1. AP Chem Get Heat HW stamped off Today: Enthalpy Cont., Hess’ Law
Unit 4 Quest: next Thurs 11/8

2. Standard enthalpy of formation, ΔHfo
enthalpy change for the formation of 1 mol of compound with all substances in their standard states (1 atm and 25°C)
Units of kJ/mol; Values will be given
The ΔHfo for the most stable form of any element is 0 (ex. ΔHfo = 0 for H2, O2 etc)
Hrxn = Hf°products – Hf° reactants

3. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)]
= [( kJ) + ( kJ)] – [( kJ) + (0 kJ)]
= ( kJ) – ( kJ) = kJ 3

4. Hrxn = Hf°products – Hf° reactants
Be careful with negative signs!
Make sure to multiply the value in the table by the coefficients in the balanced equation
1) kJ
2) kJ
3a) kJ b) kJ
4) Hf for CaC2 = kJ/mol

5. Hess’s Law:
If a reaction is carried out in a series of steps, the ΔH for the reaction is the sum of the individual ΔH for each step.
We can estimate H using published H values and the properties of enthalpy.

6. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Example
Calculate ΔH for the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Given:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ
H2O(l) → H2O(g) ΔH = +44 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH = -802 kJ
2 H2O(g) → 2 H2O(l) ΔH = 2 x - 44 = -88 kJ
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = -802 kJ kJ = -890 kJ

7. From the following heats of reaction:
2 H2 (g) + O2 (g)  2 H2O (g) ΔH = kJ
3 O2 (g)  2 O3 (g) ΔH = kJ
Calculate the heat of the reaction for:
3 H2 (g) + O3 (g)  3 H2O (g)
To get just one O3 on the reactants side, I need to flip the second equation and divide it by 2
O3 (g)  3/2 O2 (g) ΔH = / = kJ
3 H2 (g) + 3/2 O2 (g)  3 H2O (g) ΔH = x 3/2
= kJ
To get 3 H2 and 3 H2O, I need to multiply the first equation by 3/2.
ΔH = kJ kJ ΔH = kJ

8. Answers 2) -304.1 kJ 3) -2486.3 kJ 5) 156.1 kJ 6) 171.5 kJ
7b) kJ