Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. I take a closed flask of fixed volume, filled with dinitrogen tetroxide, N two O four, and I heat it. Its starting pressure, counted at that final temperature, is one atmosphere. Some of it splits into nitrogen dioxide, N O two, and at equilibrium the gauge reads one point two atmospheres. Just one pressure reading. Can you find the equilibrium constant from that alone? Sounds like too little information, right? Hold that thought. By the end of this class, you will do it in four lines. This is set one of mixed problems on equilibrium: short numericals of the kind that come up often in JEE Main. Two quick ideas first, then two full problems. So, let's begin.

0:50Idea 1: a strong acid and a weak acid together

Idea number one: two acids in one beaker. Hydrochloric acid, H C l, at zero point zero two molar, the capital M on screen, meaning moles per litre, together with acetic acid, also zero point zero two molar. Find the p H, and you are given log two equals zero point three zero. Okay? Now, both acids give H plus, the hydrogen ion, with its plus charge written up top. So the H plus concentration, written in square brackets, is zero point zero two plus zero point zero two, that is zero point zero four molar, right? No, no, careful! H C l is a strong acid. It ionises completely, so by itself it gives zero point zero two molar of H plus. With me? But acetic acid is weak, and its own ionisation also makes H plus. So the H plus from H C l is a common ion: it pushes the weak acid's equilibrium back, and the weak acid is suppressed almost completely. So the H plus concentration is approximately zero point zero two molar; the wavy equals sign means approximately. Clear? Now, p H is minus log of the H plus concentration. Zero point zero two is two times ten to the power minus two, so minus its log is two minus log two. Two minus zero point three zero: one point seven zero. So, when a strong acid is present, the p H comes from the strong acid alone. Got it?

2:23Idea 2: what temperature does to K

Idea number two, and it is a quick one. Hydrogen, H two, plus iodine, I two, gives two H I, hydrogen iodide; the small twos written below count the atoms, and the two half arrows mean equilibrium. The equilibrium constant, capital K, is sixty four at seven hundred kelvin, and fifty four at eight hundred kelvin. The capital K after each temperature is just the unit, kelvin. Okay? So, we heat it, and the equilibrium constant falls. Tell me: is the forward reaction exothermic, or endothermic? Exothermic, yes, exactly! K decreases as capital T, the temperature, rises, so the forward reaction gives out heat. By Le Chatelier's principle, heating shifts an exothermic reaction backward, giving less product and a smaller equilibrium constant. Right? And the other way round: for an endothermic forward reaction, K increases as the temperature rises. And remember, only temperature changes the equilibrium constant. Clear?

3:29Example 1: Kp from Kc for 2SO3 to 2SO2 + O2

Now, example one. Two S O three split into two S O two, plus O two; the small g in brackets after each one means gas. K c, capital K with a small c below, is four point zero times ten to the power minus three, moles per litre, at one thousand kelvin; the minus one power on the L just means per litre. Find K p, the constant in partial pressures. Pause the video and try it first. Done? Okay, let's check. To go from K c to K p, we need delta n g, the change in moles of gas; the little triangle is the Greek capital delta, and the small g means gas. So, what is delta n g here? Products: two moles of S O two, plus one of O two, three moles of gas. Reactants: two moles. So delta n g is three minus two, plus one. Students who forget the O two get zero here, and a wrong answer. Careful! So K p equals K c times R T to the power one: K c times R T, nothing more. Okay? Concentrations in moles per litre, pressures in atmospheres, so R, the gas constant, is zero point zero eight two one, litre atmosphere per kelvin per mole; again, the minus one powers mean per kelvin and per mole. Right? Now multiply: four point zero times ten to the power minus three, times zero point zero eight two one, times one thousand, the temperature in kelvin. With me? That gives zero point three two eight four, so K p is about zero point three three atmosphere. Delta n g is positive, and R T is much bigger than one, so K p comes out bigger than K c, as expected. Got it?

5:23Example 2: Kp of N2O4 from one pressure reading

Now the harder one, and it is our puzzle from the start. N two O four at one point zero atmosphere is heated in a closed vessel of fixed volume. At equilibrium, the total pressure is one point two atmospheres. Find K p for N two O four giving two N O two. Pause the video and try it. Done? Okay, let's check. The pressure went up from one point zero to one point two, a rise of zero point two atmosphere. So the N O two pressure, small p with N O two written below, is zero point two, right? No, no, careful! The volume is fixed, and, as usual in such problems, the one point zero atmosphere is the starting pressure at the reaction temperature itself. So we can work straight in pressures. Let small p atmospheres of N two O four dissociate. Then N two O four drops to one minus p. But each N two O four gives two N O two, so N O two is two p, not p. Okay? Now the total: one minus p, plus two p, is one plus p. That equals one point two, and the double arrow on screen means 'so': p equals zero point two. So the rise in pressure is p, the amount that split, not the N O two. Clear? So the partial pressures. N two O four: one minus zero point two, zero point eight atmosphere. N O two: two times zero point two, zero point four atmosphere. Right? Now we write K p. Which pressure gets squared, and why? The N O two pressure, yes! Its coefficient is two. K p equals p of N O two squared, over p of N two O four. Zero point four squared is zero point one six, over zero point eight: zero point two zero atmosphere. See? One pressure reading, and we got K p. Puzzle solved!

7:24Quick revision: common ion, K and T, Kp from Kc

So, quick revision. Strong acid with a weak acid: the strong acid sets the H plus concentration, and the weak one is suppressed. If K falls as the temperature rises, the forward reaction is exothermic. K p equals K c times R T to the power delta n g, counting gaseous moles only. And in a pressure table, use the coefficients: N two O four loses p, N O two gains two p, and the total rises by p. Now go and try the practice problems on this page. Watch the coefficients, and you will find them easy. All the best!