The Third Law and Standard Entropies

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. For an element in its standard state, like oxygen gas, the standard enthalpy of formation is zero, remember? Now look up oxygen gas in a table of standard entropies. Two hundred and five joules per kelvin per mole! Not zero. So is the table wrong? Not at all. Entropy has a real zero, given by nature itself, and today we find where it is. Then we use these tables for reactions. Questions on this come up often in JEE, so let's begin.

0:37The third law: a true zero for entropy

So, look at this graph from an earlier lesson: capital S, the entropy, going up, and capital T, the temperature, going across. The entropy climbs as the substance warms, and jumps when it melts and when it boils. Now, do you remember: at zero kelvin, where does the line start? At zero, exactly! This is the third law of thermodynamics: the entropy of a perfectly crystalline substance is zero at zero kelvin; capital K is kelvin. Every particle sits perfectly in place, so there is no disorder at all. But note the condition: perfectly crystalline. Okay? So entropy has a true starting point, and we can measure the absolute entropy of a substance, counting up from zero. We write it as capital S with a small circle on top, read S naught; the circle means the standard value. Clear? Now, delta f H naught: the little triangle is the capital Greek letter delta, meaning change in, and the small f means formation. For an element in its standard state, it is zero. So its S naught is zero too, right? No, no, careful! The zero for delta f H naught is only a choice we made. Entropy has a real zero, at zero kelvin, and by two hundred and ninety eight kelvin every substance has warmed up from there. So S naught is not zero; the equals sign with a slash means not equal to. Got it? For oxygen gas, S naught is two hundred and five joules per kelvin per mole: capital J is joules, K to the power minus one is per kelvin, and mol to the power minus one is per mole. Puzzle solved! With me?

2:18What makes a standard entropy larger?

Now, what makes S naught large? Take iodine as a solid, a liquid and a gas. Which one has the largest S naught? The gas, of course! Gas molecules move freely, so they have the most disorder. On screen, the small s, l and g in brackets mean solid, liquid and gas, and the less than signs put them in order. Right? Next, molar mass. Helium, neon and argon are all gases of single atoms. The heavier the atom, the larger S naught. Clear? Then complexity: methane, ethane and propane. More atoms in the molecule means more ways to vibrate and rotate, so a larger S naught. Got it? Now, diamond and graphite are both pure carbon, and both are solids. So the same S naught, right? No! Diamond is two point four, graphite five point seven joules per kelvin per mole. Diamond is a rigid three dimensional network; graphite has layers that can vibrate and slide more freely. Softer and less ordered means larger entropy. With me?

3:27Entropy change of a reaction

Alright, now the most useful part: the entropy change of a reaction. Delta r S naught, small r for reaction, equals the sum of nu times S naught of the products, minus the same sum for the reactants. The big symbol is the capital Greek letter sigma, meaning sum, and nu is a small Greek letter; more on it in a moment. Okay? So, nu is the coefficient in the balanced equation: two moles means multiply by two. And remember our puzzle: elements have S naught too, so they count. Right? Now, gases have the largest entropies. So if a reaction makes more moles of gas, what is the sign of delta r S naught? Positive, yes! When the moles of gas increase, delta r S naught is greater than zero. Clear? And if the moles of gas decrease, it is less than zero. Topper's trick: count the moles of gas first and predict the sign. If the moles of gas change and your final answer has the other sign, check your working. Very useful!

4:32Residual entropy: when the crystal is not perfect

Now, remember the condition: perfectly crystalline. Is the entropy of solid carbon monoxide zero at zero kelvin? Not quite! In carbon monoxide, C O, and nitrous oxide, N two O, the molecules can point either way. Okay? Carbon monoxide has a very small dipole, so each molecule sits in the crystal as C O or O C, almost at random. Two orientations remain even at zero kelvin, so the crystal is not perfect. Right? This leftover is the residual entropy, and it comes close to R times l n two, where capital R is the gas constant and l n is the natural log. R times l n two is five point seven six joules per kelvin per mole; that is the value if every molecule picks one of the two orientations fully at random.

5:23Example 1: entropy change for CaCO3 decomposition

Clear? Now, Example one. Solid calcium carbonate gives solid calcium oxide and carbon dioxide gas; the arrow means gives. The S naught values are on screen, in joules per kelvin per mole. Find delta r S naught. Pause the video, and try it. Done? Okay, let's check. First, count the moles of gas: zero on the left, and one mole of carbon dioxide on the right. Gas increases, so we expect a positive answer. Okay? Now, products minus reactants. Products: thirty nine point eight plus two hundred and thirteen point seven. Minus ninety two point nine for calcium carbonate. Every coefficient here is one. Right? The products add up to two hundred and fifty three point five. So the answer is plus one hundred and sixty point six joules per kelvin, for the reaction as written. Positive, as predicted. That's it!

6:24Example 2: entropy change for ammonia synthesis

Now Example two, a little harder. Nitrogen gas plus three hydrogen gas gives two ammonia gas. The S naught values are on screen. Find delta r S naught. Pause and try. Hint: count the gas moles first. Done? Okay, let's check. One plus three, four moles of gas on the left; two moles of ammonia on the right. Gas decreases, so we expect a negative answer. With me? Products: two times one hundred and ninety two point five, which is three hundred and eighty five point zero. Don't forget that two! Okay? Now the reactants. Nitrogen and hydrogen are elements, so their S naught is zero, and the answer is plus three hundred and eighty five, right? Oh no, careful! That is our puzzle from the start: elements do not have zero S naught. And plus three hundred and eighty five has the wrong sign anyway. So, one hundred and ninety one point six, plus three times one hundred and thirty point seven, which is three hundred and ninety two point one. Clear? So delta r S naught is three hundred and eighty five point zero, minus one hundred and ninety one point six, minus three hundred and ninety two point one. Minus one hundred and ninety eight point seven joules per kelvin. Negative, as predicted. Perfect!

7:46Quick revision: third law, standard and reaction entropy

So, quick revision. One: a perfect crystal has zero entropy at zero kelvin, so entropies are absolute, and elements do not have zero S naught. Two: S naught grows from solid to liquid to gas, and with heavier, more complex, softer substances. And number three: delta r S naught is products minus reactants, times nu, and you check the sign with the moles of gas. Now go and try the practice problems on this page. All the best!