Standard Enthalpies of Formation

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Standard Enthalpies of Formation

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Standard Enthalpies of Formation. Standard Enthalpy of Formation. the quantity of energy associated with the formation of one mole of a substance from its elements in their standard states (at SATP) Symbol: Δ H ° f Units: kJ/mol For example: - PowerPoint PPT Presentation

Transcript of Standard Enthalpies of Formation

Page 1: Standard Enthalpies of Formation

Standard Enthalpies of Formation

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Standard Enthalpy of Formation the quantity of energy associated with the formation of the quantity of energy associated with the formation of

one mole of a substance from its elements in their one mole of a substance from its elements in their standard states (at SATP)standard states (at SATP)

Symbol: Symbol: ΔΔHH°°ff Units: kJ/molUnits: kJ/mol

For example:For example:

C(s) + ½ OC(s) + ½ O2 (g)2 (g) + 2 H + 2 H2 (g)2 (g) CH CH33OHOH (l) (l) Δ Δ HH°°ff = -239.1 = -239.1 kJ/molkJ/mol

2 Al2 Al (s) (s) + 3/2 O + 3/2 O2 (g)2 (g) Al Al22OO3 (s)3 (s) Δ Δ HH°°ff = -1675.7 kJ/mol = -1675.7 kJ/mol

All of the elements on the left sides of the equations are All of the elements on the left sides of the equations are in their standard states (SATP)in their standard states (SATP)

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Writing Formation Equations Step 1Step 1: Write one mole of the product in the state that : Write one mole of the product in the state that

has been specified.has been specified.Example:Example: FeSO FeSO4 (s)4 (s)

Step 2Step 2: Write the reactant elements in their standard : Write the reactant elements in their standard states (reference periodic table. Pg 838).states (reference periodic table. Pg 838).

FeFe (s) (s) + S + S (s) (s) + O + O2 (g)2 (g) FeSO FeSO4 (s)4 (s)

Step 3Step 3: Write coefficients for the reactants to give a : Write coefficients for the reactants to give a balanced equation yielding one mole of product.balanced equation yielding one mole of product.

Fe (s) + S (s) + 2 O2 (g) FeSO4 (s)

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Calculating ΔH A third way of calculating ΔH is possible using standard A third way of calculating ΔH is possible using standard

enthalpies of formation. enthalpies of formation.

Table C.6 pg 799 gives standard enthalpies of formation Table C.6 pg 799 gives standard enthalpies of formation for common molecules.for common molecules.

According to Hess’s Law, the enthalpies of known According to Hess’s Law, the enthalpies of known equations may be used to calculate the enthalpy of an equations may be used to calculate the enthalpy of an unknown reaction.unknown reaction.

For exampleFor example: The formation of hydrogen and carbon : The formation of hydrogen and carbon monoxide gas from water and graphite:monoxide gas from water and graphite:

HH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g) ΔH = ?ΔH = ?

Use the formation equations for each of the products and Use the formation equations for each of the products and reactants to create the target equation.reactants to create the target equation.

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Calculating ΔH0, continuedHH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g)

ReactantsReactants

HH22(g) + 1/2 O(g) + 1/2 O22(g) --> H(g) --> H22O(g) O(g) HH00f H2Of H2O = - 241.8 kJ/mol = - 241.8 kJ/mol

CC (s) (s) – graphite in standard state – graphite in standard state HH00ff = 0 kJ/mol = 0 kJ/mol

ProductsProducts

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g) (g) --> CO(g) HH00f COf CO = - 110.5 kJ/mol = - 110.5 kJ/mol

HH2 (g)2 (g) – hydrogen in standard state – hydrogen in standard state HH00ff = 0 kJ/mol = 0 kJ/mol

HH22O(g) --> HO(g) --> H22(g) + 1/2 O(g) + 1/2 O22(g) (g) HH00 = 241.8 kJ/mol = 241.8 kJ/mol

C(s) + 1/2 OC(s) + 1/2 O22(g) --> CO(g) (g) --> CO(g) HH00 = - 110.5 kJ/mol = - 110.5 kJ/mol

HH22OO (g) (g) + C + C (s) (s) H H2 (g)2 (g) + CO + CO (g) (g) HH00 = 131.3 kJ/mol = 131.3 kJ/mol

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In general, when all enthalpies of formation are known, the In general, when all enthalpies of formation are known, the enthalpy of a reaction is equal to the sum of the enthalpy of a reaction is equal to the sum of the enthalpies of formation of the products minus the sum of enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants.the enthalpies of formation of the reactants.

where n = the amount in moles of each product or where n = the amount in moles of each product or reactantreactant

)(reactants Hn - (products) Hn H ffrxn

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Sample Problem

Calculate the heat of combustion of methanol.

CH3OH (l) + 3/2 O2 (g) --> CO2 (g) + 2 H2O (g) H0comb = ?

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Multistep Energy Calculations

Several energy calculations may be required to solve multistep problems

–Heat flow q = mcΔT

–Enthalpy changes ΔH = n ΔHx

–Hess’s Law ΔHtarget = ∑ΔHknown

–Enthalpies of Formation

)(reactants Hn - (products) Hn H ffrxn

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Sample Problem, pg 336

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Sample problem cont’d

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Sample problem cont’d