Download Chemistry 30 Notes - Heat of Formation February 2nd

Chemistry 30 Notes – Heat of Formation February 2nd, 2015
Review/Practice
The standard heat of formation, ΔHof, for sulfur dioxide (SO2) is -297 kJ/mol. How many kJ of energy are given off when
25.0 g of SO2 (g) is produced from its elements?
Step 1: Calculate moles SO2
molar mass of SO2 = 32.1 + 2(16.0) = 64.1 g/mol
25.0 g
moles SO2 =
= 0.390 mol
64.1 g/mol
If you have trouble converting mass into moles, better dig out your old Chemistry 20 notes or textbook and review.
Chemistry 30 will require you to remember this basic chemistry skill!
Step 2: Determine kJ for 0.390 mol
We know from the question that 297 kJ of energy is released for 1 mole of SO2 — the definition of heat of formation.
Determine how much energy will be released for 0.390 mol of SO2:
kJ released = (0.39 mol)(-297 kJ/mol) = - 116 kJ
Heat of Formation
Turn now to a special type of chemical reaction, one in which we form one mole of a compound from its elements. We
are interested in how much energy is either absorbed or released during this synthesis reaction.
For example, the formation reaction for liquid water is described by the following equation:
H2 (g) + ½O2 (g) → H2O(l) + 285.8 kJ
The amount of energy released during this reaction, 285.8 kJ, is referred to as the standard heat of formation. Since the
reaction is exothermic we would indicate the heat of formation, ΔHof, as -285.8 kJ.
Key items to note regarding this definition:


one mole of the compound is formed
from its elements
Standard Heat of Formation ΔHof
The amount of energy gained or lost when 1 mole of the substance is formed from its elements under standard
conditions (25oC, 1 atm = 101.3 kPa). Keep in mind that a heat of formation is just a special case of a heat of reaction it's just that in this particular type of reaction one mole of the compound forms from its elements.
The heat of formation for pure elements, such as H2(g), O2(g), Al(s), etc. is 0 kJ·mole-1. You'll find it useful to remember
this.
Writing heat of formation reactions is an important skill. Keep the following points in mind:



Balance the equation so that one mole of the compound is produced.
Remember the diatomic molecules and write them correctly (H2, N2, O2, F2, Cl2, Br2, I2).
The reactants must be elements, not polyatomic ions. Examples of polyatomic ions are hydroxide, OH-,
carbonate, CO32-, and ammonium, NH4+.
Write heats of formation reactions for each of the following compounds. Be sure to include the energy term with the
equation, either as part of the equation or separately as Δ H. You will need to refer to a Table of Thermochemical Data.
CO2 (g), CuCl2 (g), N2H4 (l), NH4Cl (s).
C (s) + O2 (g) → CO2 (g) + 393.5 kJ
Cu (s) + Cl2 (g) → CuCl2 (s) + 220.1 kJ
N2 (g) + 2H2 (g) + 50.6 kJ → N2H4 (l)
½N2 (g) + 2H2 (g) + ½Cl2 (g) → NH4Cl (s) + 314.4 kJ
What determines whether a reaction will actually occur spontaneously or not? In chemistry, a "spontaneous" reaction is
a reaction that will occur on its own - it may be fast or it may be slow.
Here are some examples of spontaneous chemical reactions:
Na(s) + ½Cl2(g) → NaCl(s) + 411.2 kJ
H2(g) + ½O2(g) → H2O(g) + 242 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) + 1461.9 kJ
Something that all of these reactions have in common is that they are all exothermic. Energy, like boulders, tends to
run downhill. The lower energy state of the products is more stable than the high energy state of the reactants. And
everything always likes to go towards the most stable, or lowest energy, state.
But of course some endothermic reactions that are also spontaneous:
Br2(l) + Cl2(g) + 29.3 kJ → 2 BrCl(g)
H2(s) + ½O2+ 6.01 kJ → H2O(l)
Hess’ Law
It is not always possible to conduct experiments to measure enthalpy. Chemist Hess 1840’s suggested that the addition
of chemical equations yields a net chemical equation whose enthalpy change is the sum of the individual enthalpy
changes. It has been tested and accepted as the law of additivity of enthalpies of reaction, also known as Hess’ Law.
Hess’ Law can be written as an equation using the uppercase Greek letter ∑ (sigma) to mean “the sum of”.
∆rH = ∆1H + ∆2H + ∆3H….. or ∆rH =∑∆rH or in the standard conditions ∆rHo =∑∆rHo
Hess’ discovery allowed the determination of the enthalpy change of a reaction without direct calorimetry, using two
rules for chemical equations and enthalpy changes.
1. If a chemical equation is reversed, then the sign of ∆rH changes.
2. If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the ∆rH is
altered by the same factor.