STATE FUNCTIONS and ENTHALPY

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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