Phase Changes and the Effect of Temperature

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Hot water at one hundred degrees Celsius splashes on your hand. It burns, right? Now steam, also at one hundred degrees Celsius. Same temperature. Yet a steam scald is much worse than a water scald. So why? Strange, isn't it? By the end of this video, you will know the exact reason, with numbers. Phase changes and their enthalpies come up often in JEE, so let's begin.

0:31Enthalpies of phase change

So, three phase changes. The little triangle is the capital Greek letter delta, meaning change in, and H is enthalpy. The small word below tells you which change. Fus is short for fusion, that is melting; vap is vaporisation, that is boiling; and sub is sublimation, a solid turning straight into a gas. Okay? Now tell me. When ice melts, does it take in heat, or give heat out? It takes heat in, yes! You must supply heat to melt, to boil, or to sublime. So all three of these enthalpies are positive. Right? And the reverse changes, freezing and condensing? They release exactly the same amounts of heat. Same size, opposite sign. Remember this; it is the key to our steam puzzle. Clear? Now, sublimation: solid straight to gas, in one jump. Can you get delta sub H from the other two? Yes, exactly! Think of it as two steps: melt first, then boil. For enthalpy, the path does not matter, only the start and the end. So delta sub H equals delta fus H plus delta vap H, but only when all three are at the same temperature. Got it?

1:46Why boiling needs far more heat than melting

Now a question. Per mole, which needs more heat: melting ice, or boiling water? Boiling, by far! Correct! Delta vap H is much greater than delta fus H, by a lot. This sign, like two greater than signs side by side, means much greater than. With me? But why? When a solid melts, the forces between the molecules, the intermolecular forces, are only loosened. In ice, only some of the hydrogen bonds break. Okay? When a liquid boils, the molecules must be separated completely, so those forces are fully overcome. That costs much more energy. Right? Look at water. Delta fus H is six point zero one, at zero degrees Celsius, and delta vap H is forty point seven, at one hundred degrees Celsius. Both are in k J mol to the power minus one, which means kilojoules per mole. Almost seven times as much for boiling! We will use that forty point seven soon. Clear?

2:52Effect of temperature: Kirchhoff's equation

Now, the second half of our title: the effect of temperature. Suppose you know delta H of a reaction at one temperature. At a higher temperature, will it be the same? Not in general! Kirchhoff's equation tells us how it changes. Delta H at T two equals delta H at T one, plus delta C P times, T two minus T one. T one and T two, with the small one and two below, are the two temperatures. And C P, with a small capital P below, is the molar heat capacity at constant pressure. Okay? And delta C P? It is the sum of C P of the products, minus the sum of C P of the reactants. The big zigzag is the capital Greek letter sigma, meaning sum of. Products minus reactants, like always. With me? Now, careful, here is a trap! Delta C P comes in joules per kelvin, but delta H is usually in kilojoules. So divide by one thousand before you add. Very common slip! Got it? Quick question. When will delta H not change with temperature at all? When delta C P is zero, exactly! Then the last term vanishes: the products and reactants have equal total heat capacities, so delta H is the same at every temperature. Right?

4:13Example 1: heat released when 9 g of steam condenses

Now, Example one, and it brings back our steam. How much heat is released when nine grams of steam at one hundred degrees Celsius condenses, and the water then cools to twenty degrees Celsius? The small circle and C mean degrees Celsius. Pause the video, and try it. Done? Okay, let's check. Delta vap H of water, forty point seven kilojoules per mole, and the specific heat of water, four point one eight joules per gram per kelvin. Okay? First, condensing. Nine grams of steam, and forty point seven. So the heat is nine times forty point seven, right? No, no, wait! Forty point seven is per mole, not per gram. One mole of water is eighteen grams, so n, the number of moles, is nine over eighteen, that is zero point five mole. Zero point five times forty point seven: twenty point three five kilojoules, released, because condensing is the reverse of boiling. Clear? Next, the water cools from one hundred to twenty degrees Celsius, a drop of eighty degrees, the same as eighty kelvin. Mass times specific heat times the drop: nine times four point one eight times eighty. Three thousand and nine point six joules. The wavy equals sign means approximately: about three point zero one kilojoules. With me? So the total is twenty point three five plus three point zero one. About twenty three point four kilojoules released. Perfect! Now look back at our puzzle. Out of twenty three point four kilojoules, twenty point three five come from condensing alone. Steam gives out this huge amount on your skin before it even starts to cool. Water at one hundred degrees skips that part. That is why the steam scald is so much worse. Mystery solved!

6:07Example 2: delta H of ammonia synthesis at 398 K

Now a harder one, with Kirchhoff's equation. For the reaction N two plus three H two gives two N H three, the arrow means gives, delta H is minus ninety two point two kilojoules at two hundred and ninety eight kelvin. The molar heat capacities: N two, twenty nine point one; H two, twenty eight point eight; N H three, thirty five point one, all in joules per kelvin per mole. Find delta H at three hundred and ninety eight kelvin. Pause and try. Done? Okay, let's check. First, delta C P, products minus reactants. So, thirty five point one, minus twenty nine point one, minus twenty eight point eight, right? Oh no, careful! Look at the equation again. It makes two moles of ammonia, and it uses three moles of hydrogen. So, two times thirty five point one, minus, in big square brackets, twenty nine point one plus three times twenty eight point eight. Okay? That is seventy point two, minus twenty nine point one, minus eighty six point four. So delta C P is minus forty five point three joules per kelvin. Right? Now, T two minus T one is three hundred and ninety eight minus two hundred and ninety eight: one hundred kelvin. Minus forty five point three times one hundred: minus four thousand five hundred and thirty joules. Remember the units trap? Divide by one thousand: minus four point five three kilojoules. Got it? Finally, delta H at three hundred and ninety eight kelvin is minus ninety two point two, plus minus four point five three. About minus ninety six point seven kilojoules. Delta C P was negative, so delta H became more negative as the temperature went up. Very good!

7:54Quick revision: fusion, vaporisation, Kirchhoff equation

So, quick revision. One: melting, boiling and subliming all take in heat, and the reverse changes give out the same amounts. Two: at the same temperature, delta sub H equals delta fus H plus delta vap H, and delta vap H is much greater than delta fus H, okay? Three: Kirchhoff's equation, delta H at T two equals delta H at T one, plus delta C P times, T two minus T one, with the units matched. Now go and try the practice problems on this page. All the best!