Effect of Temperature, and Industrial Equilibria

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Ammonia is made from nitrogen and hydrogen, and the reaction gives out heat. So a cooler reactor should give more ammonia, right? And yet, factories run it hot, at about seven hundred kelvin! Why choose a temperature that lowers the yield? By the end of this class, you will know. Today is about temperature, the one change that moves the equilibrium constant itself. In JEE, this comes up often. So, let's begin.

0:34What can change K itself?

So, Le Chatelier's principle: disturb an equilibrium, and it shifts to partly counteract the change. We can change concentration, pressure or temperature, or add a catalyst. Which of these changes capital K, the equilibrium constant, itself? Temperature, yes, exactly! Temperature is the only thing that changes K, nothing else. And a catalyst? It never changes K, never. Okay? Now, which way does K move? An endothermic forward reaction absorbs heat, so think of heat as a reactant. If I heat the mixture, which way does it shift? Forward, correct! Going forward uses up some of the extra heat. The little triangle is the Greek capital delta, and delta H is the enthalpy change, the heat of reaction. Delta H greater than zero means endothermic: K increases with T, the temperature, and heating shifts it forward. Right? And exothermic, delta H less than zero, is exactly the opposite. Heat comes out, like a product, so K decreases as the temperature rises, and heating shifts it backward. Got it?

1:46How much does K change? The van't Hoff equation

Now, how much does K change? That is the van't Hoff equation. Log of K two over K one equals delta H standard over two point three zero three R, times, in brackets, one over T one minus one over T two. K one is the equilibrium constant at temperature T one, and K two at T two; the small ones and twos are subscripts. The little circle means standard, R is the gas constant, and the log is to base ten. With me? Now, if we heat, T two is bigger than T one. So is the bracket positive or negative? Positive, yes! Bigger temperature, smaller one over T, so we subtract a smaller number. T two greater than T one: the bracket is positive. Clear? So the log simply takes the sign of delta H standard. Greater than zero: K two is greater than K one. Less than zero: K two is less than K one. Exactly our rule from before. Okay? Now, the numerical trap. R is eight point three one four joules per mole per kelvin; the minus ones mean per. So delta H goes in joules, not kilojoules, and every temperature in kelvin. Got it?

3:05The graph: log K against 1/T

Now the same equation as a graph. Plot log K up the side, against one over T along the bottom. You get a straight line, with slope minus delta H standard over two point three zero three R, okay? So, endothermic: delta H is positive, the slope is negative, and the orange line falls as one over T grows. Exothermic: delta H is negative, so the slope is positive, and the dark blue line rises; its label says slope greater than zero. Clear?

3:38Industry: the Haber process

Now, into a factory. The Haber process: N two plus three H two, giving two N H three; the two half arrows mean reversible. Delta H is minus ninety two kilojoules, so it is exothermic. First, pressure. Count the moles of gas. High pressure or low? High, correct! Four moles of gas give two, so high pressure shifts it towards fewer moles, towards ammonia. Industry uses about two hundred atmospheres. Okay? Now, temperature. It is exothermic, so for the biggest K, high or low? Low, yes! Low temperature gives a larger K, but here is the catch. At low temperature, the reaction is far too slow. Right? So here is the answer to our puzzle! Industry picks a compromise between yield and rate: about seven hundred kelvin, an iron catalyst, and molybdenum as a promoter. The catalyst gives speed; it never changes K at all. Clear? And the ammonia is removed by cooling. That keeps Q, the reaction quotient, less than K, so the reaction keeps going forward. With me?

4:51Two more equilibria

Next, the contact process: two S O two plus O two, giving two S O three. Delta H is about minus one hundred and ninety seven kilojoules; the wavy equals sign means approximately. Okay? The conditions: about seven hundred and twenty kelvin, V two O five, vanadium pentoxide, as the catalyst, and only about two bar of pressure, with excess air. Got it? And one endothermic case: N two plus O two, giving two N O, nitric oxide, with delta H plus one hundred and eighty kilojoules. Endothermic, so K grows with temperature, and nitric oxide forms appreciably only at very high temperature: in lightning, and in engines. Clear?

5:39Example 1: delta H when K doubles from 300 K to 310 K

Now, example one. The equilibrium constant of a reaction doubles when the temperature rises from three hundred kelvin to three hundred and ten kelvin. Find delta H standard. Pause the video, and try it first. Done? Okay, let's check. First, a topper's habit: predict the sign. K went up on heating. So is delta H standard positive or negative? Positive, exactly! K rose with temperature, so it is endothermic. Now van't Hoff. K two over K one is two, so log two equals delta H standard over two point three zero three times eight point three one four, times, one over three hundred minus one over three hundred and ten. Right? Log two is zero point three zero one, and two point three zero three times eight point three one four is nineteen point one five. The bracket is one point zero seven five times ten to the minus four. With me? Solve it, and delta H comes out in joules, since R is in joules. That is about fifty three point six kilojoules per mole, roughly plus fifty four. Positive, as predicted. Clear?

6:57Example 2: how far K for ammonia falls from 298 K to 400 K

Now a harder one, back to our ammonia. For N two plus three H two, giving two N H three, delta H standard is minus ninety two kilojoules. By what factor does K change between two hundred and ninety eight kelvin and four hundred kelvin? Pause and try it. Hint: watch the units. Alright. Exothermic, so K must fall; that is our sign check. Now, delta H is minus ninety two, so we put minus ninety two on top, right? No, no, wait! That is the trap. R is in joules, so minus ninety two kilojoules becomes minus ninety two thousand joules. Got it? Now van't Hoff, with T one as two hundred and ninety eight and T two as four hundred. Log of K two over K one equals minus ninety two thousand over nineteen point one five, times, one over two hundred and ninety eight minus one over four hundred. Okay? That is about minus four thousand eight hundred and five, times eight point five six times ten to the minus four: minus four point one one. Clear? The log is negative, so flip it over: K one over K two is ten to the power four point one one, about one point three times ten to the power four. So K falls about thirteen thousand times, and high temperature lowers the yield of ammonia. Yet the factory still runs hot, because cold is far too slow. Puzzle solved!

8:31Quick revision: temperature and K, van't Hoff, Haber

So, quick revision. Only temperature changes K: for an endothermic reaction, K rises with T; for an exothermic one, K falls. The van't Hoff equation gives the exact change, with delta H in joules. Haber: high pressure, about seven hundred kelvin and an iron catalyst, a compromise of yield and rate. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!