Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Here is a system that loses seventy five joules of heat. And yet, its internal energy goes up! Is that even possible? Hmm, sounds wrong, right? Hold on to that question. Today we open the unit problem bank for chemical thermodynamics: how to use it, the constants you need, and this exact puzzle, solved. Problems like these come up often in JEE, and this chapter is where you test yourself on the whole unit. So, let's begin.

0:34How to use this chapter

So, let's open this chapter. It has three parts. Part A: twenty solved problems, at JEE Main level. Okay? Then Part B: sixteen problems, harder, and multi-step. Multi-step means the answer of one step feeds the next one, so a small slip early on spoils everything after it. Right? And Part C: fourteen short questions that check the ideas. No long calculation here; you explain why something happens, in a line or two. Clear? Now, a question for you. When you open Part A, do you read the solutions first, or do you attempt the problems first? Attempt first, of course! The chapter says it clearly: attempt each part as a test, and only then read the solutions. Sit with a clock and a pen, and keep the solutions closed till you finish. With me?

1:29The constants for the whole chapter

Next, the constants the chapter asks you to use. R, the gas constant: eight point three one four joules per kelvin per mole. J is for joules, capital K for kelvin, and mol for mole. The small minus ones on top simply mean per. Okay? Then F, the Faraday constant: ninety six thousand five hundred coulombs per mole, with capital C for coulombs. Got it? And now a very useful one. At two hundred and ninety eight kelvin, two point three zero three R T equals five point seven one kilojoules per mole; k J is kilojoules. Here T is the temperature in kelvin. Right? But where does this five point seven one come from? Multiply it out yourself, on paper. Two point three zero three, times eight point three one four, times two hundred and ninety eight. The wavy equals sign means approximately: about five thousand seven hundred and six joules per mole. That is five point seven one kilojoules per mole. Clear? And these are the values to use unless the problem tells you otherwise. But notice the words at two hundred and ninety eight kelvin, on the third line. Remember them; Example two will test you on exactly this. Okay?

2:47Part C: one concept check

Now, here is one Part C question. Why can a process with delta H greater than zero be spontaneous? Delta, the little triangle, means a change, and H is enthalpy. This sign means greater than. So the process absorbs heat, and yet it happens on its own. How? Think of Gibbs energy. At constant temperature T and constant pressure P, delta G equals delta H, minus T delta S, where S is entropy. Right? So if T delta S is greater than delta H, then delta G is less than zero; this sign means less than. And a negative delta G, at constant temperature and pressure, means spontaneous. With me? So the answer, in one line: the gain in entropy drives it. That is how Part C works: one short reason, under the right conditions. Got it?

3:42Example 1: delta U, 75 J heat lost and 200 J work done on it

Now, our puzzle from the start. It is the very first problem of Part A, an easy one. Read it with me. The system loses seventy five joules of heat, while two hundred joules of work is done on it. Find delta U, the change in internal energy. Pause the video, and try it first. Done? Okay, let's check. The first law: delta U equals q plus w. Here q is the heat that goes into the system, and w is the work done on the system. Okay? Now, the system loses seventy five joules of heat. So is q plus seventy five, or minus seventy five? Minus, yes, exactly! Heat goes out of the system, so q is minus seventy five joules. Careful here: many students copy the number from the question and write plus seventy five. That one sign costs the whole answer. Right? And the work is done on the system, so energy comes in: w is plus two hundred joules. Clear? So, delta U equals minus seventy five, plus two hundred: plus one hundred and twenty five joules. There is our puzzle, solved! The system lost seventy five joules as heat, but gained two hundred as work, so its internal energy went up. Got it?

5:03Example 2: K for ammonia synthesis at 500 K

Now a harder one, a Part B problem. For the ammonia synthesis, delta H naught is minus ninety two point two kilojoules; the little circle, read as naught, means standard. Delta S naught is minus one hundred and ninety eight point seven joules per kelvin. Assuming both stay constant with temperature, find capital K, the equilibrium constant, at five hundred kelvin. Pause the video, and try it first. Hint: it takes two steps. Done? Okay, let's check. Step one: delta G naught equals delta H naught, minus T delta S naught. Now watch the units! Delta H naught is in kilojoules, but delta S naught is in joules per kelvin. So change delta S naught to kilojoules per kelvin: minus zero point one nine eight seven. That gives minus ninety two point two, minus five hundred times, in brackets, minus zero point one nine eight seven. Okay? Minus times minus is plus, so this is minus ninety two point two, plus ninety nine point three five. Plus seven point one five kilojoules. Positive! At five hundred kelvin, the entropy term, plus ninety nine point three five, outweighs delta H naught, so this time entropy works against the reaction. Right? Step two links delta G naught to K: delta G naught equals minus two point three zero three R T, times log K, where log means the log to base ten. With me? Now, we have that shortcut: two point three zero three R T is five point seven one kilojoules per mole. So log K is minus seven point one five over five point seven one, right? No, no, careful! This is the trap. Five point seven one is only for two hundred and ninety eight kelvin, and here T is five hundred kelvin. So work it out in full, with seven point one five kilojoules written as seven thousand one hundred and fifty joules, to match R in joules. Log K equals minus seven thousand one hundred and fifty, over two point three zero three times eight point three one four times five hundred. That is minus zero point seven four seven. Clear? So K is ten to the power minus zero point seven four seven: approximately zero point one eight. Less than one, just as a positive delta G naught tells us. That is a Part B problem: two steps, and a trap in each. Got it?

7:42Quick revision: parts as tests, q and w signs, 2.303RT

So, quick revision. Attempt Parts A, B and C as tests, and only then read the solutions. Delta U equals q plus w: heat lost gives a negative q, and work done on the system gives a positive w. And two point three zero three R T is five point seven one kilojoules per mole only at two hundred and ninety eight kelvin; at any other temperature, work it out. Now open Part A, start your clock, and begin. You already know every idea these problems need, so trust yourself. All the best!