Standard Gibbs Energies

Transcript

0:00Introduction

Hello everyone! Diamonds are forever, right? But chemistry says that diamond turning into graphite, the black stuff in your pencil, is a spontaneous change: its Gibbs energy change is negative! So, should your diamond ring slowly turn into pencil lead? Strange, isn't it? Today, with the standard Gibbs energies, we find out why diamonds last. This comes up often in JEE, so let's begin.

0:28Standard Gibbs energy of formation

So, first, the key quantity. The little triangle is the capital Greek letter delta, meaning change in, and capital G is Gibbs energy. The small f below means formation, and the little circle on top, read as naught, means standard. So delta f G naught is the Gibbs energy change when one mole of a compound is formed. Okay? Formed from what? From its elements in their standard states. In the brackets: a pressure of one bar, and the temperature, usually two hundred and ninety eight kelvin, that capital K, right? Now tell me. What is delta f G naught of oxygen gas, O two, with the g in brackets for gas? To form oxygen gas from oxygen gas, what changes? Nothing changes, exactly! So for any element in its standard state, delta f G naught equals zero. Clear? Quick test. Oxygen gas: zero. Liquid bromine, B r two, with the l in brackets for liquid: zero too, since liquid is its standard state. Now ozone, O three gas, with the question mark. It is pure oxygen too, so zero again, right? No, no, careful! The standard state of oxygen is O two gas, not ozone. So ozone has a real value: plus one hundred and sixty three, k J mol to the power minus one, that is, kilojoules per mole. Got it?

1:57Standard Gibbs energy of reaction

Now, delta r G naught, with a small r for reaction, is the standard Gibbs energy of reaction. The big zigzag is the capital Greek letter sigma, meaning sum of; the small v-shaped letter is the Greek letter nu, which stands for each coefficient in the balanced equation. So: the sum of nu times delta f G naught of the products, minus the same sum for the reactants. Okay? Now, does this look familiar? Where have you seen exactly this pattern before? Yes, exactly! Delta r H naught, from enthalpies of formation. Same pattern: products minus reactants, each times nu. Right? And there is a second route. Delta r G naught also equals delta r H naught, minus T times delta r S naught. Capital T is the temperature in kelvin, and capital S is entropy. It is our Gibbs equation, now with standard values. With me?

2:56Positive Gibbs energy of formation: unstable yet lasting

Now, what does the sign tell us? If delta f G naught is greater than zero, that is, positive, the compound is thermodynamically unstable with respect to its elements. Breaking back into its elements would lower the Gibbs energy. Okay? Look at two real cases. Nitric oxide, N O, plus eighty six point six; and ethyne, C two H two, plus two hundred and nine point two kilojoules per mole. Both positive! So they should fall apart at once, and cannot exist at all, right? No, no, wait! Both of them exist. Unstable does not mean fast: their decomposition is very slow, so they survive. We call them kinetically stable; kinetics is about speed. So Gibbs energy gives the direction, not the speed. Got it? And now, our diamond. Diamond turning into graphite, the arrow means turns into, has delta G naught of minus two point nine kilojoules per mole. Negative, so spontaneous. Clear? So now you tell me. If the change is spontaneous, why does your diamond not turn into graphite? Yes, exactly! Same idea. The change has an enormous activation energy, a huge energy barrier to climb first. So at ordinary temperatures, it simply does not occur. Puzzle solved! Diamonds last because the change is far too slow. That's it!

4:23Example 1: delta G of 2SO2 + O2 to 2SO3 from formation data

Now, Example one, the easier one. Find delta r G naught for: two S O two gas, plus O two gas, gives two S O three gas. Given: delta f G naught of S O two, minus three hundred point two, and of S O three, minus three hundred and seventy one point one, both in kilojoules per mole. Pause the video, and try it. Done? Okay, let's check. But wait, no value is given for oxygen. Is the question incomplete? No, it is complete! Oxygen gas is an element in its standard state, so its delta f G naught is zero. Right? Now, products minus reactants. Many students simply write minus three hundred and seventy one point one, minus, minus three hundred point two. Is that correct? Oh no, careful! That forgets nu, the coefficients: two moles of S O three, and two of S O two. So: two times minus three hundred and seventy one point one, minus the square bracket: two times minus three hundred point two, plus zero for oxygen. With me? Two times minus three hundred and seventy one point one is minus seven hundred and forty two point two. The bracket is minus six hundred point four, and subtracting it means adding six hundred point four. Okay? So delta r G naught is minus one hundred and forty one point eight kilojoules. Keep this number in mind for the next example. Very good!

5:57Example 2: same reaction from delta H minus T delta S

Now Example two, a harder one: check Example one by the second route, at two hundred and ninety eight kelvin. Delta f H naught: S O two, minus two hundred and ninety six point eight; S O three, minus three hundred and ninety five point seven kilojoules per mole. And delta r S naught: minus one hundred and eighty seven point nine, J K to the power minus one, that is, joules per kelvin. Pause and try. Hint: watch the units. Done? Okay, let's check. First, delta r H naught, products minus reactants: two times minus three hundred and ninety five point seven, minus the square bracket: two times minus two hundred and ninety six point eight, plus zero for oxygen. Minus one hundred and ninety seven point eight kilojoules. Okay? Now, delta r G naught equals delta r H naught, minus T times delta r S naught. So we just put in minus one hundred and eighty seven point nine, right? No, no, wait! Delta r H naught is in kilojoules, but delta r S naught is in joules per kelvin. Mix them, and the answer is completely wrong. So divide by one thousand first: minus zero point one eight seven nine kilojoules per kelvin. Clear? Now substitute T, two hundred and ninety eight kelvin. Minus one hundred and ninety seven point eight, minus two hundred and ninety eight times, in brackets, minus zero point one eight seven nine. The cross means times. With me? Now careful with the sign. This second term: does it add, or subtract? It adds, correct! Minus times minus is plus. Two hundred and ninety eight times zero point one eight seven nine is fifty six point zero. So, minus one hundred and ninety seven point eight, plus fifty six point zero: minus one hundred and forty one point eight kilojoules. Okay? And look! The same answer as Example one. Both routes agree, exactly as they must. So use whichever data the question gives you. Perfect!

8:16Quick revision: zero for elements, products minus reactants

So, quick revision. One: delta f G naught is for one mole formed from elements in their standard states; for an element in its standard state, it is zero. Two: delta r G naught is products minus reactants, each times nu; or delta r H naught minus T delta r S naught, in kilojoules. Three: positive delta f G naught means unstable, yet it may last if the change is very slow. Now go and try the practice problems on this page. All the best!