Thermodynamics - Review

Chem I | Chem II AP
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Fill in the blank:
1)
4
Al(s)
+
3
O2(g)
2
Al2O3(s)
   
ΔH =
kcal
Al(s)
+
O2(g)
Al2O3(s)
   
ΔH =
kcal

The reaction was cut in half and ΔH should also be cut in half.

2)
1/2
N2(g)
+
O2(g)
+
8.1
kcal
NO2(g)
   
ΔH =
kcal
2
NO2(g)
N2(g)
+
O2(g)
+
kcal
   
ΔH =
kcal

The reaction was doubled and reversed, so ΔH should also be doubled and the sign should be reversed.

3)
Na(s)
+
1/2
Cl2(g)
NaCl(s)
   
ΔH =
-98.2
kcal
2
Na(s)
+
Cl2(g)
NaCl(s)
   
ΔH =
kcal

The reaction was doubled but not reversed, so ΔH should be doubled but the sign should stay the same.

Calculate ΔH for the following reactions:
4)
2 CO(g) + O2(g) → 2 CO2(g)
ΔH =
kcal
[2(-94.05)] - [2(-26.42) + 0]
[-188.1] - [-52.84]
-188.1 + 52.84 = -135.26
5)
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
ΔH =
kcal
[4(-94.05) + 2(-68.32)] - [2(54.19) + 5(0)]
[-376.2 + -136.64] - [108.38]
-512.84 - 108.38 = -621.22
6)
CH4(g) + 2 Cl2(g) → CCl4(l) + 2 H2(g)
ΔH =
kcal
[-33.34 + 2(0)] - [-19.1 + 2(0)]
[-33.34] - [-19.1]
-33.34 + 19.1 = -14.24
Using bond energies, calculate ΔH:
7)
2 H2(g) + O2(g) → 2 H2O(g)
ΔH =
kcal
Bonds Breaking: (total = 327 kcal)
2 H-H bonds = 2(104) = 208 kcal
1 0=0 bond = 119 kcal
Bonds Forming: (total = - 444 kcal)
4 O-H bonds = 4 (111) = - 444 kcal (given a negative value since bonds forming is exothermic)
Total energy change in reaction:
327 + -444 = -117 kcal
8)
Given: C2H4(g) + H2(g) → C2H6(g)
   
ΔH = -32.7 kcal
Is this reaction endothermic or exothermic?

ΔH is negative for an exothermic reaction

How much heat is produced when two moles of C2H4 react?
kcal
2 mol x
32.7 kcal
1 mol C2H4
= 65.4 kcal

The heat produced is positive 65.4 (not negative) because the question is not asking for the net change of ΔH.

Which has greater enthalpy, the products or the reactants?

Since the reaction is exothermic, the reactants have more enthalpy. Exothermic means that it gives off energy, so there is more energy at the beginning of the reaction than at the end.

Choose the equation with the heat term added to the correct side:
A) C2H4(g) + H2(g) + 32.7 kcal → C2H6(g)
B) C2H4(g) + H2(g) - 32.7 kcal → C2H6(g)
C) C2H4(g) + H2(g) → C2H6(g) + 32.7 kcal
D) C2H4(g) + H2(g) → C2H6(g) - 32.7 kcal

Since the reaction is exothermic the ΔH should be placed on the products side, but it is represented as a positive number in the chemical equation.

How many kcal are given off when 168 g of C2H4 react?
kcal
168 g C2H4 x
1 mol C2H4
28 g C2H4
x
32.7 kcal
1 mol C2H4
= 196.2 kcal

Since the problem is not asking for the ΔH, the value for kcal should be a positive number.

9)
Iron reacts with the oxygen in the air to form rust (Fe2O3).
4 Fe + 3 O2 → 2 Fe2O3
How many kcal are given off when a pound of Fe reacts to form rust? (1 lb = 454 g)
kcal
ΔH = [2(-196.5)] - [4(0) + 3(0)] = -393 kcal
454 g Fe x
1 mol Fe
55.8 g Fe
x
393 kcal
4 mol Fe
= 799.8 kcal
Since the problem is not asking for ΔH, the value for the kcal given off should be a positive number.
10)
Given: Mg + 1/2 O2 → MgO
   
ΔH = -143.8 kcal
How many grams of Mg must react in order to produce 1000.0 kcal of heat?
g
1000.0 kcal x
1 mol Mg
143.8 kcal
x
24.3 g Mg
1 mol Mg
= 168.89 g
11)
Is entropy increasing or decreasing?
H2O(g) → H2O(l)

Entropy is increasing because a gas is condensing to a liquid, making the molecules more ordered.

H2O is heated from 10°C to 30°C

Entropy is increasing because temperature is increasing, making the molecules move faster and therefore have more disorder.

Ag+(aq) + Cl-(aq) → AgCl(s)

Entropy is decreasing because two ions in an aqueous solution (liquid) are forming a solid, making the molecules more ordered.

Ice melts

Entropy is increasing because as ice melts it changes from a solid to a liquid, making the molecules have more disorder.

12)
Given: 2 CH3OH(l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O(g)
a)
ΔH =
kcal
[2(-94.05) + 4(-57.8)] - [2(-57.02) + 3(0)]
[-188.1 + -231.2] - [-114.04]
-419.3 + 114.04 = -305.26 kcal
b)
ΔS =
cal/K
[2(51.07) + 4(45.1)] - [2(30.3) + 3(49)]
[102.14 + 180.4] - [60.6 + 147]
[282.54] - [207.6] = 74.94 cal/K
c)
ΔG at 25°C =
kcal
[2(-94.26) + 4(-54.64)] - [2(-39.73) + 3(0)]
[-188.52 + -218.56] - [-79.46]
-407.08 + 79.46 = -327.62 kcal
An alternate way to work this problem is to use the ΔH and ΔS calculated above in the Gibb's equation:
ΔG = -305.26 kcal - (298 K)(.07494 kcal/K) = -327.59 kcal
Remember when using the Gibb's equation to change the temperature to Kelvin and the ΔS to kcal/K.
d)
Is the reaction endothermic or exothermic?
How do you know?

ΔH tells if the reaction is endothermic or exothermic.

e)
Is entropy increasing or decreasing?
How do you know?

ΔS tells if entropy of the reaction is increasing or decreasing.

f)
Is the reaction spontaneous at 25°C?
How do you know?

ΔG tells whether or not the reaction is spontaneous. However, if ΔH is negative and ΔS is positive, the reaction will always be spontaneous (at any temperature) because the Gibb's equation will always produce a negative value for ΔG.

g)
Is the reaction spontaneous at 200°C?
ΔG = -305.26 kcal - (473 K)(.07494 kcal/K) = -269.81 kcal

Since ΔG is negative, the reaction will be spontaneous at this temperature, but since ΔH is negative and ΔS is positive, the reaction should be spontaneous at any temperature.

h)
Is the reaction spontaneous at -50°C?
ΔG = -305.26 kcal - (223 K)(07494 kcal/K) = -288.15 kcal

Since ΔG is negative, the reaction will be spontaneous at this temperature, but since ΔH is negative and ΔS is positive, the reaction should be spontaneous at any temperature