state functions and enthalpy
DESCRIPTION
STATE FUNCTIONS and ENTHALPY. heat energy transferred. work done by the system. energy change. FIRST LAW OF THERMODYNAMICS. ∆E = q + w. Energy is conserved!. Chemical Reactivity. What drives chemical reactions? How do they occur? - PowerPoint PPT PresentationTRANSCRIPT
1STATESTATEFUNCTIONSFUNCTIONS
andand
ENTHALPYENTHALPY
STATESTATEFUNCTIONSFUNCTIONS
andand
ENTHALPYENTHALPY
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FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS
FIRST LAW OF FIRST LAW OF THERMODYNAMICSTHERMODYNAMICS
∆E = q + w
heat energy transferred
energychange
work doneby the system
Energy is Energy is conserved!conserved!
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ChemicalChemical Reactivity ReactivityChemicalChemical Reactivity ReactivityWhat drives chemical reactions? How do they occur?
The first is answered by THERMODYNAMICS and the second by KINETICS.
Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
• electron transfer in a battery
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ChemicalChemical Reactivity ReactivityEnergy transfer allows us to predict reactivity.
In general, reactions that transfer energy to their
surroundings are product-favored.
So, let us consider heat transfer in chemical processes.So, let us consider heat transfer in chemical processes.
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If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
If Hfinal < Hinitial then ∆H is negative
Process is EXOTHERMIC
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
If Hfinal > Hinitial then ∆H is positive
Process is ENDOTHERMIC
ENTHALPY: ENTHALPY: ∆HENTHALPY: ENTHALPY: ∆H∆∆H = HH = Hfinalfinal - H - Hinitialinitial
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This equation is valid because ∆H is a STATE FUNCTION
Depend ONLY on the state of the system
NOT how it got there.
Can NOT measure absolute H,
ONLY measure ∆H.
∆∆H along one path = H along one path = ∆H along another ∆H along another pathpath
∆∆H along one path = H along one path = ∆H along another ∆H along another pathpath
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8Standard Enthalpy Values
Standard Enthalpy Values
Most ∆H values are labeled ∆Ho
Measured under standard conditions
P = 1 bar
Concentration = 1 mol/L
T = usually 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas
9Standard Enthalpy Standard Enthalpy ValuesValues
∆Hfo = standard molar
enthalpy of formation
— — the enthalpy change when 1 mol of compound is formed from elements under standard conditions.
See Table 6.2 and Appendix LSee Table 6.2 and Appendix L
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11∆∆HHffoo, standard molar , standard molar
enthalpy of formationenthalpy of formation
H2(g) + 1/2 O2(g) --> H2O(g)
∆Hfo (H2O, g)= -241.8 kJ/mol
By definition,
∆Hfo
= 0 for elements in their
standard states.
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HESS’S LAWHESS’S LAW
In general, when ALL
enthalpies of formation are known,
Calculate ∆H of reaction?
∆Horxn = ∆Hf
o (products)
- ∆Hfo
(reactants)
∆Horxn = ∆Hf
o (products)
- ∆Hfo
(reactants)
Remember that ∆ always = final – initial
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18Using Standard Enthalpy Values
Using Standard Enthalpy Values
Calculate the heat of combustion of
methanol, i.e., ∆Horxn for
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
∆Horxn = ∆Hf
o (prod) - ∆Hf
o (react)
19Using Standard Enthalpy Using Standard Enthalpy ValuesValues
Using Standard Enthalpy Using Standard Enthalpy ValuesValues
∆Horxn = ∆Hf
o (CO2) + 2 ∆Hf
o (H2O)
- {3/2 ∆Hfo
(O2) + ∆Hfo
(CH3OH)}
= (-393.5 kJ) + 2 (-241.8 kJ)
- {0 + (-201.5 kJ)}
∆Horxn = -675.6 kJ per mol of methanol
CHCH33OH(g) + 3/2 OOH(g) + 3/2 O22(g) --> CO(g) --> CO22(g) + 2 H(g) + 2 H22O(g)O(g)
∆Horxn = ∆Hf
o (prod) - ∆Hf
o (react)
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