Review Sheet – Chemistry, Level 3 –
Ch. 12 & 23: Thermochemistry & Thermodynamics
Name: ____________________
Date:___________
Period:___
The student should be able:
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Communicate effectively with thermochemistry and thermodynamics vocabulary, including
enthalpy, entropy, heat, specific heat, heat, thermal energy, and temperature.
Compare exothermic to endothermic reactions, including by using energy diagrams.
Calculate heats given off by reactions when provided with enthalpies for a reaction
Calculate enthalpies of reaction from calorimetry data, Hess’s Law situations, and heats of
formations.
Describe the factors that cause entropy to increase or decrease.
Analyze the spontaneity of a chemical reaction, including Gibbs free energy.
1. Define each of the following terms clearly and completely.
(a) enthalpy –
(h) Heats of formation
(b) exothermic –
(i) entropy –
(c) endothermic –
(j) spontaneous chemical reaction –
(d) standard state –
(k) Gibbs Free Energy –
(e) Hess’s law –
(l) heat –
(f) Specific heat –
(m) temperature –
(g) Calorimetry –
(n) thermal energy –
2. How much heat is required to change 63.4 grams of liquid water to steam (vaporize)? The Hvap of water is
40.7 kJ/mol.
 1mole  40.7kJ 

q(heat)  63.4g
  143 kJ
 18.0g  1mole 
3. Using the balanced equation below, how much heat is given off when the following react:
2 C2H6(g) + 7 O2(g) 
4 CO2(g)
+ 6 H2O(l)
H = - 2860 kJ
A. 5.60 g of C2H6
 1 mole  - 2860 kJ 

q(heat)  5.60g
  -267 kJ
 30.0g  2 mole 
B. 7.04 L of O2 (@ STP)
 1 mole  - 2860 kJ 
q(heat)  7.04L

  -128 kJ
 22.4 L  7 mole 
4. Using Hess’s law, calculate the H for the reaction:
CaCO3 
CO2 + CaO
from the following enthalpy changes:
Ca + C + 3/2 O2  CaCO3
H = - 1206.9 kJ
2 Ca + O2  2CaO
H = - 1270.2 kJ
C + O2  CO2
H = - 393.5 kJ
CaCO3 Ca + C + 3/2 O2
H = + 1206.9 kJ
Ca + ½ O2  CaO
H = - 635.1 kJ
C + O2  CO2
H = - 393.5 kJ
__________________________________________________
CaCO3 
H = + 178.3 kJ
CO2 + CaO
5. What is the specific heat of carbon tetrachloride (CCl4) if it requires 77 J of heat to raise the temperature of
6.5-g sample from 21C to 35C?
C
q
77J

 0.85J/g * C
mT (6.5)(35 - 21)
6. Using heats of formation, calculate the H for the reaction:
C2H4(g) + 3 O2(g) 
H (kJ)
+52.3
0
2 CO2(g)
+ 2 H2O(l)
-393.5
-285.8
H = [2(-393.5)+2(-285.8)] – [1(52.3)] = - 1410 kJ
7. Propane burns in oxygen according to the following equation: 2 C2H6 + 7 O2  4 CO2 + 6 H2O. Draw an
enthalpy diagram for this reaction and label the enthalpies of the products, the reactants, and the change in
enthalpy. How does this diagram tell you the reaction is exothermic?
See Slide # 17 on the Ch. 12 & 23 Powerpoint. It is very similar to that one.
8. Some reaction has a H = -106 kJ and S = + 58 J/K. Calculate its Gibbs free energy (G) at 500 C. Is this
reaction spontaneous at this temperature?
ΔG  (106kJ)  (773 * 0.058kJ/K)  -151 kJ
G is negative, so this reaction is spontaneous at this temperature.
9. Calculate the Gibbs free energy (G) at 25C for
3 H2(g) + CO(g) 
CH4(g) + H2O(l)
A. Given that H = -206 kJ and S = - 214 J/K for this reaction. Is this reaction spontaneous at this
temperature?
ΔG  (206kJ)  (298 * 0.214kJ/K)  -142 kJ
G is negative, so this reaction is spontaneous at this temperature.
B. At what temperature would it go from being spontaneous to not-spontaneous?
G =0 when the reaction changes from spontaneous to non-spontaneous, so solving 0=G-TS for T, we get:
T
H
 206

 963K
S  0.214
10. Using the standard tables, calculate the H, S, and G at 25C for the reaction:
A. 2 CCl4(l) + H2(g) 
2 CHCl3(l)
+ 2 Cl2(g)
NOT POSSIBLE (G and S for CHCl3 are not available).
11. The explosion of dynamite, a mixture of nitroglycerin (liquid) and diatomaceous earth (solid), gives gaseous
products such as water and carbon dioxide. Much heat is given off as well. Give the algebraic (+,-) signs of H,
S, and G. Explain your answers.
H is negative because the explosion of dynamite is exothermic (heat is given off). S is positive because
gaseous products are produced from a liquid and solid reactant. G is negative, because the reaction happens
without constant outside intervention.
12. Why does Gibbs free energy determine a reaction’s spontaneity?
G takes into account the three factors that determine spontaneity: enthalpy(H), entropy(S), and temperature.