Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle from the lab. We mix one hundred millilitres of dilute hydrochloric acid with one hundred millilitres of dilute sodium hydroxide, both zero point five molar. The heat given out is enough to warm this solution by about three point four kelvin. But the thermometer shows a rise of only three point one kelvin. So, where has the missing heat gone? Hold that thought; a problem in this chapter answers it. This is the unit problem bank for practical chemistry, and calorimetry questions like this one are a familiar JEE pattern, so let's begin.

0:39How to use this chapter

So, first, how to use this chapter. This problem bank has three parts. Part A is at the JEE Main level, so that is where you start. Okay? Part B is harder, and its problems are multi-step, with one result feeding the next. Right? And Part C is the concept checks. It asks for the reasons behind the procedures: not only what we do, but why. Clear? Now, the most important line of all. Attempt each part as a test, and read the solutions only after that. Okay?

1:10Part C asks why

Let us take the Part C question first. Why is no indicator added in permanganate titrations? Here is a hint: think about the colours. The answer is in the colours, exactly. M n O four minus, the permanganate ion, is intensely purple; the four is written low, and the minus sign written high is its charge. And M n two plus, with the two plus written high, is the manganese ion it becomes, and it is almost colourless. With me? So as long as there is something left to react with, each purple drop turns almost colourless. Then the first drop in excess has nothing left to react with, and it gives a visible pink colour. Right? That is why permanganate is called a self-indicator: its own colour shows the end point. This is the kind of reason Part C wants, in one or two clear lines. Got it?

2:04Two values for the whole chapter

Now, the chapter asks us to take two values in every problem. First, for a strong acid with a strong base, delta H equals minus fifty seven point one kilojoules per mole. The little triangle is the capital Greek letter delta, which means change in, so delta H is the enthalpy change. And the raised minus one on mol means per mole. Okay? Now tell me, what does that minus sign mean? Is the heat given out, or taken in? Yes, exactly, the minus sign means heat is given out, and the value is per mole of water formed: H plus and O H minus, in aqueous solution, give H two O, a liquid. The raised plus and minus are charges, the a q in brackets means aqueous, the l means liquid, and the arrow means gives. Right? The second value is the specific heat of dilute solutions. Small c equals four point one eight joules per gram per kelvin; each raised minus one again means per. Clear? And we join them with the calorimeter equation from Chapter seven. Small q, the heat given out, equals m times c, plus capital C with a small cal written below it, all in brackets, times delta T, the rise in temperature. Here m is the mass of the solution, and C cal is the calorimeter constant, the heat capacity of the vessel. With me?

3:27Example 1: heat from 0.5 mol HCl and 0.3 mol NaOH

Now Example one, taken from Part A of the problem bank. How much heat is evolved when solutions containing zero point five moles of hydrochloric acid and zero point three moles of sodium hydroxide are mixed? Here mol is short for moles. Pause the video and try it first. Done? Okay, let's check. Hydrochloric acid, H C l, gives zero point five moles of H plus, and sodium hydroxide, N a O H, gives zero point three moles of O H minus. With me? Now, there is more acid than base. So the heat is zero point five times fifty seven point one kilojoules, right? The cross on screen means times. No, that is the trap, so be careful. Each water needs one H plus and one O H minus. There are only zero point three moles of O H minus, so O H minus is limiting, and zero point three moles of water form. The other zero point two moles of H plus are left over. Clear? So the heat evolved is zero point three times fifty seven point one, which is seventeen point one three, or about seventeen point one kilojoules. The wavy equals sign means approximately. Always count the water from the limiting ion. Okay?

4:45Example 2: calorimeter constant from a 3.1 K rise

Now Example two, from Part B, and this is our opening puzzle. One hundred millilitres of zero point five molar hydrochloric acid and one hundred millilitres of zero point five molar sodium hydroxide are mixed in a calorimeter, and the temperature rises by three point one kelvin. Find the calorimeter constant. On screen, m L is millilitres, capital M is molar, moles per litre, and capital K is kelvin. Pause the video and try it first. Done? Okay, let's check. One hundred millilitres is zero point one zero zero litres, and the capital L stands for litres. Times zero point five molar, that gives zero point zero five zero moles each of H plus and of O H minus. Got it? Now, zero point zero five zero moles of acid plus zero point zero five zero moles of base. So that makes zero point one zero zero moles of water, right? No, careful, that is the trap. One H plus and one O H minus make one water, so only zero point zero five zero moles of water form; that is n, with H two O written below it. Fifty seven point one kilojoules is fifty seven thousand one hundred joules. So q equals zero point zero five zero times fifty seven thousand one hundred, which gives two thousand eight hundred and fifty five joules. With me? For the mass, though, we do add, because both solutions are in the calorimeter. Taking one millilitre of dilute solution as one gram, m is one hundred plus one hundred, which is two hundred grams. Clear? Now into the calorimeter equation. Two thousand eight hundred and fifty five equals two hundred times four point one eight, plus capital C, in brackets, all times three point one. Here capital C is the calorimeter constant we want. Right? Two hundred times four point one eight is eight hundred and thirty six. So eight hundred and thirty six plus capital C equals two thousand eight hundred and fifty five over three point one, which gives about nine hundred and twenty one. Take away eight hundred and thirty six, and capital C is about eighty five joules per kelvin. Got it? And now our puzzle is solved. With no vessel, delta T would be two thousand eight hundred and fifty five over eight hundred and thirty six, about three point four kelvin. But the vessel itself takes eighty five joules for every kelvin, so the rise is only three point one. That is where the rest of the heat went. Okay?

7:13Quick revision: limiting ion, 57.1 kJ, q = (mc + C) x dT

So, quick revision. Part A is Main level, Part B is multi-step, and Part C asks for the reasons; take each one as a test, then read the solutions. For a strong acid with a strong base, fifty seven point one kilojoules are given out per mole of water, counted from the limiting ion. And in the calorimeter, q equals m times c plus C cal, times delta T, with small c equal to four point one eight; and remember, the vessel takes heat too. Now open Part A and attempt it as a test. All the best!