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themochemistry notes

# themochemistry notes - THERMOCHEMISTRY Chemistry CHEM 213W...

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THERMOCHEMISTRY Chemistry CHEM 213W David Ronis McGill University 1. Enthalpy Calculations: Chemical Reactions and Hess’ Law The enthalpy change for a process, H , is equal to the heat absorbed by the system if that process is done under constant pressure conditions (and assuming that only P-V work is possi- ble). Since the enthalpy of a system, H = E + PV, is a state function, we can systematize enthalpy calculations by considering a path whereby the compounds first turn into their con- stituent elements in their standard states (by convention at 25 o C and 1 atm pressure) and then recombine to form the products. The enthalpy change in the latter step is just the enthalpy of for- mation of the products and the former is the enthalpy of destruction (i.e., the negative of the enthalpy of formation) of the reactants. Hence, H = Σ H 0 f ( products ) - ∆ H 0 f ( reactants ) . (1) Since we are interested in calculating a difference, the absolute enthalpy of the elements in their standard states is unimportant [it cancels out of Eq. (1)], and we adopt the convention that the enthalpy of formation of an element in its standard state is zero. Consider the following example (reduction of iron oxide): Fe 2 O 3 ( s ) + 3 H 2 ( g ) 25 o C , 1 atm 2 Fe ( s ) + 3 H 2 O ( l ). A table of thermochemical data gives: Enthalpies of Formation at 1 atm and 25 C Compound H 0 f (kJ/mole) Fe 2 O 3 (s) -824.2 H 2 (g) 0.0 Fe(s) 0.0 H 2 O (l) -285.830 By using these in Eq. (1), we find that H = [3( - 285. 830) - ( - 824. 2)]kJ/mole = - 33. 29kJ/mole. Winter Term 2001-2002

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Thermochemistry -2- Chemistry CHEM 213W Note that the calculated enthalpy change depends on how the reaction was written. For example, if we wrote 1 2 Fe 2 O 3 ( s ) + 3 2 H 2 ( g ) 25 o C , 1 atm Fe ( s ) + 3 2 H 2 O ( l ), then H = - 16. 65 kJ/mole. 2. Measuring H 0 f There are a number of ways in which to measure the enthalpy of formation of a compound; here are two. The most obvious is to simply carry out the formation reaction from the con- stituent elements in their standard states in a constant pressure calorimeter (recall that H = Q p ). For example, consider the combustion of graphite to form carbon dioxide C ( graphite ) + O 2 ( g ) 25 o C , 1 atm CO 2 ( g ). The heat released in this reaction is -∆ H 0 f ( CO 2 ), since the standard enthalpy of formation of the reactants is zero. For this method to work, two conditions must be met: 1) the reaction goes to completion and 2) only one product is formed. Thus, the reaction C ( graphite ) + 2 H 2 ( g ) 25 o C , 1 atm CH 4 ( g ) is not suitable for this method since it doesn’t readily go to completion and we get a complicated mixture of hydrocarbons.
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