Transcript Calculating ΔH using molar heats of formation

Calculating ΔH using molar
heats of formation
Chem 12
• If 1 mol of compound is formed from its constituent
elements, then the enthalpy change for the reaction is
called the enthalpy of formation, Hof .
• Standard conditions (standard state): 1 atm and 25 oC
(298 K).
• Standard enthalpy of formation: 1 mol of compound is
formed from substances in their standard states.
Examples – write formation reactions for each:
*Remember the compounds are formed directly
from their elements.
1. H2SO4
2. NH4Cl
•
•
You want only one mole of the product
being formed.
Look up the ΔHf on the table
1. H2 + S + 2O2  H2SO4
ΔHf = -814 kJ/ mol
2. ½ N2 + 2H2 + ½ Cl2  NH4Cl
ΔHf = -314.4 kJ/ mol
Enthalpies of Formation
Using Enthalpies of Formation for Calculating
Enthalpies of Reaction
• For a reaction
H rxn   nH f products   nH f reactants
By definition, the enthalpy of formation of an
element in its standard state is zero.
Example, oxygen (O2) and chlorine (Cl2)
both have Hof of zero.
Sample Problem 1
• Calculate ΔH for the following reaction
using standard molar heats of formation,
ΔH°f .
• 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g)
ΔH = ?


H rxn   nH f products   mH f reactants
• ΔH°f for NH3(g) = -45.9 kJ/mol
• ΔH°f for HCl(g) = -92.3 kJ/mol
• ΔH°f for Cl2(g) and N2(g) is 0
• ΔHrxn = Σ nΔH°f(product) - Σ nΔH°f(reactant)
• ΔHrxn= (0 + 6(-92.3 kJ)) - (2(-45.9 kJ) + 0)
= (-553 kJ) - (-91.8 kJ)
= -461 kJ
Examples
Use the tables in the back of your book to calculate
ΔH for the following reactions:
1. 4 CuO (s)  2 Cu2O (s) + O2 (g)
2. C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
3. NH3 (g) + HCl (g)  NH4Cl (s)
1. +292 kJ
2. -2220. kJ
3. -176.2 kJ
Practice:
• Page 685 #15
• Page 687 #19-21
• Page 691 # 5