Gibbs Energy and the Equilibrium Constant

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Here is a reaction with delta G naught equal to minus five kilojoules per mole. Negative! So it must go forward, right? Hmm, not always! Today you will see a mixture of this very reaction that runs backward. The answer lies in Q, K, and the difference between delta G and delta G naught. Questions linking delta G naught and K come up often in JEE, so let's begin.

0:32Gibbs energy is lowest at equilibrium

So, look at this graph. Capital G, the Gibbs energy of the reaction mixture, goes up; the extent of reaction goes across, from pure reactants to pure products. Where is G the lowest? At pure products? No! At this orange point, the minimum: the equilibrium mixture. Why a mixture? Mixing reactants and products adds a large entropy of mixing, so the lowest point is a mixture. And note the condition: constant temperature and pressure, capital T and capital P, okay? Now, suppose the mixture sits to the left of the minimum. Which way will it move? Forward, downhill! Here capital Q, the reaction quotient for the mixture right now, is less than capital K, the equilibrium constant, the value Q takes at equilibrium. And delta G is less than zero; the little triangle is the capital Greek letter delta, meaning change in. So the forward reaction proceeds. Clear? And to the right of the minimum, Q is greater than K, and delta G is positive, so the reaction runs backward, again downhill. With me? At the minimum, delta G is zero and Q equals K, okay? And this orange arrow is delta r G naught: small r for reaction, and the small circle on top, read naught, means standard. It is the gap between pure reactants and pure products.

1:59The link between delta G naught and K

Now, the key equation: delta G equals delta G naught, plus R T times the natural log of Q, okay? Delta G is for the mixture right now; delta G naught, for standard states. Capital R is the gas constant, capital T the temperature in kelvin, and l n the natural log. So, what happens at equilibrium? Think back to the bottom of that graph. Yes, exactly! Delta G becomes zero and Q becomes K, so zero equals delta G naught plus R T times the natural log of K, okay? Move it across: delta G naught equals minus R T times the natural log of K, or, with log to base ten, minus two point three zero three R T log K, as you see. This is the heart of the lesson. Right? Now a topper's trick. At two hundred and ninety eight kelvin, two point three zero three R T is five point seven one kilojoules per mole. Careful: this capital K is kelvin, the unit, and mol to the power minus one means per mole. With me?

3:06What the sign of delta G naught tells you

Now, read the sign. If delta G naught is negative, is K bigger than one, or smaller? Bigger, correct! Minus R T times the natural log of K is negative only when K is more than one. The double headed arrow means it works both ways. Clear? And if delta G naught is exactly zero, K equals one. Got it? Now a trap. At equilibrium, delta G is zero. So delta G naught is zero there too, right? No, no, careful! At equilibrium delta G is zero, but delta G naught stays fixed at minus R T times the natural log of K, okay? Why? Look at the orange arrow. Delta G naught refers to every species in its standard state, a fixed reference, not the equilibrium mixture; on this graph it is the gap between pure reactants and pure products. At a given temperature it does not depend on the mixture, so it stays fixed. Right?

4:05Temperature and K: the van't Hoff equation

Now, temperature. From the last lesson, delta G naught equals delta H naught minus T delta S naught, the standard enthalpy and entropy changes. Equate it with minus R T times the natural log of K, as on screen. Okay? Divide both sides by minus R T, and the natural log of K equals minus delta H naught over R T, plus delta S naught over R, as shown. Right? So a graph of log K against one over T is a straight line: slope minus delta H naught over two point three zero three R, and intercept delta S naught over two point three zero three R, okay? Write it at T one and T two, with equilibrium constants K one and K two, and subtract. The van't Hoff equation: log of K two over K one equals delta H naught over two point three zero three R, times, in brackets, one over T one minus one over T two. Clear? For an exothermic reaction, delta H naught is negative. We raise the temperature. Does K rise, or fall? It falls! With delta H naught negative, the first term, minus delta H naught over R T, is positive, and it shrinks as T rises. For an endothermic reaction, the opposite: K rises with temperature. Got it?

5:30Example 1: delta G naught when K = 100 at 298 K

Your turn. Find delta G naught at two hundred and ninety eight kelvin, for a reaction with K equal to one hundred. Pause the video and try it. Hint: use the five point seven one trick. Done? Let's check. At two hundred and ninety eight kelvin, delta G naught is minus five point seven one times log K, in kilojoules per mole. And log of one hundred, to base ten, is two. Right? So, minus five point seven one times two: minus eleven point four two, about minus eleven point four kilojoules per mole; the wavy equals sign means approximately. And see: K above one, delta G naught negative, as our rule says. Clear?

6:16Example 2: delta G from Q, which way the reaction runs

Now the harder one: our puzzle from the start! Delta G naught is minus five point zero kilojoules per mole, at three hundred kelvin. At a certain moment, Q is twenty. Find delta G, and the direction of reaction. Given: the natural log of twenty is three point zero. Pause the video and try it. Done? Okay, let's check. Delta G equals delta G naught plus R T times the natural log of Q, as before. But delta G naught is in kilojoules, and the R T term comes out in joules. Can we add them as they are? No, no! Convert first: minus five point zero kilojoules per mole is minus five thousand joules per mole. This slip is very common. Okay? R is eight point three one four joules per kelvin per mole. So, eight point three one four, times three hundred, times three point zero, is about seven thousand four hundred and eighty joules per mole. Right? So delta G is minus five thousand plus seven thousand four hundred and eighty: plus two thousand four hundred and eighty joules per mole, about plus two point five kilojoules per mole. Now, delta G naught was negative, so the reaction runs forward, right? No, careful! The direction right now is decided by delta G, not by delta G naught. Delta G is positive, which means Q is greater than K, so the reaction runs backward, as on screen. Puzzle solved! With me?

7:51Quick revision: delta G, Q, K and the van't Hoff equation

So, quick revision. One: delta G equals delta G naught plus R T times the natural log of Q, and at equilibrium, delta G is zero and Q equals K, okay? Two: delta G naught equals minus two point three zero three R T log K, and at two hundred and ninety eight kelvin, two point three zero three R T is five point seven one kilojoules per mole. Three: the direction now comes from the sign of delta G; how K moves with temperature, from the sign of delta H naught. Now try the practice problems on this page. All the best!