Thermochemical equations
• A typical chemical equation is S + O2  SO2
• It is called a “thermochemical equation” when
we add information about H …
S + O2  SO2 H = – 296.9 kJ
• If we change the equation, then the H also
changes …
SO2  S + O2 H = + 296.9 kJ
• If the reaction is reversed the sign is reversed
• Also, if numbers in the equation change, so
will the amount of energy produced/absorbed:
2S + 2O2  2SO2 H = – 593.8 kJ
• Read 1st part of handout. Do Q 1-3
Answers
Q1 C + O2  CO2
C + O2
CO2
Q2 H2 + I2  2HI
2HI
H2 + I2
H = – 393.5 kJ
393.5 kJ
H = + 53.2 kJ
53.2 kJ
Q3 1/2H2 + 1/2I2  HI H = + 26.6 kJ
Answers pg. 175, Q 5.45-5.50
5.45 - At STP (25°C and 1 atm)
5.46 - The H° value
5.47 - Moles. (You can’t have 1/2 an atom)
5.48 - answer in back of book
4Al(s) + Fe2O3(s)  2Al2O3(s) + 4Fe(s)
H° = -1708 kJ
Answers pg. 175, Q 5.45-5.50
5.49 - If 2 = 6542 kJ then 1.5 = 4907 kJ
(6542/2 x 1.5)
Actually, it should be -4907 kJ since it is
exothermic
5.50 10CaO(s) + 10H2O(l)  10Ca(OH)2(s)
H° = -653 kJ
Equation is reversed, thus H° sign changes
Equation is multiplied by 10, thus so is H°
Standard heats of reaction
5.45 ° indicates 1 atm (and is usually associated
with a temperature of 25°C)
5.46 An energy term is added (e.g. H or H°)
5.47 Moles
5.48
Al(s) + 1/2Fe2O3 (s)Fe(s) + 1/2Al2O3 (s) H°= – 426.9 kJ
4Al(s) + 2Fe2O3(s)  4Fe(s) + 2Al2O3 (s) H° = – 1708 kJ
5.49 1.5 is 3/4 of 2. 3/4 of 6542 kJ is 4906.5 kJ
5.50
10CaO(s) + 10H2O(l) 10Ca(OH)2(s) H°=–653kJ
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