Set 1

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Two students solve the same question with the same data: the enthalpies of combustion of propane, graphite and hydrogen. One gets plus one hundred and three point seven kilojoules per mole. The other gets minus one hundred and three point seven. Same numbers, opposite signs! So who is right? Hold that thought. This is our first mixed problem set in chemical thermodynamics. Today, the tools that crack it, and the traps inside. Problems like these come up often in JEE Main. So, let's begin.

0:39Tool 1: first law, fix the signs first

So, first tool: the first law. Delta U equals q plus w. The little triangle is the Greek letter delta, meaning change in. U is the internal energy, q the heat, and w the work, okay? Now the signs. q is greater than zero, see the greater than sign, when the system absorbs heat. And w is greater than zero when work is done on the system. So heat given out, and work done by the system, are negative. Clear? Now problem one of this set. The system absorbs two point five kilojoules of heat and does one point two kilojoules of work; k J means kilojoules. Delta U is the question mark. So just add them: three point seven, right? No, no, careful! The system does the work, so w is minus one point two. It absorbs the heat, so q is plus two point five. Delta U is plus one point three kilojoules. Fix the signs first, then add. Got it?

1:42Tool 2: heating an ideal gas, Cv or Cp?

Second tool: heating an ideal gas. Its internal energy depends only on temperature. So for any process, delta U equals n C V delta T, where n is the number of moles. C V, the C with a small V below it, is the molar heat capacity at constant volume. Okay? And at constant pressure, capital P, the heat q P equals delta H, which equals n C P delta T, with C P the molar heat capacity at constant pressure. Right? Now problem three. One mole of a monatomic ideal gas is heated by forty kelvin at constant pressure, and we need q, and delta U as well. For a monatomic gas, C P is five by two R, twenty point seven nine, and C V is three by two R, twelve point four seven, in J K to the minus one mol to the minus one: joules per kelvin per mole. With me? Constant pressure, so q is twenty point seven nine times forty: about eight hundred and thirty two joules. The wavy equals sign means approximately. Now, eight hundred and thirty two joules went in, so delta U is also eight hundred and thirty two, right? Oh no, careful! At constant pressure the gas expands and does work, so part of that heat leaves as work. Delta U uses C V: twelve point four seven times forty, about four hundred and ninety nine joules. Clear?

3:01Tool 3: which way do we subtract?

Third tool, and it decides our puzzle. With enthalpies of formation, we subtract products minus reactants. The small r means reaction, the small f formation, and the little circle, naught, means standard. The big sigma means sum, and nu, the Greek letter, is each coefficient in the equation. Okay? Now, bond enthalpies. Is it products minus reactants again? No, the opposite! Breaking bonds costs energy, and forming bonds gives it out. So delta r H equals bonds broken minus bonds formed: reactants minus products. And this method is for gaseous reactions. Clear? Take problem eight. Methane burns: C H four plus two O two gives C O two plus two H two O, all gases. The arrow means gives, the small numbers below count atoms, and g in brackets means gas, okay? Broken: four C H bonds at four hundred and fourteen, and two O O double bonds at four hundred and ninety eight; the number in front multiplies each bracket. That makes two thousand six hundred and fifty two. Formed: two C O double bonds at seven hundred and ninety nine, and four O H bonds at four hundred and sixty three: three thousand four hundred and fifty. Right? So delta r H is two thousand six hundred and fifty two minus three thousand four hundred and fifty: minus seven hundred and ninety eight kilojoules. Negative, so heat is given out: the reaction is exothermic. Got it?

4:34Example 1: delta H minus delta U for propane combustion

Now, example one: problem five of the set. Propane burns at three hundred kelvin, as in the equation on screen: C three H eight plus five O two gives three C O two plus four H two O, where the brackets give the states: g for gas, and l for liquid. Find delta H minus delta U for this reaction. Pause the video, and try it first. Done? Okay, let's check. Delta H equals delta U plus delta n g R T, where R is the gas constant and T the temperature in kelvin. So delta H minus delta U is just delta n g R T, nothing more. Clear? Delta n g, the n with a small g below it, is the moles of gaseous products minus the moles of gaseous reactants. So, what is delta n g here? If you said seven minus six, plus one, careful! Water here is a liquid, the small l. Only gases count. Gas products: three. Gas reactants: one plus five, six. So delta n g is three minus six: minus three. Right? Now, minus three, times eight point three one four, times three hundred, in joules, since R is in joules. That is about minus seven point four eight kilojoules. So delta H is lower than delta U here. With me?

5:57Example 2: enthalpy of formation of propane from combustion

Now the harder one, and it is our opening puzzle: problem seven. The enthalpies of combustion of propane, graphite and hydrogen are minus two thousand two hundred and twenty, minus three hundred and ninety three point five, and minus two hundred and eighty five point eight kilojoules per mole. Find delta f H naught of propane. Pause and try. Hint: write the formation equation first. Done? Okay, let's check. Formation means one mole of propane from its elements in their standard states. So three carbons, as graphite, plus four H two gas, give C three H eight gas. Now, formation questions use products minus reactants, right? So minus two thousand two hundred and twenty, minus the values for carbon and hydrogen. That gives plus one hundred and three point seven. Is that our answer? Oh no, careful! These are combustion values, not formation values. Combustion data, just like bond enthalpies, go the other way: reactants minus products. The small c means combustion. Clear? So delta f H naught equals three times minus three hundred and ninety three point five, plus four times minus two hundred and eighty five point eight, minus, in brackets, minus two thousand two hundred and twenty. Okay? That is minus one thousand one hundred and eighty point five, minus one thousand one hundred and forty three point two, plus two thousand two hundred and twenty. Got it? So delta f H naught of propane is minus one hundred and three point seven kilojoules per mole. Puzzle solved! The second student is right. The first used the wrong order, and got the right number with the wrong sign. Always check which data you are given!

7:44Quick revision: signs, Cv, delta ng, which way to subtract

So, quick revision. First law: delta U equals q plus w, and work done by the system is negative. For an ideal gas, delta U is n C V delta T, and delta n g counts gases only. Formation data: products minus reactants. Bond enthalpies and combustion data: reactants minus products. Now try all fifteen problems of this set on this page, then check the solutions. You can do it. All the best!