Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Here is a reaction that gives out forty kilojoules of heat. Giving out heat helps a reaction, so it should go at any temperature, right? But heat it above five hundred kelvin, and it refuses to go on its own! How can warming it up stop it? We will crack this today, in example one. This is Part A of our unit problem bank: JEE Main level problems from all of chemical thermodynamics. Questions like these come up often. So, let's begin.

0:33Tool 1: the first law and its signs

So, first tool: the first law. For a closed system, delta U equals q plus w. The capital delta, this little triangle, means change. U is the internal energy, q is the heat, and w is the work. Okay? Now the signs, because this is where marks are lost. In chemistry, q is greater than zero, positive, when the system absorbs heat. And w is positive when work is done on the system. Heat given out, or work done by the system, is negative. Clear? Now a problem from the bank. Say a system loses seventy five joules of heat, while two hundred joules of work is done on it; J is for joules. We want delta U, the change in internal energy. It is losing heat, so its internal energy must go down, right? No, no, wait! Read the signs one by one. Heat is lost, so q is minus seventy five joules. Work is done on it, so w is plus two hundred joules. With me? So, delta U equals minus seventy five, plus two hundred: plus one hundred and twenty five joules! The work pumped in is more than the heat that leaked out, so the energy goes up. Got it?

1:50Tool 2: delta H versus delta U

Alright, second tool. H is the enthalpy, and delta H equals delta U plus delta n g R T, the link between the two. R is the gas constant, T is the temperature in kelvin, and delta n g, with the small subscript g, is the change in the moles of gas. Right? And how do we get delta n g? Moles of gaseous products, minus moles of gaseous reactants. Gaseous: hold on to that word. Okay? Now this one from the bank. Carbon burns in oxygen to give carbon dioxide. The arrow means gives, the small twos are subscripts counting atoms, and the s and g in brackets mean solid and gas. How do delta H and delta U compare? Two moles on the left, one on the right, so delta n g is minus one, right? Oh no, careful! Carbon is a solid, and only gases count. One mole of gas on each side, so delta n g is one minus one: zero. So delta H and delta U are equal. That's it! Got it?

2:57Tool 3: Gibbs energy decides

Now the third tool: Gibbs energy. G is the Gibbs energy, and S is the entropy. Delta G equals delta H minus T delta S, and that is the whole formula. Clear? And the rule: at constant temperature T and pressure P, if delta G is less than zero, the process is spontaneous. Now look at the graph: delta G up the side, temperature T along the bottom, and one coloured line for each sign pattern of delta H and delta S, okay? Now the interesting case: delta H and delta S both negative. Delta H helps, but minus T delta S is now positive, and it grows as the temperature rises. So, spontaneous at low temperature, or at high? At low temperature, yes! It is spontaneous only below T equals delta H over delta S, where the dark blue line crosses zero. Remember this case; our puzzle lives right here. With me?

3:59Example 1: spontaneity range from delta H and delta S

Now, example one, and this is our puzzle. So, a reaction has delta H equal to minus forty kilojoules, and delta S equal to minus eighty joules per kelvin. The k J means kilojoules, and J K to the minus one means joules per kelvin. In what temperature range is it spontaneous? Pause the video, and try it first. Done? Okay, let's check. Delta H and delta S are both negative: the special case from our third tool. So it is spontaneous below T equals delta H over delta S, clear? So, minus forty over minus eighty: zero point five kelvin. Done, right? No, no, careful! This is the commonest mistake with this formula. Delta H is in kilojoules, but delta S is in joules per kelvin. Convert first: minus forty kilojoules is minus forty thousand joules. Okay? Now, minus forty thousand over minus eighty: five hundred kelvin. The two minus signs cancel, so the temperature comes out positive, as it must. Got it? So it is spontaneous below five hundred kelvin. Puzzle solved! Giving out heat helps, but the entropy falls, and minus T delta S grows with temperature. Above five hundred kelvin, that term wins, and the reaction stops going on its own. Perfect!

5:26Example 2: work in reversible isothermal compression

Now the harder one. Zero point five zero moles of an ideal gas at three hundred kelvin is compressed reversibly and isothermally, from twenty litres to two litres. Mol is moles, and L is litres. Find the work done on it. Pause and try. Hint: reversible and isothermal means the log formula. Done? Okay, let's check. For an ideal gas, changed reversibly and isothermally, w rev equals minus two point three zero three, n R T, log of V two over V one. The subscript rev means reversible, n is the number of moles, V one is the starting volume, and V two the final one. And log here is the log to base ten. Clear? Now the volumes, twenty and two. Twenty over two is ten, log of ten is one, so w comes out negative, right? Oh no, careful! It is V two over V one: final over initial. Two over twenty is zero point one, and log of zero point one is minus one. With me? Now put everything in: minus two point three zero three, times zero point five zero, times eight point three one four for R, times three hundred, times minus one. Two minus signs, so the answer is positive. Okay? So w is plus two thousand eight hundred and seventy two joules, about plus two point eight seven kilojoules. And it makes sense: we squeezed the gas, so work was done on it, and w is positive, just like in our first tool. Very good!

7:04Quick revision: first law, delta H, Gibbs energy, work

So, quick revision. Delta U equals q plus w, with w positive when work is done on the system; and delta H equals delta U plus delta n g R T, counting gases only. Delta G equals delta H minus T delta S, and negative means spontaneous; convert kilojoules to joules first. And reversible isothermal work is minus two point three zero three n R T, log of V two over V one, final over initial. Now go through all twenty problems of Part A on this page. Try each one yourself first, then check the answer. You can do it. All the best!