1. How much heat energy is required (at constant pressure) to convert 50g of ice at 100K to liquid water at 315K given the following data: Cwater = J/gK, Cice = 2.1 J/gK, Hfusion = 6.01 kJ/mol.
315 K
Temperature (K)
273 K
100 K
Energy (J)
Heating ice up to melting point: q = cice  mice  Tice = (2.1 J/gK)(50g)(273K-100K) = J
Converting ice to liquid water: qp = n  Hfus,water = (50g)(1 mol/18g)(6.01 kJ/mol) = kJ
Heating up liquid water: q = cwater  mwater  Twater = (4.184J/gK)(50g)(315K-273K) = 8786 J
Total Heat Required = q1 + q2 + q3 = kJ kJ kJ = kJ

2. A certain piece of metal with a mass of 140. 0 g is heated to 135° C
A certain piece of metal with a mass of g is heated to 135° C. The metal is then placed into a coffee cup calorimeter containing g of water initially at 25 ° C. Determine the specific heat of the metal if the calorimeter and metal reach thermal equilibrium at 37° C.
q metal = q water
140.0g ·c (37° C ° C = g ·4.18 J/gC · (37 ° C ° C )
c= 5016 J = .366 J/gC
140 g · 98 ° C

3. Enthalpy- heat evolved or absorbed by the reaction
Thermochemical equation – balanced chemical equations that show the associated enthalpy change.
Heat of reaction- enthalpy change that accompanies a reaction
Standard heat of formation- the change in enthalpy for the reaction that forms 1 mol of the compound (usually reported at 298 K)

4. Guidelines for Thermochemical Equations :
Enthalpy is an extensive property.
The magnitude of ΔH is directly proportional to the amount of reactant consumed by the process.

5. How much heat is released when 4
How much heat is released when 4.50 g of methane gas is burned in a constant – pressure system according to the equation below?
CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ

6. Guidelines for Thermochemical Equations :
2. The enthalpy change for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction.
CH4 (g) + 2O2(g)  CO2 (g) + 2 H2O (l) ΔH= -890 kJ
CO2 (g) + 2 H2O (l)  CH4 (g) + 2O2(g) ΔH= 890 kJ

7. Guidelines for Thermochemical Equations :
3. The enthalpy change for a reaction depends on the state of the reactants and products.

8. Hess’s Law
Hess's Law of Constant Heat Summation states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. 
Germain Henri Hess

9. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

10. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

11. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
Step #3: Multiply reaction #2 by 2

12. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
CH4 + 2O2  CO2 + 2H2O kJ
Step #4: Sum up reaction and H

13. Hess’s Law
“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

14. Calculation of Heat of Reaction
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  n Hf(products) -  n  Hf(reactants)
    Substance
Hf  
CH4 kJ O2
0 kJ
CO2 kJ
H2O kJ
Hrxn = [ kJ + 2( kJ)] – [-74.80kJ]
Hrxn = kJ