ΔH and ΔU of Reactions

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Burn one mole of methane in a sealed steel container: about eight hundred and eighty five kilojoules of heat come out. Burn it in an open vessel, at constant pressure: about eight hundred and ninety. Same reaction. So where do the five extra kilojoules come from? Hold that thought. This idea comes up often in JEE, so let's begin.

0:27Two calorimeters, two different heats

So, how do we measure the heat of a reaction? With a calorimeter. In the figure, the weighed sample sits in a steel bomb with excess oxygen, O two, and the bomb sits in water in an insulated jacket. Ignition wires fire it; a thermometer and stirrer follow the water, okay? Now think. The steel bomb is sealed and rigid. Its volume cannot change. So, can the gases inside do any expansion work? No, none at all! No expansion, no work, so the heat exchanged equals the change in internal energy. At constant volume, capital V, the heat, q with a small subscript V, equals delta U, okay? Delta, the little triangle, means change in, and U is the internal energy. Right? Now, a coffee cup calorimeter is open to the air, so the pressure, capital P, stays constant. There, the heat, q with a small subscript P, equals delta H, the change in enthalpy. Clear? And the number? The heat released by the reaction equals C cal times delta T, the rise in temperature. Capital C with the small subscript cal is the heat capacity of the calorimeter. The reaction gives this heat out, so its q is negative: that is the minus sign in the figure's formula, okay? And C cal itself? We first burn a standard, like benzoic acid, whose heat of combustion is known. The figure also links delta H and delta U; that is next. Let's go.

2:01Linking ΔH and ΔU

Now, how are delta H and delta U related? From the enthalpy lesson: H equals U plus P V, pressure times volume. So delta H equals delta U plus delta of P V in brackets, the change in P V, okay? Now tell me. In a reaction, which substances can change P V by a big amount: solids, liquids, or gases? Gases, exactly! Solids and liquids hardly change their volume. For ideal gases, P V equals n R T: n moles, R the gas constant, T the temperature in kelvin. At the same temperature, delta of P V becomes delta n g times R T: that is n with a small subscript g, the change in moles of gas. With me? And that gives the key result: delta H equals delta U plus delta n g R T, okay? Write this down; it is the heart of today's lesson. Got it? And delta n g is the moles of gaseous products minus the moles of gaseous reactants, read from the balanced equation. Clear?

3:05Equal or not? Count only the gases

Let's use it. Hydrogen gas plus iodine gas gives two moles of hydrogen iodide gas; the arrow means gives, and the small g in brackets means gas. Two moles of gas each side: delta n g is two minus two, zero. Simple, right? Now carbon burns in oxygen to give carbon dioxide. One mole of product, two of reactants, so delta n g is one minus two, minus one, isn't it? No, no, careful! Carbon has a small s in brackets: it is a solid, and solids do not count. The only gaseous reactant is oxygen, so delta n g is one minus one, zero. Classic trap: count only the gases. Okay? So, delta H equals delta U whenever delta n g is zero, or when only solids and liquids take part. Why? Their volumes change very little, so P V hardly changes. Got it? And when delta n g is negative, delta H is less than delta U, okay? This sign, pointing left, means less than. For an exothermic reaction, delta H is more negative. Remember this for the puzzle, right? Last, the signs. Exothermic, heat given out: delta H is less than zero. Endothermic, heat taken in: delta H is greater than zero. Clear?

4:24Topper's trick: keep RT ready at 298 K

Now, a trick toppers use. Many questions are set at two hundred and ninety eight kelvin, capital K for kelvin. So keep R T ready: eight point three one four times two hundred and ninety eight, two thousand four hundred and seventy eight joules per mole; the little minus one means per. The wavy sign means approximately: about two point four eight kilojoules per mole, okay? So at two hundred and ninety eight kelvin, delta H minus delta U is just delta n g times two point four eight kilojoules per mole. One multiplication. Nice, isn't it?

5:00Example 1: delta U for N2O4 to 2NO2

Okay, Example one. N two O four gas gives two moles of N O two gas. Delta H is plus fifty seven point two kilojoules per mole, at two hundred and ninety eight kelvin. Find delta U, okay? Pause the video, and try it first. Done? Okay, let's check. Gaseous products: two moles of N O two. Gaseous reactants: one mole of N two O four. So delta n g equals two minus one, plus one. With me? Now, R T is two thousand four hundred and seventy eight. So delta U is fifty seven point two minus two thousand four hundred and seventy eight, right? Oh no, careful! That R T is in joules, but delta H is in kilojoules: convert first, two point four eight. So delta U equals delta H minus delta n g R T: fifty seven point two minus one times two point four eight. Okay? That gives plus fifty four point seven kilojoules per mole. Gas moles go up, so delta U comes out smaller than delta H here. Perfect!

6:05Example 2: delta U for combustion of methane

Now Example two, a little harder: our puzzle. Delta c H of methane, small subscript c for combustion, is minus eight hundred and ninety point three kilojoules per mole, with liquid water formed. Find delta c U at two hundred and ninety eight kelvin. Pause and try. Hint: write the equation first. Done? Okay, let's check. Methane plus two moles of oxygen gives carbon dioxide plus two moles of water. The small l in brackets means the water is liquid. Okay? Now delta n g. Three moles on the left, one plus two, three on the right. So zero, and delta H equals delta U, right? No, no, wait! The water is a liquid, so it does not count. Gaseous products: one carbon dioxide. Gaseous reactants: three. So delta n g is one minus three, minus two. Clear? Now rearrange: delta U equals delta H minus delta n g R T, okay? Delta n g is minus two, and minus times minus gives plus, so delta U equals delta H plus two R T, got it? Use our trick: R T is two point four eight. So minus eight hundred and ninety point three, plus two times two point four eight, that is four point nine six. Right? So delta c U equals minus eight hundred and eighty five point three kilojoules per mole. Puzzle solved! The open vessel gives delta H, eight hundred and ninety point three released; the sealed bomb gives delta U, eight hundred and eighty five point three. The extra, two R T, about five kilojoules, comes because three moles of gas shrink to one. That's it!

7:49Quick revision: bomb calorimeter, delta ng, RT at 298 K

So, quick revision. One: a bomb calorimeter, at constant volume, gives delta U; at constant pressure, the heat is delta H, okay? Two: delta H equals delta U plus delta n g R T; count only gaseous moles, and R T is two point four eight kilojoules per mole at two hundred and ninety eight kelvin. Three: delta n g zero, they are equal; negative, delta H is smaller. Now try the practice problems on this page. All the best!