Standard Enthalpies of Formation

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Oxygen gas, O two, and ozone, O three. Both are nothing but oxygen atoms, right? Yet the standard enthalpy of formation of oxygen gas is zero, and of ozone, plus one hundred and forty two point seven kilojoules per mole. Same element, two different answers? Strange, isn't it? This idea comes up often in JEE, so let's begin.

0:28Standard state and enthalpy of formation

So, first, what does standard mean? The standard state of a substance is its pure form at one bar and the stated temperature, usually two hundred and ninety eight kelvin, that capital K, okay? Now the key quantity. The little triangle is the capital Greek letter delta, meaning change in, and H is enthalpy. The small f means formation, and the little circle on top, read as naught, means standard. So delta f H naught is the enthalpy change when one mole of a compound is formed. Right? Formed from what? From its elements in their standard states, also called reference states. So, two conditions: exactly one mole, and elements in their reference states. Clear? Now tell me. N two gas, plus three H two gas, gives two N H three gas. The arrow means gives; the g in brackets means gas. Is this the formation equation of ammonia? No, no, careful! That forms two moles of ammonia, not one. The correct one: half N two gas, plus three halves H two gas, gives N H three gas. One mole, from the elements in their reference states. Got it?

1:38Which substances have zero enthalpy of formation?

Now, which substances have delta f H naught equal to zero? Think about oxygen gas. To form O two gas from O two gas, what changes? Nothing changes, exactly! So delta f H naught is zero for every element in its reference form: H two gas, O two gas, N two gas, C l two gas, B r two liquid, that is the l in brackets, and I two solid, the s in brackets. With me? For carbon, the reference form is graphite; for sulfur, rhombic sulfur; for phosphorus, white phosphorus. Easy to forget, so learn them, okay? One more, by convention: delta f H naught of H plus, aq, is zero. H plus is the hydrogen ion; aq means aqueous, in water. Right? Now a quick test. Ozone gas, diamond, liquid bromine, iodine gas, nitrogen gas, and H gas, that is single hydrogen atoms. Which of these have delta f H naught equal to zero? Only liquid bromine and nitrogen gas, correct! Diamond? No, carbon's reference form is graphite. Iodine gas? No, iodine's reference form is the solid. And single hydrogen atoms are not H two gas. Clear? And ozone? The reference form of oxygen is O two gas, not ozone. So ozone has plus one hundred and forty two point seven, k J mol to the power minus one, that is, kilojoules per mole. Puzzle solved: only one form is the reference. That's it!

3:13Enthalpy of reaction from enthalpies of formation

Now the most useful result. Delta r H naught, with a small r for reaction, is the standard enthalpy of reaction. It equals the sum of nu times delta f H naught of the products, minus the same sum for the reactants. The big zigzag is the capital Greek letter sigma, meaning sum of; the small v-shaped letter is the Greek letter nu, each substance's coefficient in the balanced equation. Okay? Now, a quick check. Is it products minus reactants, or reactants minus products? Products minus reactants, yes! Final minus initial, like any change. Flip it, and the sign comes out wrong. So never flip it. Got it?

3:54Reverse, multiply, and write the states

Two more rules, really simple. Reverse an equation, and delta H changes sign. Multiply an equation by a number, and delta H is multiplied by that number. Got it? Now let's use our ozone. Its delta f H naught is plus one hundred and forty two point seven kilojoules per mole. What is delta H for three O two gas giving two O three gas? Two moles of ozone: two times plus one hundred and forty two point seven, minus three times zero for oxygen. Plus two hundred and eighty five point four kilojoules. Right? And the reverse, two O three giving three O two? Just flip the sign: minus two hundred and eighty five point four kilojoules. Clear? Now look at water. Delta f H naught of liquid water is minus two hundred and eighty five point eight; of water vapour, the g, it is minus two hundred and forty one point eight kilojoules per mole. With me? Different numbers for the same formula! The gap, forty four point zero kilojoules per mole, is the enthalpy of vaporisation of water at two hundred and ninety eight kelvin. So always write the physical states. Okay?

5:07Example 1: enthalpy of decomposition of limestone

Now, Example one. Find delta r H naught for the decomposition of limestone. Given delta f H naught values: calcium carbonate, C a C O three, minus one thousand two hundred and six point nine; calcium oxide, C a O, minus six hundred and thirty five point one; carbon dioxide, minus three hundred and ninety three point five, all in kilojoules per mole. Pause the video, and try it. Done? Okay, let's check. First, the equation: solid calcium carbonate breaks into solid calcium oxide, plus carbon dioxide gas. Every coefficient is one. Right? Now, products minus reactants. Inside the square brackets, the products: minus six hundred and thirty five point one, plus minus three hundred and ninety three point five. Then subtract the reactant, minus one thousand two hundred and six point nine, in its own brackets. With me? The products add to minus one thousand and twenty eight point six. So the answer is that, minus one thousand two hundred and six point nine, right? Oh no, careful! That is the trap. Subtracting a negative number means adding it. So, minus one thousand and twenty eight point six, plus one thousand two hundred and six point nine: plus one hundred and seventy eight point three kilojoules. Positive, so heat is absorbed. Very good!

6:32Example 2: H2S and SO2 reaction from formation data

Now a harder one. Find delta r H naught for: two H two S gas, plus S O two gas, gives three S solid, plus two H two O liquid. Given delta f H naught values: hydrogen sulfide, minus twenty point six; sulfur dioxide, minus two hundred and ninety six point eight; liquid water, minus two hundred and eighty five point eight, all in kilojoules per mole. Pause and try. Done? Okay, let's check. But wait, there is no value given for sulfur. Is the question incomplete? No, it is complete! Solid sulfur is an element in its reference form, so its delta f H naught is zero. You are expected to know it. Right? Products: three times zero, plus two times minus two hundred and eighty five point eight. That gives minus five hundred and seventy one point six. Okay? Reactants: two times minus twenty point six, plus minus two hundred and ninety six point eight. Minus forty one point two, minus two hundred and ninety six point eight: minus three hundred and thirty eight point zero. Don't drop that two in front of H two S! With me? Finally, products minus reactants: minus five hundred and seventy one point six, minus, minus three hundred and thirty eight point zero. That is minus two hundred and thirty three point six kilojoules. Heat is released. Perfect!

8:02Quick revision: reference states, products minus reactants

So, quick revision. One: delta f H naught is for one mole, formed from elements in their reference states, and for those elements it is zero. Two: delta r H naught is products minus reactants, each times its coefficient. Three: reverse, change the sign; multiply, multiply delta H; and always write the states. Now try the practice problems on this page. All the best!