Post on 19-Mar-2016
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Thermochemisty (Enthalpy) and Hess’s Law
Chapter 10Sections 10.6-10.7
Enthalpy• Change in enthalpy (ΔHp = qp): the
amount of heat exchanged when heat exchange occurs under conditions of constant pressure
• Enthalpy is a state function – State function is defined as a property that changes
independent of pathway
• ΔH is independent of the path taken
At constant pressure, 890kJ of energyper mole of CH4 is produced as heat
Qp = ΔHp = -890 kJ exothermic
If 5.8 g of methane is burned, how much heat will be released?
5.8g CH4 x 1mol CH4 = 0.36 mol CH4 16.0 g CH4
0.36 mol CH4 x -890 kJ = -320 kJ mol CH4
Calorimeter• Calorimetry the
process of measuring heat flow in the form of temperature changes in a system
• Calorimeter an insulated device used to measure temperature changes
Hess’s Law• Hess’s Law
– ΔH for a multi-step process is the sum of ΔH for the individual steps
– Hess’s law allows us to calculate ΔH for new reactions from previously determined values
Germain Henri Hess
Altitude
San Francisco, CA0 ft
Salt Lake City, UT4,266 ft
Denver, CO5,280 ft
Flight 1
Flight 2
Flight 1: Δ Alt = -4266Flight 2: Δ Alt = +5280SL City-Denver Δ Alt = +1014
What is the change in altitude (Δ Alt) between Salt Lake City and Denver?
?
4266 ft
1014 ft
5280 ft
Hess’s Law
• Applying Hess’s Law– When reversing a reaction, change the
sign of ΔH. – When multiplying a reaction by a new
coefficient, multiply ΔH
Hess’s LawHess’s Law“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”
(products) (reactants)o oreaction p f r fH n H n H
Calculate the enthalpy change for a partial combustion of 1 mol
of graphite to diamond C(s, graphite) C(s, diamond) ΔH = ?
Given:C(s, graphite) + O2(g) CO2(g) ΔH = -393.5 KJC(s, diamond) + O2(g) CO2(g) ΔH = -395.4 KJ
Answer:C(s, graphite) + O2(g) CO2(g) ΔH = -393.5 kJCO2(g) C(s, diamond) + O2(g) ΔH = +395.4 kJC(s, graphite) C(s, diamond) ΔH = +1.9 kJ
Hess’s Law
Hess’s Law Example Hess’s Law Example ProblemProblemCalculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.
CH4 C + 2H2 +74.80 kJ
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
CH4 C + 2H2 +74.80 kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side
C + O2 CO2 -393.50 kJ
Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2O Reaction Ho
C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ CH4 C + 2H2 +74.80 kJ
C + O2 CO2 -393.50 kJ
Step #3: Multiply reaction #2 by 2
2H2 + O2 2 H2O -571.66 kJ
Calculate H for the combustion of methane, CH4: CH4 + 2O2 CO2 + 2H2O
Reaction Ho C + 2H2 CH4 -74.80 kJ C + O2 CO2 -393.50 kJ
H2 + ½ O2 H2O -285.83 kJ
CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ
2H2 + O2 2 H2O -571.66 kJ
Step #4: Sum up reaction and HCH4 + 2O2 CO2 + 2H2O
-890.36 kJ
Calculation of Heat of Reaction Calculation of Heat of Reaction Calculate H for the combustion of methane, CH4:
CH4 + 2O2 CO2 + 2H2OHrxn = Hf(products) - Hf(reactants)
Substance Hf
CH4 -74.80 kJO2 0 kJ
CO2 -393.50 kJ
H2O -285.83 kJ
Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]Hrxn = -890.36 kJ