Calculating Equilibrium Concentrations

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. One mole of hydrogen and one mole of iodine go into a one litre flask, and we heat it. The equilibrium constant is sixty four, a big number. So, how much hydrogen is left? Three concentrations to find, and the equilibrium constant gives only one equation. Three unknowns, one equation? Sounds impossible, right? Today, one simple table turns all three unknowns into one letter, x. And these calculations come up often in JEE problems. So, let's begin.

0:38The equilibrium table: one unknown, x

So, the method, step by step. Step one: the balanced equation, and the expression for capital K, the equilibrium constant. Use the value of K for the temperature of the experiment, because K changes with temperature. Okay? Step two: under each species, three rows: the initial amount, the change, and the equilibrium amount. Got it? Now think. Every species changes, so again we have many unknowns. How do we get down to one? From the balanced equation, yes, exactly! Call one change x, and every other change follows from the stoichiometric ratios, the numbers in front of the formulas. One unknown for the whole table. Clear? Let's test it. N two O four gives two N O two; the small numbers below are subscripts, and the two half arrows mean equilibrium. If x of N two O four is used up, how much N O two forms? Two x, correct! So the change is minus x for N two O four, and plus two x for N O two. Minus for what is used up, plus for what forms. With me?

1:50From the table to K

Step four. K c, capital K with a small c below, uses concentrations, so divide each amount by capital V, the volume in litres. For K p, with a small p, use partial pressures instead. Okay? Now, a common trap. One point zero mole in a two point zero litre vessel; mol is short for mole, and capital L is litre. So the initial value is one point zero, right? No, no, careful! One point zero over two point zero: the initial concentration of P C l five, phosphorus pentachloride, is zero point five zero moles per litre. Square brackets mean concentration, the small zero below means initial, and capital M means molar, moles per litre. Got it? Step five: substitute the equilibrium row into K, and solve for x. Right? Often it is a quadratic with two roots. Keep the root that leaves every amount positive; the other is just algebra, not chemistry.

2:53Shortcut: when K is very small

Okay? Now, a time saving shortcut. If K is very small, the reaction hardly goes forward, so x is much smaller than the initial amounts. The double arrow means implies, and the double less than sign means much less than. Okay? So in the brackets, initial minus x is almost the initial amount; the wavy equals sign means approximately. Drop the x, and the algebra becomes really simple. Nice, right? But wait. Is this shortcut always safe? Not always! So we check afterwards. Here x must be below about five percent of the initial amount, the percent sign on screen. Clear? Now, a trap from the course examples. N two O four starts at zero point one zero moles per litre, and K c is four point six four times ten to the power minus three, moles per litre, at twenty five degrees Celsius; the minus one power means per litre, the little circle means degrees. K looks small, so drop x, right? Oh no, careful! Solved exactly, x is zero point zero one zero two. That is about ten percent of zero point one zero, well above five percent. So here, solve the full quadratic. Got it?

4:12Example 1: H2 and I2 with Kc = 64, find concentrations

Now, our puzzle. One point zero mole each of hydrogen, H two, and iodine, I two, is heated in a one litre flask. K c is sixty four for H two plus I two giving two H I, hydrogen iodide. Find the equilibrium concentrations. Pause the video, and make the table first. Done? Okay, let's check. One litre, so moles and moles per litre are the same number here. Initially: one of hydrogen, one of iodine, and zero H I, right? Now x of hydrogen reacts. Iodine reacts one to one, so minus x too, and H I goes up by two x, because of the two in front. Okay? So at equilibrium: one minus x, one minus x, and two x. See? Three unknowns, one letter. With me? Now substitute. K c is H I squared, over H two times I two; the little two up top means squared. So two x, whole squared, over one minus x, whole squared, equals sixty four. Do you need the quadratic formula here? Something faster, yes! Both sides are perfect squares, so take the square root. Two x over one minus x equals eight; the positive root, because concentrations are positive. Then ten x equals eight, so x is zero point eight. Clear? So, H I at equilibrium is zero point eight moles per litre, right? No, no, wait! H I is two x: one point six moles per litre. Hydrogen and iodine are each one minus zero point eight: zero point two moles per litre. So K is big, but zero point two of hydrogen is still left. Puzzle solved!

6:11Example 2: NO from air at 2000 K, small x shortcut

Now a harder one. Nitrogen plus oxygen gives two N O, nitric oxide. K c is four point zero times ten to the power minus four, at two thousand kelvin; that capital K after the number is kelvin. Air at this temperature is taken as zero point eight zero moles per litre of nitrogen, and zero point two zero of oxygen. Estimate the concentration of N O at equilibrium. Pause and try. Hint: look at the size of K first. Done? Okay, let's check. Same table as before. Nitrogen and oxygen each lose x, and N O gains two x. So at equilibrium: zero point eight zero minus x, zero point two zero minus x, and two x. With me? Now, would you solve the full quadratic here? The shortcut, exactly! K is very small, so x is much less than zero point two zero, the smaller starting amount. Drop x in the brackets: two x, whole squared, over zero point eight zero times zero point two zero, equals four point zero times ten to the power minus four. Okay? Multiply across: four x squared is six point four times ten to the power minus five. So x squared is one point six times ten to the power minus five, and x is four point zero times ten to the power minus three. Right? And N O is two x, not x, remember: eight point zero times ten to the power minus three moles per litre. So, are we done? Not yet! Always check the shortcut. Here x is two percent of zero point two zero, below five percent, so it is fine. The exact answer is seven point nine times ten to the power minus three; really close. Perfect!

8:07Quick revision: ICE table, solve for x, 5 percent check

So, quick revision. Make the table: initial, change, equilibrium, everything in one unknown x. Divide by capital V for concentrations, substitute in K, and solve. And when K is very small, drop x in the brackets, but always check it is under five percent. Now go and try the practice problems on this page. Make the table first every time, and you will find them easy. All the best!