1. DH = qp Internal Energy, E Enthalpy, H H = E + PV = + q w heat + _
Chapter A.P. Chemistry
Thermochemistry
CHANGES IN...
WHAT THEY BOTH HAVE IN COMMON
CHANGES IN...
Internal Energy, E
Total amount of K.E. and P.E.
Extensive
Total capacity for heat exchange
Enthalpy, H
State Functions
H = E + PV DE = q w
DH = DE + DPV
PERSPECTIVE IS IMPORTANT!!
heat
…flow of energy due to temp. diff.’s
DH = qp + _
Flowing in
work
…force x distance (moving things)
FINDING THIS
q = m c DT
System
Surroundings
Flowing out +
Done to
VS.
Calorimetry _
(Everything Else) 1.
(Your Focus)
Done by
qsys = -(qrxn + qcal)
w = -PDV
1 mole of a substance formed from its elements in std. states 2.
Given the eq.’s
AN EXPANDING GAS
Heats of Formation.
HESS’S LAW
DH = Snprod.DHf – Snreact.DHf
Write your own 3.

2. positive negative positive
2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn
The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below.
Process
DHº q w
Na(s) → Na(g) v
Na(g) → Na+(g) + e-
-----
O2(g) → 2O(g) x
O(g) + e- → O-(g) y
O-(g) + e- → O2-(g) z
positive
negative
positive
How much heat energy is released or absorbed when mol of solid sodium is produced from solid sodium oxide?
A.) 213 kJ is released
B.) 104 kJ is absorbed
C.) 104 kJ is released
D.) 213 kJ is absorbed

3. positive negative positive
2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn
The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below.
Process
DHº q w
Na(s) → Na(g) v
Na(g) → Na+(g) + e-
-----
O2(g) → 2O(g) x
O(g) + e- → O-(g) y
O-(g) + e- → O2-(g) z
positive
negative
positive
With regards to the production of sodium oxide the DH value of kJ refers to
A.) the heat of combustion of sodium oxide
B.) the heat of formation of sodium oxide
C.) the heat of neutralization of sodium oxide
D.) the heat of solution of sodium oxide

4. positive negative positive
2Na(s) + ½ O2(g) → Na2O(s) DHº = -416 kJ/molrxn
The elements Na and O react directly to produce the compound Na2O according to the equation above. Related information is found in the table below.
Process
DHº q w
Na(s) → Na(g) v
Na(g) → Na+(g) + e-
-----
O2(g) → 2O(g) x
O(g) + e- → O-(g) y
O-(g) + e- → O2-(g) z
positive
negative
positive
O2(g) + 4e- → 2O2-(g)
Which of the following represents DHº for the reaction above?
A.) x + y
B.) x – 2y + z
C.) x + z
D.) x + 2y +2z

5. DH = -68.1 kJ CsOH (0.125 L) (0.250 M) = 0.0313 mol CsOH
HF ( L) (0.625 M) = mol HF
Both L.R. 1:1 ratios in terms of what you need and what you have.
qrxn= (175 g)(4.20 J/g°C)(2.90°C) = 2130 J
qsys= J DH J/ mol = J/mol

6. C2H6 (g) + 7/2 O2 (g) 2 CO2 (g) + 3 H2O (g)
1.) Find DH for this reaction by using the heats of formation from the back of the book.
2.) Use the DH you found to determine the amount of heat given off when 5.00 g of ethane reacts with 5.00 g of oxygen.
3.) What mass of each substance would be needed to release kJ of heat?
1.) DH = [2 mol (-393.5kJ/mol) + 3 mol (-242 kJ/mol)] – [1 mol (-84.7 kJ/mol)]
DH = kJ (associated with the reaction equation above)
2.) (5.00 g C2H6)(1 mol C2H6/30.0g) = mol C2H6
(5.00 g O2)(1 mol O2/32.0g) = mol O2
◄ L.R. (because the balanced reaction requires more oxygen than ethane and you don’t have it!)
0.156 mol O2 (-1428 kJ/3.5 mol) = kJ
3.) O2 ► x (-1428kJ/3.5 mole) = -650 kJ x = 1.59 moles (50.9 g)
C2H6 ► x (-1428kJ/1 mole) = -650 kJ x = moles (13.65 g)