Chapter 5
Thermochemistry
ENTHALPY
H, Enthalpy, is the change in the HEAT CONTENT of a reaction.
H = Hproducts – Hreactants
H > 0 (positive) ENDOTHERMIC, Heat is absorbed.
H < 0 (negative) EXOTHERMIC, Heat is evolved.
Example:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -890.4 kJ
1 mol CH4 produces 890.4 kJ of energy
1 mol O2 produces 445.2 kJ of energy
Stoichiometry and Enthalpy
Question: For the combustion of 100. g of methane (CH4), how much heat would be generated?
Answer:
5570 kJ
Solution:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -890.4 kJ
(100. g CH4)/(16.0 g/mol) = 6.25 mol CH4
(6.25 mol CH4)(890.4 kJ) = 5570 kJ of heat produced
CALORIMETRY
Heat Capacity … heat required to raise the temperature of an object 1-degree Celsius.
C = ms
or
C = Energy/oC
Specific Heat … heat required to raise the temperature of 1-gm of an object, 1-degree Celsius.
q = msT
T = Tfinal-Tinitial
Question: 120. gm of metal X is at 98oC. It is lowered in to a Styrofoam cup calorimeter (heat
capacity = 0 J/ oC) that has 1200. gm of water at 20. oC. The final temperature is 34oC. What is
the specific heat of metal X?
Answer:
9.15 J/g oC
Solution:
Heat lost by metal X = Heat gain by the water
(mass)(specific heat)( T)metal= (mass)(specific heat)( T)water
(120. g)(specific heat)metal (98 oC -34 oC) = (1200. g)(4.184 Jg-1 oC -1)(34 oC -20 oC)
specific heat metal =9.15 Jg-1 oC -1
Question: 1.9862 g of benzoic acid (C6H5COOH) is heated in a constant-volume calorimeter
(C = 853 J/oC). What is the expected change in temperature of the calorimeter?
(Hobenzoic acid = -3226.7 kJ/mol)
Answer:
61.6 oC temperature increase
Solution:
(1.9862 g)/(122.12467 g/mol) = .016264 mol benzoic acid
(.016264 mol)x(3226.6 kJ/mol) = 52.476 kJ of heat produced
C = Energy/oC
o
C = Energy/C
(52.476 kJ)/(.853 kJ/ oC) = 61.519 oC
Question: 1.9862 g of benzoic acid (C6H5COOH) is heated in a constant-volume calorimeter
(C = 853 J/ oC) that contains .8 liters of H2O. What is the expected change in temperature of the
calorimeter? (Hobenzoic acid = -3226.7 kJ/mol)
Answer:
12.49oC temperature increase
Solution:
(1.9862 g)/(122.12467 g/mol) = .016264 mol benzoic acid
(.016264 mol)(3226.6 kJ/mol) = 52.476 kJ of heat produced
52,476 J of heat = heat absorbed by calorimeter + heat absorbed by water
= [(C)(T)] + [(mass)(spec. heat)( T)]water
= (853 J/ oC)( T) + (800 gm)(4.184 Jg-1 oC -1)( T)
T = 12.5 oC
STANDARD STATE AND ENTHALPY
Hof is the heat change that results when a compound is formed from its elements at a pressure
of 1 atm and at a temperature of 25oC.
Convention: Hof for an element in its most stable form is zero.(Hint: Look at the periodic table)
HESS’S LAW
The Horxn may be determined from a list of the Hof for the reactants and products.
(Hint: Watch the stoichiometry)
Question: What is the Ho for the combustion of C3H8 ?
Answer:
-2226.9 kJ/mol
Solution:
C3H8(g) + 5O2(g)  3CO2(g) +
4H2O(l)
Ho = Hoproducts - Horeactants
Ho = [3HoCO2(g) + 4HoH2O(l)] – [HoC3H8(g)
+
5HoO2(g)]
Ho = [3(-393.5 kJ/mol) + 4(-285.8 kJ/mol)] – [ (-96.8 kJ/mol ) + 5(0 kJ/mol)]
HEAT OF SOLUTION
The heat associated with the dissolving of a solute in a given solution.
Examples: (Solvent is water)
Endothermic
NaCl
KCl
NH4Cl
NH4NO3
+4 kJ/mol
+17.2 kJ/mol
+15.2 kJ/mol
+26.2 kJ/mol
Exothermic
LiCl
CaCl2
-37.1 kJ/mol
-82.8 kJ/mol