Hess's Law

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Burn carbon completely, and you get carbon dioxide in one step. Now do it in two steps: first carbon to carbon monoxide, then carbon monoxide to carbon dioxide. Which route gives out more heat? The one step? The two steps? Or exactly the same? Keep your guess in mind. Today's idea, Hess's law, comes up often in JEE thermochemistry. So, let's begin.

0:32Same ΔH by every route

So, look at this diagram. Enthalpy, capital H, goes up the side, and each flat line is one state. At the top: carbon as graphite, plus oxygen gas. The small g in brackets means gas, and the two in O two is a small subscript, okay? At the bottom: carbon dioxide gas, much lower in enthalpy. The orange arrow on the left is the one step route. Delta H equals minus three hundred and ninety three point five. The little triangle is the capital Greek letter delta, and delta H means the change in enthalpy. The unit, k J mol to the power minus one, is read as kilojoules per mole. Right? Now the green route, in two steps. First, down to the middle line: carbon monoxide, with half a mole of oxygen left over, the one half on screen. Delta H one, with a small subscript one, is minus one hundred and ten point five. Then carbon monoxide burns to carbon dioxide: delta H two is minus two hundred and eighty three point zero. Add the two. What do you get? Minus three hundred and ninety three point five. Exactly the same! So, the answer to our puzzle: one step or two, the same heat comes out. Clear? This is Hess's law of constant heat summation. The enthalpy change of a reaction is the same, whether it takes place in one step or in several. Okay? But why? Why must every route agree? Because enthalpy is a state function, yes! It depends only on where you start and where you finish, not on the path. If two routes gave different values, you could go out by one and come back by the other, and create energy from nothing. The first law of thermodynamics does not allow that. With me?

2:22Thermochemical equations work like algebra

So, in practice: thermochemical equations, the ones with delta H written beside them, work just like algebra. Rule one: add two equations, and you add their delta H values. The double arrow on screen is read as: it follows that. That is exactly what the green route did. Okay? Now think. You run a reaction backwards. What happens to its delta H? The sign changes, correct! That's rule two. If going forward gives out heat, coming back must take in the same amount. And subtracting an equation is just adding its reverse. Got it? Rule three: multiply an equation by a number n, and multiply its delta H by the same n. Oh no, careful here! Students often double the equation, but forget to double its delta H value. Don't do that! Quick test with our figure. Suppose you know only the orange arrow and the second green arrow. How would you get delta H one? Subtract, yes! The orange route equals the two green steps together. So, delta H one equals delta H, minus delta H two. Clear? Minus three hundred and ninety three point five, minus, in brackets, minus two hundred and eighty three point zero. Careful with those two minus signs: together they make a plus. So, minus one hundred and ten point five kilojoules per mole, the same as the figure shows. Perfect!

3:49From combustion data: note the order

Now, questions often give you enthalpies of combustion: the enthalpy change when one mole of a substance burns completely in oxygen. With formation enthalpies, we take products minus reactants. So with combustion enthalpies also, products minus reactants, right? No, no, careful! Here it is the other way round: reactants minus products. Delta r H, small subscript r for reaction, equals the sum of nu times delta c H of the reactants, minus the sum of nu times delta c H of the products. The capital sigma means add them all up. Nu, the Greek letter that looks like a v, is the coefficient from the balanced equation. And the small c stands for combustion. Okay? Why this order? Hess's law again! Take this route: burn the reactants completely, then reverse the burning of the products. So the reactants come in with a plus sign, and the products, reversed, with a minus sign. With me? So, remember the pair. Formation data: products minus reactants. Combustion data: reactants minus products. Mix them up, and you get the right number with the wrong sign. Got it?

5:01Example 1: enthalpy of formation of SO3

Alright, a course question. Solid sulfur, the small s in brackets, burns in oxygen to give sulfur dioxide; the arrow means gives. Delta H is minus two hundred and ninety seven kilojoules, the k J on screen. Then sulfur dioxide plus half a mole of oxygen gives sulfur trioxide: delta H is minus ninety eight kilojoules. Find delta f H naught of sulfur trioxide. The small f is for formation, and the little circle, read as naught, means standard. Pause the video, and try it first. Done? Okay, let's check. The second equation makes sulfur trioxide, so its formation enthalpy is just minus ninety eight, right? Oh no, careful! Formation means one mole made from its elements, in their standard states. Sulfur dioxide is not an element. So our target is: solid sulfur, plus three halves of a mole of oxygen, gives one mole of sulfur trioxide. With me? Now just add the two equations. Sulfur dioxide appears on both sides, so it cancels. And one mole of oxygen plus half a mole gives three halves of a mole. That is exactly our target! Right? So, add the delta H values too: minus two hundred and ninety seven, plus minus ninety eight. Minus three hundred and ninety five kilojoules per mole. That's it!

6:28Example 2: enthalpy of formation of CS2 from combustion

Now a harder one. Find delta f H naught of carbon disulfide, C S two, a liquid; the small l in brackets means liquid. You are given enthalpies of combustion. Carbon disulfide burns to carbon dioxide and sulfur dioxide: minus one thousand one hundred and ten. Carbon: minus three hundred and ninety three point five. Sulfur: minus two hundred and ninety seven. All in kilojoules per mole. Pause and try. Hint: which formula fits combustion data? Done? Okay, first things first. Which reaction are we actually after? Formation, so from the elements, yes! One carbon plus two sulfur gives one C S two. Notice that two in front of the sulfur. Okay? Now the formula. Combustion data, so which comes first: reactants or products? Reactants first, correct! Carbon gives minus three hundred and ninety three point five. Sulfur gives two times minus two hundred and ninety seven, because of that two. Together, minus nine hundred and eighty seven point five. Right? The product side has only carbon disulfide: minus one thousand one hundred and ten. Clear? So, delta f H naught equals minus nine hundred and eighty seven point five, minus, in brackets, minus one thousand one hundred and ten. Plus one hundred and twenty two point five kilojoules per mole! And if you did products minus reactants? Minus one hundred and twenty two point five: the right number, the wrong sign. That is the trap!

8:04Quick revision: add, reverse, multiply; combustion data rule

So, quick revision. One: Hess's law, the same delta H by every route, because enthalpy is a state function. Two: add, reverse or multiply the equations, and do exactly the same to delta H; reversing changes the sign. Three: with combustion data, it is reactants minus products. Now go and try the practice problems on this page. All the best!