Wednesday, March 07, 2007

Move Over ,Buddha

One thing I've absolutely been struggling with in chemistry is the study of kinetics. Part of the problem is that it's riddled with higher mathematics. I'm no slouch when it comes to algebra but, much to my disgrace, I've never taken calculus. I want to, mind you. I've just not had to do it, yet.

Now, according to the class syllabus, and the text book, I can use algebra to figure this stuff out. It's harder and weirder and they're not going to explain where it all comes from (they all derive from calculus), but I should be "just fine."

Yeah. Right.

Because I was sick all last week (and still have a cough), I missed the classes where we covered all of this, so I've been playing catch-up with the text book to figure it all out. One of the equations that's just been giving me fits is this:


f is the frequency of molecules in a sample at temperature T, having the species required energy of activation (Ea) to start a reaction. R is the gas constant (8.314 J/mol-K).

So far so good, right? So just what the heck is this lower case "e" all about?

Racking my brains, and the text, for far too long, I skipped that bit and went on to study the rest of the chapter.

When I went back to class last night, I was met with quiz on, you guessed it, the kinetics chapter! I actually did very well (I only got one wrong - stupid trick equation), in spite of my misgivings. When I was looking at the assigned homework questions, though, I was stuck with something that completely threw me:

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

Say what, now? How the heck am I supposed to figure this out? I'm guessing it has something to do with the equation above, but how? I asked one of my classmates, a bright young kid destined for medical school.

"Oh! Yeah, that one threw me for a minute, too. It's really simple, though. You just plug the numbers into this equation." he said.

"What? But how do you know what the energy of activation is?" I asked.

"It's given right here, in the problem." he replied.

"How do you know that's the energy of activation?" I persisted.

"Well, I guess you don't. It's just the number given." he said, sheepishly.

Finally Dr. OH stepped in. "It may not be the energy of activation, but you can still use the equation. You're just finding a percentage of molecules with a certain level of energy, it doesn't have to be the energy of activation."

"Okay. I get that, now. But what the heck is this 'e' in the equation?" I asked.

"That's the energy of activation." Dr. OH tried to explain.

"No, that's 'Ea.' What's this 'e' thing?"

"That?" he asked, giving me a quizzical look. "That's just the inverse log function."

I sat down. Fast.

I couldn't believe it. I knew what an inverse log was, but I had been looking at the bloody equation and thinking it was a variable all this time.

My built up world crumbled around me and I saw the face of God.

Okay, not really, but the light did go on for me. Finally, I got it.

"I can't believe I'm that stupid." I told Dr. OH.

He smiled and said, "Don't worry. Bring it up in class next time. I can't believe you're the only one that was confused."

And now I am enlightened.

You can find the solution to this problem, here.

1 comment:

Anonymous said... how would you solve that problem? i still don't get it..