Dinitrogen and Ammonia

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. You have made some ammonia gas in the lab, and it is wet. So you dry it with concentrated sulphuric acid, or over anhydrous calcium chloride. And your ammonia is lost! So why does this happen, and what do we use instead? We will solve it today, with dinitrogen and the Haber process. This topic comes up often in JEE, so let's begin.

0:30Making dinitrogen, N₂

First, dinitrogen, N two, the small two below counting nitrogen atoms. In the lab, we mix ammonium chloride, N H four C l, and sodium nitrite, N a N O two, both aqueous, the a q in brackets. The arrow means gives: N two, plus two H two O, plus N a C l. Okay? Or heat ammonium dichromate: N H four in brackets, taken twice, then C r two O seven. The Greek delta over the arrow means heat. It gives N two, plus C r two O three, plus four H two O, right? For very pure dinitrogen, we use barium azide, B a, then N three in brackets, taken twice. It gives barium, plus three N two. Commercially, it comes from the fractional distillation of liquid air. Got it?

1:17Why dinitrogen is so unreactive

Now, dinitrogen hardly reacts at room temperature. Why is it so lazy? Because of its triple bond, N triple bond N, the three lines on screen. Its bond enthalpy, delta with bond written below, then H, is nine hundred and forty one kilojoules per mole; mol to the minus one means per mole. That is huge, so it is inert at room temperature. With me? Still, lithium reacts with it at room temperature: six L i plus N two gives two L i three N, a nitride. Magnesium needs heating: three M g plus N two gives M g three N two, right? And oxygen reacts only at about two thousand kelvin, the capital K on screen, as in lightning. N two plus O two gives two N O; the double half arrows mean the reaction is reversible. It is endothermic, so it takes in heat. Clear?

2:12Ammonia: the Haber process

Now ammonia, N H three, by the Haber process. N two plus three H two gives two N H three, again reversible. Delta, with a small f below for formation, then H nought, the little circle meaning standard, with N H three in brackets, is the standard enthalpy of formation of ammonia. It is minus forty six point one kilojoules per mole, that is, per mole of ammonia formed; the equation as written makes two moles, so twice this. It is negative, so the reaction is exothermic. Okay? The conditions are about two hundred atmospheres, written a t m, and seven hundred kelvin, with an iron oxide catalyst promoted with K two O and A l two O three. Now count the gas molecules: one plus three makes four on the left, just two on the right. So for more ammonia, high pressure or low? High pressure, yes, exactly! Four gas molecules become two, so by Le Chatelier's principle, high pressure favours ammonia. The reaction is exothermic, so low temperature favours ammonia too. Then why not run it really cold? Because then the reaction is far too slow. So seven hundred kelvin, with the catalyst, is a compromise between yield and rate. Got it?

3:25Ammonia in the laboratory

In the lab, two N H four C l react with calcium hydroxide, C a, then O H in brackets, taken twice. This gives two N H three, plus two H two O, plus C a C l two, calcium chloride. Okay? And the gas is dried over quicklime, C a O, calcium oxide. Why we choose quicklime is our puzzle, and example one solves it, right?

3:50Ammonia: a Lewis base and a ligand

The ammonia molecule is pyramidal and hydrogen bonded, and very soluble, giving a weakly basic solution. So tell me, what makes ammonia a base? Yes, exactly, the lone pair on nitrogen, the N on screen. Ammonia donates this pair, so it is a Lewis base, and a ligand too. With me? So copper two ions, C u two plus, the raised two plus being the charge, take four N H three, giving the complex ion in square brackets, C u with four N H three, two plus. It is deep blue. Okay? Silver chloride takes up two N H three, giving silver with two N H three in square brackets, then C l. Clear? And ammonium hydroxide, N H four O H, precipitates metal hydroxides, or hydrated oxides. F e C l three gives brown F e two O three, dot, x H two O, where x, a number, counts the water molecules. Z n S O four gives white Z n, O H in brackets, taken twice. Got it?

4:51Ammonia with chlorine and hypochlorite

Next, chlorine: it depends on which is in excess. With excess ammonia, eight N H three plus three C l two gives six N H four C l, plus N two. But there is no H C l among the products. Where did it go? The extra ammonia captures it as ammonium chloride, N H four C l. With excess chlorine, all three hydrogens are replaced: N H three plus three C l two gives N C l three, plus three H C l. Careful, nitrogen trichloride is explosive. Okay? With sodium hypochlorite, N a O C l, the Raschig process, ammonia first gives N H two C l, plus N a O H, with me? Then N H two C l, with more ammonia and N a O H, gives N two H four, hydrazine, plus N a C l and water.

5:38Example 1: why dry NH3 over quicklime, not H2SO4 or CaCl2

Clear? Now, example one is our puzzle from the start. Why is ammonia dried over quicklime, and not over concentrated sulphuric acid or anhydrous calcium chloride? Pause the video and try it first. Done? Okay, let's check. First, tell me, is ammonia acidic or basic? It is basic, yes! So concentrated sulphuric acid, H two S O four, conc for short, reacts with it, giving ammonium sulphate. The ammonia is gone. Now, anhydrous calcium chloride, C a C l two, is not an acid. So it is safe, right? Oh no, careful, it is not safe at all! It forms an adduct, C a C l two, dot, eight N H three. So again, the ammonia is lost. Got it? But quicklime, C a O, is basic, so it does not react with ammonia. And that solves our puzzle. Clear?

6:33Example 2: ammonia mass from 28 g N2, 20 percent converted

Now example two, a numerical. In the Haber process, twenty eight grams of N two, the g on screen meaning grams, are mixed with excess H two, and twenty percent of the nitrogen is converted. What mass of ammonia forms? Pause the video and try it first. Done? Okay, let's check. Twenty eight grams over the molar mass, twenty eight grams per mole, the division sign on screen, gives exactly one mole. Twenty percent of it reacts, which is zero point two zero moles of N two. So we get zero point two zero moles of ammonia as well, right? No, no, wait, look at the equation! One N two gives two N H three. So zero point two zero moles of N two give zero point four zero moles of ammonia, double. That factor of two is the trap. With me? Now the mass. The molar mass of ammonia is seventeen point zero grams per mole, so zero point four zero times seventeen point zero gives six point eight grams. Forget the factor of two, and you get only half of this.

7:41Quick revision: inert N2, Haber conditions, drying NH3

Clear? So, quick revision. Dinitrogen is inert, thanks to its triple bond of nine hundred and forty one kilojoules per mole, but it reacts with lithium, heated magnesium, and oxygen near two thousand kelvin. Haber process: about two hundred atmospheres, seven hundred kelvin, an iron oxide catalyst, a compromise between yield and rate. And ammonia is a Lewis base: dry it over quicklime, not concentrated sulphuric acid or calcium chloride. Now try the practice problems on this page. All the best!