Tuesday, October 05, 2010

The Buddha and Colission Theory - How to Solve the Equation

In the past three years I've gotten a lot of traffic to my “Move Over Buddha” post. Some of the traffic, undoubtedly, trying to figure out the answer to the very question I was faced with:

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.

I know this only because I discovered that someone had left me a comment, asking for clarification.

It may help you if I explain a bit about where this comes from, and then rewrite the equation for you. Keep in mind that I haven't dealt with this kind of thing for about three years so, I'm a bit rusty. If you notice anything seriously wrong, leave a comment and let me know.

Collision Theory

With few exceptions, reaction rates increase with increasing temperature. A rule of thumb in chemistry is that an increase in temperature by 10 degrees C doubles the reaction rate. The rate constant's dependence on temperature is explained by collision theory. Collision theory assumes that, for a reaction to occur, reactant molecules must collide with an energy greater than some minimum value and with the proper orientation. This minimum energy of collision required for two reactant molecules to react is called the activation energy, or Ea.

Raising the temperature increases the fraction of molecules having high kinetic energies. These are the ones most likely to react when they collide. The higher the temperature, the larger the fraction of molecules that can provide the activation energy needed for reaction.

The fraction of reactant collisions having energy greater than the activation energy can change rapidly with even small temperature changes. It can be shown that f is related to the activation energy, Ea, by the equation

Let's define the parts of the equation:

f = frequency of molecules having a high enough kinetic energy to react (i.e. greater than the activation energy)

Ea = the energy of activation for the kind of molecule in question (i.e. argon molecules)

R = the universal gas constant (8.31 J/mol.K )

T = the absolute temperature (Kelvin)

e = the inverse natural logarithm (inverse ln).

Now, let's rewrite the equation slightly, just to help me clarify the math.

Let's add the values and units from the original problem.

The top uses kilojoules, while the bottom uses joules. Let's the top part by 1000/ k in order to bring the units on the top and bottom of the equation into agreement.  We'll also rewrite the temperature (400K) to make the math more clear.

Multiplying the top part of the equation, and the bottom part of the equation, the temperature unit (K) cancels, and the equation is somewhat simplified.

We can now invert the denominator to simplify the equation further, and make sure we deal with the units properly.

Multiplying across, the energy units (J - Joules) cancel and the equation if further simplified.

Dividing we end up with this:

Now, we take the inverse natural log of 3.0 mol and find the solution: 20.1 mol of argon.

I hope that explanation helped. If you see anything wrong with my math, or anything else. Let me know.


BigSis.Luvs.LilBro said...

The back of the book says the answer is 2.33 x 10^-2, and I thought the equation was f = e ^ -(Ea / RT). Explain! I've tried both yours and my equation, and neither gives me the right answer.

John Newman said...

Apparently, there is something wrong with my math.