1. Natural gas consists primarily of methane, CH 4.It is used in a process called stream reforming to prepare a gaseous mixture of carbon monoxide and hydrogen for industrial use. CH 4 (g) + H 2 O(g)  CO(g) + 3 H 2 (g)  H = 206 kJ The reverse reaction has been explored as a way to prepare methane. Which of the following are exothermic? Of these, which one is the most exothermic? 1.CH 4 (g) + H 2 O(g)  CO(g) + 3 H 2 (g) 2.2 CH 4 (g) + 2 H 2 O(g)  2 CO(g) + 6 H 2 (g) 3.CO(g) + 3 H 2 (g)  CH 4 (g) + H 2 O(g) 4.2 CO(g) + 6 H 2 (g)  2 CH 4 (g) + 2 H 2 O(g) exothermic The most exothermic

2. Plan for Wed, 22 Oct 08 Today’s lab will be written up as a formal report…we will talk about the format on Friday. Lecture –Standard Enthalpies of Formation (6.4) Quiz 4

3. Standardizing Enthalpy Calculations As we saw with the Hess’s Law examples, it is pretty handy to use combustion reactions to calculate  H values for other, more-difficult- to-study reactions. Wouldn’t it be great if there were some kind of standardized list of enthalpy values that we could use in a Hess’s Law calculation? Luckily, there is… Standard Enthalpy of Formation (  H f o ) –The enthalpy change that accompanies the formation of one mole of a compound from its elements with all substances in their standard states. H 2 (g) + ½ O 2 (g)  H 2 O(l)  H f o = -286 kJ/mol Mg(s) + O 2 (g) + H 2 (g)  Mg(OH) 2 (s)  H f o = -925 kJ/mol Ag(s) + C(s) + N 2 (g)  AgCN(s)  H f o = +146 kJ/mol Cl 2 (g)  Cl 2 (aq)  H f o = -23 kJ/mol

4. Definitions of “Standard State” compounds:pure substance in its normal phase at P = 1 atm, and T of interest (gases assumed to behave ideally) solutions:P=1 atm, conc. of exactly 1 M, and T of interest element:the form in which the element exists at P = 1 atm and T = 25 o C. (Most thermodynamic tables are tabulated at 25 o C.) STANDARD STATE is not STP!!! It is crucial that the phases of all substances be indicated in a chemical equation.

5. Why do we need to define “Standard State”? Many thermodynamic functions depend on the concentrations (or pressures) and temperature of the substances involved. Furthermore, while we can fairly easily measure enthalpy changes from heat flow experiments, we have no way to measure absolute enthalpy values. To calculate thermodynamic properties correctly, we need a common reference point.

6.  H f o For Elements By definition,  H f ° = 0 for elements in their normal phases at standard conditions. Typically, an element in a nonstandard state has  H f o > 0:  H f o (graphite) = 0;  H f o (diamond) = 1.9 kJ/mol  H f o (O 2 (g)) = 0;  H f o (O 3 (g)) = 143 kJ/mol  H f o (Na(s)) = 0;  H f o (Na(g)) = 107 kJ/mol Some exceptions in your book’s appendix 4:  H f o (red P(s)) = -18 kJ/mol  H f o (black P(s)) = -39 kJ/mol

7. Which of the following are formation reactions? 4 Al(s) + 3 O 2 (g)  2 Al 2 O 3 (s) 2 Al(s) + 3 / 2 O 2 (g)  Al 2 O 3 (s) H 2 (g) + ½ O 2 (g)  H 2 O(g) H 2 (g) + ½ O 2 (g)  H 2 O(l) CH 4 (g)  C(s) + 2 H 2 (g) C(s) + 2 H 2 (g)  CH 4 (g) Cl - (aq) + Ag + (aq)  AgCl(s) ½ Cl 2 (g) + Ag(s)  AgCl(s) None of them! But these are…

8. Using  H f o Values Consider the combustion of ammonia: 4 NH 3 (g) + 7 O 2 (g)  4 NO 2 (g) + 6 H 2 O(l) We can think of this reaction as a two step process: Since  H is a state function, we can add up the enthalpies for processes a, b, c, and d to obtain  H o rxn.