Standard Enthalpy of Formation

EQ: Why does the Hfº for a free element equal zero?

A. Standard Enthalpy of formation (Hfº)

Hfº : the energy change that occurs when 1 mole of a compound is formed (most are negative)

A. Standard conditions = 25ºC & 1 atm (gases) & 1 Molar (aqueous solutions)

B. values found in table or appendix C (pg 921)C. All free elements in their most stable form

Hf º= zero

Since you can not make elementsThis includes diatomics (Cl2, H2, O2, etc.)

A. Standard Enthalpy of formation (Hfº)

Ex: reaction for the heat of formation for H2O (g) Elements in water = H and O Stable form = H2 and O2

H2 (g) + ½ O2 (g) H2O (g) Hfº= -241.8 kJ/mol

B. Calculating Hrxn

The Hºrxn can also be found using the Hfº valuesuse coefficients in the balanced reactionpay attention to states (s, l, g, aq)

Hºrxn = nHfº(products) - nHfº (reactants)

n= moles (sigma) = sum of

Example 1Use the standard enthalpies of formation

on pg 510 to calculate Hºrxn for: 4NH3 (g) + 7O2 (g) 4NO2 (g) + 6H2O (l)

Hºrxn = [ (4 mol x 33 kJ/mol) + (6 mol x -286 kJ/mol)]

- [(4 mol x -46 kJ/mol) + (7 mol x 0 kJ/mol)]

Hºrxn = [132 kJ + -1716 kJ] - [-184 kJ + 0kJ]= -1400 kJ . Round to the least number of

decimal places out of all of the table values

-46 0 (free element) 33 -286From table

Example 2Use the table on pg 921 and the reaction

below to determine the standard enthalpy of formation for CCl4 gas.

CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g) ∆H= –397 kJ

∆Hrxn =[(1 mol x ∆HfCCl4 (g)] + (4 mol x –92.307kJ/mol)]

- [(1 mol x -74.81kJ/mol)+ (4 mol x 0 kJ/mol)]

[1 mol ∆Hf+ -369.228 kJ ] - [-74.81 kJ]-397 kJ =

-74.81 0 ? (gas) -92.307 From table

Example 2 (continued)

[1 mol ∆Hf+ -369.228 kJ ] - [-74.81 kJ]-397 kJ = 1 mol ∆Hf - 294.418 kJ -397 kJ =

-102.6 kJ = 1 mol ∆Hf1 mol1 mol

∆Hf= -103 kJ/mol

Ex 3: Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH 0 (products)f= nH 0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x– 286 ] – [ 2x49.04 ] = -6536 kJ

-6536 kJ2 mol

= - 3268 kJ/mol C6H6