Overview

Transcript

0:00Introduction

Hello everyone! Let us begin with a small experiment. Take one beaker of nitric acid and one beaker of potassium hydroxide solution, both at room temperature. Pour them together, and the mixture gets warm. But by how much? Will it warm up by ten degrees, or by fifty degrees? Hmm, hard to guess, right? By the end of this video, we will work it out on paper, with just two values from this chapter. Calorimetry questions like this do appear in JEE Main, so let's begin.

0:33How to use this chapter

So, first, how to use this chapter. It has three timed sets, with fifteen original problems in each set. Okay? The problems are in the style of recent JEE Main papers, and each set mixes all the chapters of this unit. Right? Give yourself twenty five minutes per set, and only then check the solutions. Do not peep at them while the timer runs; that is what trains you for the exam. Clear?

1:01Value 1: the enthalpy of neutralisation

Now, every set asks you to take two values. Value one is the enthalpy of neutralisation. When an acid meets a base in dilute solution, H plus from the acid and O H minus from the base give H two O, water. The raised plus and minus are the charges, the small two written low is a subscript, the letters A Q in brackets mean aqueous, dissolved in water, the small letter L means liquid, and the arrow means gives. Okay? For a strong acid with a strong base, take delta H as minus fifty seven point one kilojoules per mole. The triangle is the Greek capital delta, so delta H is the enthalpy change. And k J mol to the power minus one means kilojoules per mole, that is, per mole of water formed. The minus sign tells you that heat is given out. Clear? Now let me ask you this. Hydrochloric acid with sodium hydroxide, or nitric acid with potassium hydroxide: will this value change from one pair to the other? No, it stays the same, very good! Both the acid and the base are fully ionised, so the only reaction is the formation of water, which gives the same heat every time. Got it? But what if the acid is weak, like acetic acid? Will we get more heat, or less? Yes, exactly, we usually get less heat. With a weak acid or a weak base, the value is usually numerically smaller, which means the number without its minus sign is smaller. I say usually, because hydrofluoric acid, H F, is a known exception, explained in the calorimetry chapter. With me? And the reason is simple: part of the heat is used to ionise the weak electrolyte first. Keep this line in mind, because it is the trap in our first example. Okay?

2:54Value 2: the specific heat, and the heat formula

Value two is the specific heat of dilute solutions. Take c as four point one eight joules per gram per kelvin. On screen, capital J is joules, small g is grams, capital K is kelvin, and each raised minus one means per. So we treat a dilute solution just like water. Right? Now, how do we use it? The heat, q, equals m times c, plus capital C cal, all times delta T, the change in temperature. Here m is the mass of the solution, and capital C with the small cal written low is the calorimeter constant, the heat capacity of the vessel. Clear? Many questions say the vessel is insulated and has negligible heat capacity. So what happens to capital C cal in this formula? That is correct, it simply becomes zero. So the formula is just q equals m times c times delta T, okay? And here is the link between heat and enthalpy. Delta H equals minus q, over n, the moles of water formed; the long line is a fraction bar, and the small H two O written low says which moles we mean. Got it? So, for a strong acid with a strong base, the heat given out is simply the moles of water, times fifty seven point one kilojoules; the cross means times. That is all our examples need. With me?

4:18Example 1: heat from neutralising 0.10 mol acetic acid

Now Example one, from Set three, okay? How much heat is evolved when zero point one zero mole of acetic acid is neutralised by sodium hydroxide? The question also gives the enthalpy of ionisation of acetic acid as plus one point nine kilojoules per mole. The plus sign means this heat is taken in. Pause the video and try it first. Done? Okay, let's check. Zero point one zero mole of acid, times fifty seven point one kilojoules, gives five point seven one kilojoules. Easy, isn't it? No, that is exactly the trap, so be careful here. Fifty seven point one is only for a strong acid with a strong base. Acetic acid is weak, so part of the heat is used to ionise it. Right? Ionisation takes in one point nine kilojoules per mole. So each mole gives out fifty seven point one minus one point nine, and that gives fifty five point two kilojoules. Clear? Now for zero point one zero mole, the heat is zero point one zero times fifty five point two, which gives five point five two kilojoules. See, it is less than five point seven one, just as we said for a weak acid. Very good if you got that!

5:37Example 2: temperature rise on mixing HNO3 with KOH

And now Example two, from Set one, and this is our beaker puzzle from the start. Two hundred and fifty millilitres of zero point four molar nitric acid, HNO three, is mixed with two hundred and fifty millilitres of zero point four molar potassium hydroxide, KOH, in an insulated vessel of negligible heat capacity. Find the rise in temperature. On screen, m L means millilitres, and capital M means molar. Pause the video and try it first. Done? Okay, let's check. Both are strong, so first we need the moles of water formed. Two hundred and fifty times zero point four gives one hundred millimoles of acid, and the same of alkali, so neither is left over. That is one hundred millimoles of water, which is zero point one zero mole; m m o l means millimoles. Okay? Now value one. Fifty seven point one kilojoules is fifty seven thousand one hundred joules. So q equals zero point one zero times fifty seven thousand one hundred, which gives five thousand seven hundred and ten joules. Clear? For value two, we need the mass of the solution. The acid is two hundred and fifty millilitres, so the mass is two hundred and fifty grams, right? Oh no, careful, both solutions are in the vessel, and the heat warms all of it. Two hundred and fifty plus two hundred and fifty gives five hundred millilitres. It is a dilute solution, so we take it as five hundred grams, just like water. Got it? Finally, delta T equals q, over m times c. Five hundred times four point one eight gives two thousand and ninety. And five thousand seven hundred and ten over two thousand and ninety gives about two point seven kelvin; the wavy equals sign means approximately. So our beaker warms by less than three degrees, and the puzzle is solved with just our two values. With me?

7:38Quick revision: 57.1 kJ, q = mc delta T and timed sets

So, quick revision. For a strong acid with a strong base, take minus fifty seven point one kilojoules per mole of water; with a weak acid or a weak base, the value is usually smaller. For the heat, q equals m times c times delta T, with c equal to four point one eight joules per gram per kelvin, and m for both solutions together. And use this chapter well: three sets of fifteen problems, twenty five minutes each, then check the solutions. Now open Set one, start your timer, and try the problems. All the best!