Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take a closed flask of fixed volume, with N two O four gas. At the working temperature, before any of it splits, it would exert one atmosphere. There, N two O four splits into N O two. At equilibrium, the total pressure is one point two atmospheres. The pressure went up by zero point two, so the flask has zero point two atmosphere of N O two, right? Hmm, are you sure? Today: this chapter of mixed JEE Main pattern problems, its seven kinds of question, the constants, and this puzzle, solved. So, let's begin.

0:44How this chapter works

So, let's open this chapter. The real JEE Main questions, from twenty twenty one to twenty twenty six, sit at the end of Chapters one to five. Okay? This chapter is different. Its problems are original, in the same style, and mostly numerical. Right? There are three sets: nine solved problems in Set one, eight in Set two, and eight in Set three. Each set ends with three practice problems.

1:13Seven kinds of question: three for gases

Clear? Now, seven kinds of question, and three are about gases. First: K c, the K with a small c, uses concentrations; K p, with a small p, uses partial pressures. K p equals K c times R T, raised to the power delta n g, with R the gas constant, and T the temperature in kelvin. Okay? Delta n g, the little triangle for change, with a small g for gas, is the moles of gaseous products minus the moles of gaseous reactants. Gases only! Solids and liquids never count. Got it? Second: equilibrium tables. Under each species, write the initial amount, the change in terms of one unknown, x, and the equilibrium amount. Then put them into K, and solve for x. Our flask puzzle needs this table. With me? Third: the degree of dissociation, alpha, the Greek letter. It is the fraction of the initial moles that has dissociated at equilibrium. Clear?

2:17Four more kinds, in solution

Now the other four kinds, all in solution. Fourth: p H of every kind of solution. p H is minus log of the concentration of H plus, the hydrogen ion, with the small plus sign for its charge; square brackets mean concentration in moles per litre. Okay? For a weak acid of concentration c, with K a its ionisation constant, the hydrogen ion concentration is the square root of K a times c. But only while alpha, the degree of ionisation, is below about five percent; that curly sign means less than about. Right? So what if alpha comes out above five percent? Use the square root anyway? No, no, careful! Then the approximation fails. Solve the quadratic: x squared over c minus x equals K a, where x is the hydrogen ion concentration. Set three has a problem just like this. Got it? Fifth: buffers. For an acidic buffer, a weak acid with its salt, the Henderson Hasselbalch equation: p H equals p K a, plus log of salt over acid, where p K a is minus log of K a. With me? Sixth: hydrolysis. In a salt, the ion of a weak acid or a weak base reacts with water, so the solution turns basic or acidic. Okay? And seventh: K s p, the solubility product. For a salt A x B y, the small x and y below the letters count the ions in one formula unit, K s p equals x to the power x, times y to the power y, times s to the power x plus y, with s the molar solubility in moles per litre. Clear?

4:05Constants to use, unless a question gives its own

Alright, the constants. Unless a question says otherwise, the problems are at two hundred and ninety eight kelvin. There, K w, the ionic product of water, is ten to the minus fourteen, so p H plus p O H equals fourteen, where p O H is minus log of the hydroxide ion concentration. Okay? And R, the gas constant: zero point zero eight two one litre atmosphere per kelvin per mole; the minus one powers are read as per. Use it with pressures in atmospheres and concentrations in moles per litre. Right? Quick question. Is neutral water always at p H seven? No! Neutral means H plus and O H minus, the hydroxide ion, have equal concentrations. p H seven is neutral only at two hundred and ninety eight kelvin. With me? At three hundred and seventy three kelvin, K w is about five point five times ten to the minus thirteen; the wavy equals sign means about. So neutral water there has p H about six point one, and it is still neutral. Got it?

5:15Example 1: Kp from Kc for 2SO3 to 2SO2 + O2

Now, Example one, an easy problem from Set one, okay? Two S O three gas is in equilibrium with two S O two gas plus O two gas. The small g in brackets means gas; the two half arrows mean both directions. K c is four point zero times ten to the minus three moles per litre, at one thousand kelvin. Find K p. Pause the video, and try it first. Done? Okay, let's check. First, delta n g. What is it here? One, correct! Gaseous products: two moles of S O two plus one of O two, that is three. Gaseous reactants: two moles of S O three. Three minus two is one. Right? So K p equals K c times R T to the power one: just K c times R T, okay? This question gives its own temperature, one thousand kelvin, so use that, not two hundred and ninety eight. And R is zero point zero eight two one, for K p in atmospheres. With me? Four point zero times ten to the minus three, times zero point zero eight two one, times one thousand: zero point three two eight four, about zero point three three atmosphere. Clear?

6:37Example 2: Kp of N2O4 from the total pressure

And now, our puzzle, a medium problem from Set one. 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 going to two N O two. Pause the video and try it. Hint: make the equilibrium table. Done? Okay, let's check. Start with one point zero atmosphere of N two O four and no N O two. As usual in such questions, one point zero is the N two O four pressure at the final temperature, before any of it splits. And work in atmospheres throughout. Okay? Now, say small p atmosphere of N two O four dissociates. How much N O two forms? Two p, yes! One molecule of N two O four gives two of N O two. So at equilibrium, N two O four is one minus p, and N O two is two p. Right? The total: one minus p, plus two p, is one plus p. One plus p equals one point two; the double arrow means this gives: p equals zero point two. With me? So, back to our puzzle. p N O two, the small p with N O two below it, is the partial pressure of N O two. It is zero point two atmosphere, right? No, no, careful! That is the trap from the start. Zero point two is p, the N two O four that split, and each one gave two N O two. So p N O two is two p: zero point four atmosphere. And p N two O four is zero point eight. Puzzle solved! Got it? Now K p: p N O two squared, over p N two O four. Squared, because the equation has two N O two. Zero point four squared is zero point one six, over zero point eight: zero point two zero atmosphere. Clear?

8:43Quick revision: 7 question types and default constants

So, quick revision. Seven kinds: K p and K c, tables, alpha, p H, buffers, hydrolysis, and K s p. Unless stated: two hundred and ninety eight kelvin, K w ten to the minus fourteen, and R zero point zero eight two one. Delta n g counts gases only. And follow the equation: one N two O four gives two N O two. Now open Set one, try each problem before reading its solution, then do the practice problems. All the best!