(©2002, François G. Amar, All rights reserved)
For this week: Chapter 5.5-5.8
A look ahead: Chapter 6
Beginning of Chapter 5: 2, 5, 8, 12, 14, 15,16, 18, 19, 22, 28, 31
End of Chapter 5: 33, 37, 39, 42, 46, 47, 50, 53, 56, 58, 67, 70
Use tabulated values to predict the heat of reaction
Some examples (1 and 2 are formation reactions, 3 is a combustion reaction):
1) H2 (g) ---> H2 (g) D H°f = D Hrxn = Hfinal - Hinitial = 0 2) H2 (g) + 1/2 O2 (g) ---> H2O (l) (1 mol)x(0 kJ/mol) + (1/2 mol)x(0 kJ/mol) ---> (1 mol)x(-285.83 kJ/mol) reactants (initial) ---> products (final) D H°rxn = Sproducts(molesxDH°f) - Sreactants(molesxDH°f) D H°rxn = -285.83 kJ for reaction (2) as written 3) C2H6 (g) + 7/2 O2 (g) ---> 2 CO2 (g) + 3 H2O (l) (1 mol)x(-84.68 kJ/mol) + (7/2 mol)x(0 kJ/mol) ---> (2 mol)x(-393.5 kJ/mol) + (3 mol)x(-285.83 kJ/mol) reactants (initial) ---> products (final) D H°rxn = Sproducts(molesxDH°f) - Sreactants(molesxDH°f) D H°rxn = -1559.81 kJ for reaction (3) as written
- at constant P (qP=DHsystem)
we can often approximate the specific heat of dilute aqueous solutions by
the value for pure water.
but
Example: if a reaction performed in a coffee cup calorimeter is exothermic then the
heat absorbed by the solution, qsolution is positive (temperature of solution rises)
But this is an exothermic reaction (DHrxn<0) in which no work is done so qrxn = - qsolution.
but
Example: if a reaction performed in a bomb calorimeter is exothermic then the
heat absorbed by the calorimeter assembly, qcalorimeter is positive (temperature of calorimeter rises)
But this is an exothermic reaction (DErxn<0) in which no work is done so qrxn = - qcalorimeter.