The Arrhenius Equation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. Milk left out on a hot summer day turns sour in hours, but in the fridge it lasts for days. Room temperature is about three hundred kelvin. Warm it by ten degrees, to three hundred and ten kelvin: only about three percent hotter. And yet, near room temperature, the rate constant usually becomes two to three times bigger! Three percent hotter, double the speed? How is that possible? Today the Arrhenius equation will answer this. It comes up often in JEE, so let's begin.

0:37The Arrhenius equation

So, here it is, the Arrhenius equation. k, the rate constant, equals capital A times e to the power minus capital E a over R T, where e is the base of natural logs. Capital E with a small a below it is the activation energy. Okay? R is the gas constant, eight point three one four joules per kelvin per mole; the minus one powers mean per. And T is the temperature, always in kelvin, never in Celsius. Right? Now think. If the temperature T goes up, what happens to k? k goes up, yes! The up arrow means rises, the down arrow means falls, and the double arrow means so. Bigger T, smaller E a over R T, a less negative power, and a bigger k. With me? But why such a big effect? e to the power minus E a over R T is the fraction of collisions with energy at least E a. T sits in the power, so a small change in T changes this fraction a lot. Clear? Last, capital A, the pre-exponential factor, or frequency factor. It is the value of k as T goes to infinity; the arrow means tends to, and the sideways eight is infinity. It has the same units as k. Got it?

2:01The Arrhenius plot

Now take the natural log of both sides; l n means natural log. The natural log of e to a power is just the power, so natural log of k equals natural log of capital A, minus E a over R T, with me? Compare with y equals m x plus c, where y is natural log of k and x is one over the temperature. So this plot, like the graph on the right, is a straight line; v s means versus. Okay? The slope is minus E a over R, so the line falls, as you see. And it cuts the vertical axis at natural log of capital A, the intercept. Clear? Many questions use log without the n, meaning log to the base ten. Divide the whole equation by two point three zero three: log k equals log capital A, minus E a over two point three zero three R T, got it? Now, if a question plots log k against one over T, the slope is still minus E a over R, right? No, no, careful here! The slope is minus E a over two point three zero three R, with the extra two point three zero three below. Before using any slope, check the axis: natural log, or log to the base ten.

3:17Two temperatures

Clear? Now the form used most in problems. Write the natural log equation at T one and at T two, and subtract; natural log of capital A cancels. Natural log of k two over k one equals E a over R, times one over T one, minus one over T two. The small one and two are just labels. Right? Now tell me. If T two is hotter than T one, is this bracket positive or negative? Positive, exactly! One over T one is bigger. So T two greater than T one, with the greater than sign, gives k two greater than k one. Heat it, and it speeds up. Clear? And a shortcut for first order reactions. t half, written t with a small one over two, is the half-life: zero point six nine three over k. So k two over k one equals the half-life at T one over the half-life at T two: flipped! Got it?

4:16How sensitive is k to temperature?

Now, how big is the effect? The temperature coefficient is k at T plus ten, over k at T: the effect of a ten degree rise. Near room temperature, it is usually two to three, just like our hook. Okay? Now think. Two reactions: one with a high activation energy, one with a low one. Which rate constant changes more when we heat them? The one with the larger E a, exactly! E a multiplies the bracket, so the same change in temperature moves natural log of k more. Bigger barrier, bigger effect. Right? And one more. The same ten degree rise changes k more at a low temperature, because one over T changes faster there. With me?

5:01Example 1: Ea 50 kJ/mol, 300 K to 310 K, k nearly doubles

Now let's solve one. The activation energy is fifty kilojoules per mole; k J means kilojoules. By what factor does k rise from three hundred to three hundred and ten kelvin? Capital K means kelvin. Pause the video and try it first. Done? Okay, let's check. Many students put fifty straight into the formula. Is that okay? No, no, careful here! R is in joules, so E a must be in joules too. Fifty kilojoules per mole is fifty thousand joules per mole. Clear? Now the bracket. One over three hundred, minus one over three hundred and ten, is ten over ninety three thousand. That is one over nine thousand three hundred, per kelvin. Okay? So natural log of k two over k one equals fifty thousand over eight point three one four, times one over nine thousand three hundred. That comes to zero point six four seven. Right? So k two over k one is e to the power zero point six four seven, about one point nine; the wavy equals sign means about. That answers our puzzle! About three percent hotter, and k almost doubles, because T sits in the power. Perfect!

6:21Example 2: Ea from half-life 20 min to 5 min, 300 K to 310 K

Now a harder one. This time, a first order reaction has a half-life of twenty minutes at three hundred kelvin, and five minutes at three hundred and ten kelvin; m i n means minutes. Find the activation energy. Pause the video and try it first. Done? Okay, let's check. The half-life goes from twenty to five, so k two over k one is five over twenty, one fourth, right? Oh no, careful here, that is the trap! The half-life fell, so the reaction got faster, and k must rise. Remember the flip: k two over k one equals the first half-life over the second, twenty over five, that is four. Okay? Now rearrange: E a equals R times natural log of four, over the same bracket, one over nine thousand three hundred. With natural log of four as one point three eight six, that is eight point three one four, times one point three eight six, times nine thousand three hundred. With me? That is about one point zero seven times ten to the power five joules per mole: one hundred and seven kilojoules per mole. See, four times faster for the same ten degrees needs a much bigger activation energy. Larger E a, more sensitive k.

7:44Quick revision: ln vs log slopes, two-T formula, half-lives

Perfect! So, quick revision. k equals capital A, e to the power minus E a over R T, with T in kelvin and E a in joules per mole. Natural log of k against one over T has slope minus E a over R; with log to the base ten, divide that by two point three zero three. For two temperatures, use natural log of k two over k one; for first order, flip the half-lives. Now try the practice problems on this page. All the best!