Entropy of the Surroundings and of the Universe

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. When water freezes, its molecules settle into an ordered solid, so the entropy of the water goes down. And the second law says that in a spontaneous process, entropy increases, right? But take supercooled water, that is, liquid water cooled below its freezing point. It can freeze all by itself! So is the second law broken? Of course not. Something is missing from our accounts, and today we find it: the surroundings. Questions on this come up often in JEE, so let's begin.

0:39The surroundings: a huge reservoir

So, first, what are the surroundings? Everything outside our system. And compared with the system, the surroundings are huge, so huge that their temperature, capital T, stays constant. Think of them as a giant reservoir at constant temperature. Okay? Now, earlier in this chapter we learnt delta S equals q rev over T, the formula in the badge. The little triangle is the capital Greek letter delta, meaning change in, and small q is heat. The small word rev written below q means this heat must be transferred reversibly. But a real reaction is usually irreversible. So can we use this formula for the surroundings at all? Yes, we can! The surroundings are a huge reservoir at constant temperature, so whatever heat they receive, they receive reversibly. So delta S surr, with the small word surr below for surroundings, equals q surr over T: the heat received by the surroundings, divided by their temperature. Clear? Now, where does that heat come from? From the system, of course. The heat gained by the surroundings is the heat lost by the system, so q surr equals minus q sys; the small word sys means system. The double arrow on screen means so. Delta S surr equals minus q sys over T, got it? And at constant pressure, the heat of the system is just its enthalpy change, delta H sys; capital H is enthalpy, remember? So, delta S surr equals minus delta H sys over T, okay? Write this down. This one line is the heart of today's lesson. With me?

2:15Exothermic helps the surroundings, more when cold

Now, let's use it. Take an exothermic reaction: it releases heat. What sign does delta S surr get, positive or negative? Positive, yes! Exothermic means delta H sys is less than zero, that is, negative. And a minus sign in front of a negative number gives a positive number, so delta S surr is greater than zero. The released heat goes into the surroundings and raises their entropy. Right? And endothermic is just the opposite. Delta H sys is greater than zero, so delta S surr is less than zero. The system takes heat from the surroundings, and the surroundings lose entropy. Clear? Now a deeper question. The same amount of heat is released, once into cold surroundings, and once into hot surroundings. Where does the entropy go up more? Into the cold surroundings, exactly! Look at the formula: T sits in the denominator. Smaller T, bigger delta S surr. The same heat causes a larger relative increase in disorder in cold surroundings. So an exothermic reaction helps more at low temperature. Remember this; we will need it in Example two, okay?

3:27The verdict: the entropy of the universe

Now we put both pieces together. The system and its surroundings make up the universe. So delta S total, the entropy change of the universe, equals delta S sys plus delta S surr. Easy, isn't it? And at constant temperature and constant pressure, capital P, we replace delta S surr by minus delta H sys over T, as we just found. So delta S total equals delta S sys, minus delta H sys over T, right? Now you need just two numbers about the system: its delta S and its delta H, nothing more. Really simple! And now the verdict. If delta S total is greater than zero, the process is spontaneous. Clear? If delta S total equals zero, the system is at equilibrium. Got it? Now the last case. Suppose delta S total comes out less than zero. So the process simply does not happen, and that is the end of the story, right? No, no, careful! This direction is not spontaneous, but the reverse process is spontaneous. Run it backwards, and every sign flips, so delta S total becomes positive. This point comes up often in multiple choice questions. With me?

4:40Example 1: entropy of surroundings when 150 kJ is released

Now, Example one, an easy one. Some reaction releases one hundred and fifty kilojoules at three hundred kelvin and constant pressure. On screen, k J means kilojoules, and capital K means kelvin. Find the entropy change of the surroundings. Pause the video, and try it. Done? Okay, let's check. The reaction releases heat, so it is exothermic, and delta H sys is negative: minus one hundred and fifty kilojoules. Entropy change is measured in joules per kelvin, so change it to joules first, capital J: minus one hundred and fifty thousand joules. Okay? Now, many students simply divide delta H by T here. Minus one hundred and fifty thousand over three hundred gives minus five hundred. So the answer is minus five hundred, right? Oh no, careful! You forgot the minus sign in the formula. Delta S surr equals minus delta H sys over T, remember? So it is minus, times minus one hundred and fifty thousand in brackets, over three hundred. Clear? Two minus signs make a plus: plus five hundred joules per kelvin. The K to the power minus one on screen is read as per kelvin. And it is positive, just as it should be, because an exothermic reaction raises the entropy of the surroundings. That's it!

6:02Example 2: is supercooled water freezing spontaneous?

Now Example two, and this is our puzzle. Supercooled water freezes at two hundred and sixty three kelvin, releasing five point six two kilojoules per mole. The mol to the power minus one means per mole. And delta S sys is minus twenty point five joules per kelvin per mole. Is the freezing spontaneous? Pause and try. Hint: include the surroundings. Done? Okay, let's check. The water becomes an ordered solid, so delta S sys is negative: minus twenty point five. The entropy goes down, so the freezing cannot be spontaneous, right? No, no, wait! That is exactly the trap from our puzzle. Never judge by the system alone. Freezing releases five point six two kilojoules per mole into the surroundings. First change it to joules: five thousand six hundred and twenty joules per mole. Delta S sys is in joules per kelvin per mole, so if you keep kilojoules here, the total comes out wrong. Okay? The surroundings receive this heat at two hundred and sixty three kelvin. So delta S surr is plus five thousand six hundred and twenty over two hundred and sixty three: plus twenty one point four joules per kelvin per mole. Positive, because the heat goes into the surroundings. Clear? So delta S total is minus twenty point five, plus twenty one point four: plus zero point nine joules per kelvin per mole. Small, but greater than zero! Right? So yes, the freezing is spontaneous. Puzzle solved! The water does lose entropy, but the surroundings gain a little more. And see the low temperature, two hundred and sixty three kelvin, in the denominator: the cold surroundings gain more entropy from the same heat. The entropy of the universe goes up, and the second law is perfectly safe. Perfect!

7:56Quick revision: entropy of surroundings and universe

So, quick revision. One: delta S surr equals minus delta H sys over T, so an exothermic reaction helps, and more so at low temperature. Two: delta S total is delta S sys plus delta S surr; greater than zero means spontaneous, and zero means equilibrium. Three: convert kilojoules to joules, and never judge by the system alone. Now go and try the practice problems on this page. All the best!