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Unformatted text preview: 2 H2O(g) ΔH° = −803.05 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Enthalpy Changes for Chemical Reactions
Calculate the heat generated when 500. g of propane burns in excess O2.
C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ΔH° = – 2220. kJ ΔH° = −890.36 kJ ΔH° is the standard enthalpy change
P = 1 bar.
T must be stated (if it isn’t, assume 25°C).
ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to form 2 mol of liquid water and release 890 kJ of heat
Change a physical state, change ΔH° : H2O(l) vs. H2O(g) Molar mass of C3H8 = 44.097 g/mol.
nC3H8 = (500. g) / (44.097 g/mol)
= 11.34 mol C3H8
Since ΔH° = qp = –2220. kJ/(1 mol C3H8)
q = (11.34 mol C3H8) (–2220. kJ / mol C3H8 )
= –2.52 x 104 kJ Bond Enthalpies Where Does the Energy Come From?
Bond Enthalpy (bond energy)
• Equals the strength of 1 mole of bonds
• Always positive
It takes E to break a bond
Separated parts are less stable than the molecule.
Less stable = higher E During a chemical reaction:
Old bonds break: requires E (endothermic)...
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This note was uploaded on 03/02/2014 for the course CHEM 101 taught by Professor Multipleprofs during the Spring '07 term at Drexel.
- Spring '07
- States Of Matter