Dougherty Valley HS AP Chemistry
Second Law Worksheet
Name:
Date:
Period:
[1] For each system below, indicate whether ∆S and ∆H is a positive or negative value. Then indicate if the reaction is
entropy driven, enthalpy driven, or neither. Will the reaction be spontaneous at high temperatures, low temperatures,
always or never? Qualitative, explain to your table group.
A) NaCl (s)
+
H2O (l)
+
heat

NaCl (aq)
∆S = Positive (more disorder [s  aq])
∆H = Positive (endothermic)
driven? Entropy – (solid  aqueous =
more randomness and disorder which is spontaneous). Spontaneous at high temperatures
B) O2 (g)
+

H2O (l)
O2 (aq)
+
heat
∆S = Negative (less disorder [g  aq])
∆H = Negative (exothermic)
driven? Enthalpy – (gas  aqueous =
∆H
less disorder, more order, not spontaneous. Heat released to surroundings (pos. ∆Ssurr due to - , where ∆H is negative).
𝑇
Spontaneous at low temperatures
C) CO2 (s)
+
heat

CO2 (g)
∆S = Positive (more disorder [s  g])
∆H = Positive (endothermic)
disorder and randomness Spontaneous at high temperatures
driven? Entropy – (solid  gas = more
[2] Calculate the ∆Hrxn, ∆Srxn, ∆Suniverse, ∆Grxn, For each system below, indicate whether ∆S and ∆H is a positive or
negative value. Then indicate of the reaction is entropy driven, enthalpy driven, or neither. Quantitative
∆H˚formation (KJ/mole)
-103.8
0
-393.5
-241.8
-939.7
-804.2
0
0
Substance
C3H8 (l)
O2 (g)
CO2 (g)
H2O (g)
TiO2 (s)
TiCl4 (l)
C (s)
Cl2 (g)
A)
C3H8 (g)
+

2O2 (g)
∆G˚formation (KJ/mole)
-23.5
0
-394.4
-228.6
-884.5
-737.2
0
0
∆S˚formation (J/moleK)
269.9
205.1
213.7
188.8
49.9
252.3
5.7
223.1
3CO2 (g)
+
4H2O (g)
∆Hrxn = np∆H˚f - nr∆H˚f = [3(-393.5) + 4(-241.8)] – [1(-103.8) + 2(0)] = -2043.9 KJ
∆Srxn = np∆S˚f - nr∆S˚f = [3(213.7) + 4(188.8)] – [1(269.9) + 2(205.1)] = 716.2 J/K
∆Grxn = Hrxn - TSrxn = np∆G˚f - nr∆G˚f = -2043.9 KJ – 298(.7162 KJ/K) = -2257.3 KJ/mole
∆Suniverse = ∆Ssys + ∆Ssurr = .7162 KJ/moleK + (2043.9 KJ)/298 = 7.575 KJ/K

Note in this system both enthalpy and entropy favor the products. This is true because the products have more
entropy than the reactants and the products have less enthalpy than the reactants. Both factors drive the reaction
towards the products.
B) TiO2 (s)
+
C (s)
+
2Cl2 (g)

TiCl4 (l)
+
CO2 (g)
∆Hrxn = np∆H˚f - nr∆H˚f = (3(-393.5) + 804.2) – (939.7 – 0 – 0) = -1316 KJ/mole
∆Srxn = np∆S˚f - nr∆S˚f = (252.3 + 213.7) – (49.9 – 5.7 – 223.1) = 187.3 J/moleK
∆Grxn = Hrxn - TSrxn = np∆G˚f - nr∆G˚f = -1316 KJ/mole – 298(.1873) = -1371.8 KJ/mole
∆Suniverse = ∆Ssys + ∆Ssurr = .1873 KJ/moleK + (1316 KJ/mole/298) = 4.60 KJ/mole