exponents are.

 Experiment [A] [B] Reaction RATE 1 0.10 0.10 0.20 2 0.20 0.10 0.40 3 0.30 0.10 0.60 4 0.30 0.20 2.40 5 0.30 0.30 5.40

Look at the table.  Notice that if the same amount of [B] is used, doubling [A] doubles the rate, tripling [A] triples the rate.  Fixing [A] and varying [B] is different.  Doubling causes a four fold increase and tripling causes a nine fold increase.  On the exam, use a calculator if needed.

Looking at experiments 1 and 2, Rate [A] m since [B] was purposely kept the same.  Between two experiments we could also say:

Ratio of the Rates = (Ratio of the concentrations [A]) m

Or in this case, 2 =  m

m could only equal 1

Looking at experiments 3 and 5, Rate [B] m since [A] was purposely kept the same.  Between two experiments we could also say:

Ratio of the Rates = (Ratio of the concentrations [B]) n

Or in this case, 9 =  n

n could only equal 2

(The same result would also occur for n if we compared experiments 4 & 5. Try it.)

So, Rate [A] 1 [B] 2 or Rate [A] 1 [B] 2. To get , you can chose any of the experiments and plug the numbers in the equation and solve for .  All five experiments will give the same value.

Concentration of a reactant over time.

Consider the simplest chemical reaction possible;

A à C

If the reaction rate is first order then Rate [A] 1

If the reaction rate is second order then Rate [A] 2

From calculus it can be shown for the first order reaction:

Rate [A] 1can produce this equation  à  [A]t = [A]0 e-kt

where [A]t is the concentration at time t

[A]0 is the starting concentration and is .

This basically will tell you the concentration of A remaining after the reaction was allowed to continue for a given amount of time.

You could do a little rearranging.  Take the natural log of both sides of the equation like so:

[A]t = [A]0 e-kt

ln([A]t) = ln([A]0) +ln(e-kt)

ln([A]t) = ln([A]0)  -kt(ln(e),  but  ln(e) = 1

à ln([A]t) = ln([A]0) -kt

This equation is GIVEN to you on the exam.  (Review log rules so that you can change it back and forth!)  The equation in beautified rearranged form is:

à ln([A]t) = -kt + ln([A]0) because it looks like

y   = mx+  b

If presented with a graph where the axis is in regular time and the axis is ln([A]t) NOT plain ([A]t) then the slope of the straight line IS k!  (In the real world [A]t is measured and then the natural log of the number is plotted on the graph.)

As a footnote, does this look familiar? This is the same equation that was used in radioactive decay and it works the same way.  (That’s right, radioactive decay is first order.)

Let’s algebraically rearrange a little further:

ln([A]t) = ln([A]0) -kt

ln([A]0) - ln([A]t) = kt

ln([A]o/[A]t) = kt

At the half life:

t=t1/2

[A]t=½[A]0 à [A]o/[A]t = 2

Substitute, ln(2)=k t1/2

Rearrange and t1/2 = ln(2)/k  (This equation they don’t give you.)

From calculus it can be shown for the second order reaction:

Rate [A] 2 can produce this equation: Where just like before:

[A]t is the concentration at time t

[A]0 is the starting concentration, and is .

Don’t bother to memorize it, you WILL be given this equation on the exam too.

The equation in beautified rearranged form is:

à because it also looks like

y    = mx + b

If presented with a graph where the axis is in regular time and the axis is NOT plain ([A]t) then the slope of the straight line IS like before k!

In either case, if you are told that if a reaction is first or second order, then use the correct equation and solve for the unknown.

Calculating Activation Energy

The activation energy (Ea) for a given reaction is unique and doesn’t change.  As long as the molecules have the minimum required amount of kinetic energy, the reaction will occur.  Kinetic energy depends on temperature as more and more molecules will move fast enough to participate in the reaction as temperature rises.  Though Ea doesn’t change with temperature, k does, and it will be reflected as the reaction is faster.  The Arrhenius equation relates temperature activation and k in the following equation: ,

where Ea is activation energy, T is temperature, R is the gas constant (in SI units),  k is k and A is the frequency factor (or in less scientific terms a fudge factor so the mathematical framework doesn’t collapse.)

Taking the natural log gets à Algebraically rearrange gets à Which looks like (you guessed it)              y    =   m  x     +    b

The algebraic form above is given to you on the exam.  Memorize it if you have nothing better to do with your time.

The x-axis is (1/T), the y-axis is ln(k) and the slope is - Ea/R.  If given two sets of T and k, invert T, natural log the k, get a slope then set it equal to - Ea/R and solve for Ea.  If given a graph, just determine the slope.  Remember to use R=8.31J mol -1K-1. (This big K is for degrees Kelvin!)

F.Y.I.

Calculus Derivation for First order kinetics equation. Yes, the minus sign does belong there.  It is slightly different from the one originally given.  The negative sign indicates that the change in [A] is negative, as in decreasing, since it’s being used up.  Without it, k values would be negative.  Chemistry already has enough negativity and doesn’t need anymore.  The same minus sign is needed for second order kinetics too.

Calculus Derivation for Second order kinetics equation. 