Advanced Chemistry Thermochemistry
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Student Worksheet for Thermochemistry
Attempt to work the following practice problems after working through the sample problems in
the videos. Answers are given on the last page(s).
Relevant Equations/Information
Specific Heat: Q=mc∆T
∆T= Final Temperature - Initial Temperature
Where Q= Energy in J, m= mass in grams, c= Specific Heat in J/g˚C, and T= ˚C
Hess’ Law: H = ∑ of H (energy) absorbed and released by all reactions/bonds broken or made
- Can also be calculated as ∆H = ∑ Hproducts – ∑ Hreactants
Influence of ∆H on Reaction Type (see energy diagram below)
- Negative ∆H= Exothermic Reaction (releases energy into the surroundings)
- Positive ∆H= Endothermic Reaction (requires energy to be put in before it can proceed)
Endothermic Reaction Exothermic Reaction
Hints for using H to determine endothermic versus exothermic reactions.
The interpretation of the H value depends on how it is used. For examples:
a) In your video, you are only looking at the H value of the bonds. Thus, if the reactants
have more energy than the products (a + H), energy was added to the system (left
graph above) and is endothermic.
b) When using Hess’ law, you are looking at the amount of energy released and
absorbed for the entire molecule (not individual bonds). This is also known as Hf.
Breaking a bond absorbs energy, and making a bond releases energy. Consider the
following equation and scenario:
A + B C
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If 10 kJ (total) of energy are absorbed to break the bonds in Molecules A & B, the 10
kJ are now available for the formation of molecule C. Remember that energy is
conserved. When C is formed, it uses 5 kJ of energy. This leaves 5 kJ of energy (such
as heat) still in the surroundings. Using Hess’ Law to determine if the reaction is
endothermic or exothermic,
- ∆H = ∑ Hproducts – ∑ Hreactants
- ∆H = 5 kJ – 10 kJ = -5 kJ of energy are released (making it exothermic) like the right
graph above.
c) Atoms in their elemental form (O2, H2, C, etc) have an Enthalpy of formation
equal to 0 kJ/mol.
Advanced Chemistry Thermochemistry
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1. Gold has a specific heat of 0.129 J/g°C. How many joules of heat energy are required to raise
the temperature of 40 grams of gold from 20°C to 70°C?
2. An unknown substance with an unknown mass absorbs 1500J while undergoing a temperature
increase of 30°C. It’s specific heat is 0.49 J/g°C. What is the mass of the substance?
3. If the temperature of 50.1 g of ethanol increases from 35°C to 87.8°C, how much heat has
been absorbed by the ethanol? The specific heat of ethanol is 2.44J/g°C.
4. If 425 g of water at 75°C loses 6570 J of heat, what is the final temperature of the water?
Liquid water has a specific heat of 4.18 J/g°C.
5. Tea or coffee is drank in cold weather because it warms us. The beverage is 40°C and the
body is normally 37.2°C, but warms to 39°C after consumption. If you drink a cup of tea with a
mass of 300g of water, how much heat energy was transferred to the body? The specific heat of
water is 4.184J/g°C.
Advanced Chemistry Thermochemistry
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6. When testing to see if a solid is pure Titanium, the temperature is raised from 25˚C to 57˚C
after the absorption of 154.2 J of heat. If the mass of the solid is 2.7 g and the specific heat of
Titanium is known to be 0.54 J/g˚C. What is the specific heat of the solid and is it pure
Titanium?
For questions 7-11, use the bond energies provided to classify them as Endothermic or
Exothermic.
7.
Bond Energy per Bond (kJ per mole)
H-H 432
O-O 204
O-H 467
2 H2 + O2 2 H2O
8.
Bond Energy per Bond (kJ per mole)
C-H 413
N-H 391
C-N 305
H-H 432
CH4 + NH3 CH3NH2 + H2
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9.
Bond Energy per Bond (kJ per mole)
H-Cl 427
S-Cl 271
H-S 347
Cl-Cl 243
2 HCl + SCl2 H2S + 2 Cl2
10.
Bond Energy per Bond (kJ per mole)
N-N 160
H-H 432
N-H 391
N2 + 3 H2 2 NH3
11.
Bond Energy per Bond (kJ per mole)
C-Cl 339
O-H 467
H-Cl 427
C-O 358
CHCl3 + H2O HCl + CHCl2OH
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For questions 12-15, use the Enthalpy formations provided and Hess’ Law to classify them as
Endothermic or Exothermic.
12.
Molecule Hformation (kJ)
H2O -286
CO -111
H2 0
CO2 -394
H2O (l) + CO (g) H2 (g) + CO2 (g)
13.
Molecule Hformation (kJ)
NH3 -368
N2O4 27
N2 0
H2O -242
8 NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g)
Advanced Chemistry Thermochemistry
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14.
Molecule Hformation (kJ)
Fe 0
O2 0
Fe2O3 -824
4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)
15.
Molecule Hformation (kJ)
CaO -635
CO2 -394
CaCO3 -1,207
CaO (s) + CO2 (g) CaCO3 (s)
Advanced Chemistry Thermochemistry
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1. Gold has a specific heat of 0.129 J/g°C. How many joules of heat energy are required to raise
the temperature of 40 grams of gold from 20°C to 70°C?
Q = mc∆T You are solving for Q
= 40*0.129*(70-20)
= 40*0.129*50
= 258 J of energy was absorbed
2. An unknown substance with an unknown mass absorbs 1500J while undergoing a temperature
increase of 30°C. It’s specific heat is 0.49 J/g°C. What is the mass of the substance?
Q = mc∆T You are solving for m.
m = 𝑄
𝑐∆𝑇
= 1500
(0.49∗30)
= 102 grams
3. If the temperature of 50.1 g of ethanol increases from 35°C to 87.8°C, how much heat energy
has been absorbed by the ethanol? The specific heat of ethanol is 2.44J/g°C.
Q = mc∆T You are solving for Q
= 50.1*2.44*(87.8-35)
= 50.1*2.44*52.8
= 6,454.5 J or ~6.5 kJ of energy was absorbed
4. If 425 g of water at 75°C loses 6570 J of heat, what is the final temperature of the water?
Liquid water has a specific heat of 4.18 J/g°C.
This one is a little different because you are not solving for an entire variable of the specific heat
equation. Instead, you are solving for the final temperature in the ∆T equation. NOTE: The water
is losing (has less) heat, so a negative sign goes in front of the Q value. There are two ways to
approach this problem, depending on which method you are most comfortable with.
Option A:
Q = mc∆T
∆T = 𝑄
𝑚𝑐
= −6570
(425∗4.184)
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= −6570
1778.2
= -3.69 ˚C
The problem tells you that the initial temperature was 75 ˚C. To solve for the final temperature,
∆T = Final Temperature-Initial Temperature
- 3.69 ˚C = Final Temperature – 75 ˚C (Add 75 ˚C to both sides)
71.31 ˚C= Final Temperature
Option B:
Q = mc(Final T- Initial T) Minor manipulation of the equation
Final T- Initial T= 𝑄
𝑚𝑐 Add Initial T to both sides
Final T= 𝑄
𝑚𝑐 + Initial T
= −6570
(425∗4.184) + 75 ˚C
= -3.69 ˚C + 75 ˚C
= 71.31 ˚C
5. Tea or coffee is drank in cold weather because it warms us. The beverage is 40°C and the
body is normally 37.2°C, but warms to 39°C after consumption. If you drink a cup of tea with a
mass of 300g of water, how much heat energy was transferred to the body? The specific heat of
water is 4.184J/g°C.
Q = mc∆T You are solving for Q
Q = 300*4.184*(39-37.2)
= 300*4.184*1.8
= 2,259.4 J or 2.26 kJ of energy was absorbed
6. When testing to see if a solid is pure Titanium, the temperature is raised from 25˚C to 57˚C
after the absorption of 154.2 J of heat. If the mass of the solid is 2.7 g and the specific heat of
Titanium is known to be 0.54 J/g˚C. What is the specific heat of the solid and is it pure
Titanium?
Q = mc∆T You are solving for c
c= 𝑄
𝑚∆𝑇 =
154.2
(2.7∗32) =
154.2
86.4 = 1.78 J/g˚C
Because Titanium has a heat capacity of 0.54 and the solid has a c of 1.78, it is NOT Titanium.
Advanced Chemistry Thermochemistry
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For questions 7-11, use the bond energies provided to classify them as Endothermic or
Exothermic.
7.
Bond Energy per Bond (kJ per mole)
H-H 432
O-O 204
O-H 467
2 H2 + O2 2 H2O
Reactants
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
H2 2 432 864
O2 1 204 204
Total 1,068 kJ
Products
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
OH 4 467 1,868
Total 1,868 kJ
1,068 kJ – 1,868 kJ= -800 kJ = Exothermic
Note: You should recognize this problem from the end of the video. In the video, the reaction
proceeded in the reverse direction. It is important to recognize that if a reaction is endothermic in
one direction, it is always exothermic in the opposite direction.
8.
Bond Energy per Bond (kJ per mole)
C-H 413
N-H 391
C-N 305
H-H 432
CH4 + NH3 CH3NH2 + H2
Reactants
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
CH 1 413 413
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NH 1 391 391
Total 804
Products
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
CN 1 305 305
H2 1 432 432
Total 737
804 kJ- 737 kJ= + 67 kJ = Endothermic
Note: In this example, there are 4 CH bonds, but when reviewing the equation, only 1
participates in the reaction. Hence, you could have accounted for them all (as well as all 3 NH
bonds), but the answer would be the same regardless. You only need to account for bonds that
participate in the reaction.
9.
Bond Energy per Bond (kJ per Bond)
H-Cl 427
S-Cl 271
H-S 347
Cl-Cl 243
2 HCl + SCl2 H2S + 2 Cl2
Reactants
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
HCl 2 427 854
SCl 2 271 542
Total 1396
Products
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
HS 2 347 694
Cl2 2 243 486
Total 1180
1,396 kJ – 1,180 kJ = + 216 kJ = Endothermic
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10.
Bond Energy per Bond (kJ per mole)
N-N 160
H-H 432
N-H 391
N2 + 3 H2 2 NH3
Reactants
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
N2 1 160 160
H2 3 432 1296
Total 1456
Products
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
NH 6 391 2346
Total 2346
1456 kJ – 2346 kJ = -890 kJ = Exothermic
11.
Bond Energy per Bond (kJ per mole)
C-Cl 339
O-H 467
H-Cl 427
C-O 358
CHCl3 + H2O HCl + CHCl2OH
Reactants
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
CCl 1 339 339
OH 1 467 467
Total 806
Products
Bond Number of Bonds/Coefficient Bond Energy (each) Total Energy
HCl 1 427 427
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CO 1 358 358
Total 785
806 kJ – 785 kJ = + 21 kJ = Endothermic
Use the Enthalpy of Formation Table on Page 2 and Hess’ Law to classify each of the reactions
in 12-15 as Endothermic or Exothermic.
12.
Molecule Hformation (kJ)
H2O -286
CO -111
H2 0
CO2 -394
H2O (l) + CO (g) H2 (g) + CO2 (g)
Reactants Products
Molecule Hformation Molecule Hformation
H2O -286 H2 0
CO -111 CO2 -394
Total/Sum -397 Total/Sum -394
∆H = ∑ Hproducts – ∑ Hreactants
= -394- (-397)
= -394 + 397
= +3 kJ = Endothermic
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13.
Molecule Hformation (kJ)
NH3 -368
N2O4 27
N2 0
H2O -242
8 NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g)
Reactants Products
Molecule Hformation Molecule Hformation
NH3 -46 * 8 = -368 N2 0 * 7 = 0
N2O4 +9 * 3 = 27 H2O -242 * 12 = -2,904
Total/Sum -341 Total/Sum -2,904
∆H = ∑ Hproducts – ∑ Hreactants
= -2,904 – (-341)
= -2,904 + 341
= -2,563 kJ = Exothermic
For this question (in comparison to # 12), there are different values for H2O. Remember to make
note of the phase of the molecule. The enthalpy of formation values are different because #12
addresses water in liquid phase whereas #13 addresses water in gas phase.
14.
Molecule Hformation (kJ)
Fe 0
O2 0
Fe2O3 -824
4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)
Reactants Products
Molecule Hformation Molecule Hformation
Fe 0 * 4 = 0 Fe2O3 2 * - 824 = -1,648
O2 0 * 3 = 0
Total/Sum 0 Total/Sum -1,648
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∆H = ∑ Hproducts – ∑ Hreactants
= -1,648 – 0
= -1,648 kJ = Exothermic
15.
Molecule Hformation (kJ)
CaO -635
CO2 -394
CaCO3 -1,207
CaO (s) + CO2 (g) CaCO3 (s)
Reactants Products
Molecule Hformation Molecule Hformation
CaO -635 CaCO3 -1,207
CO2 -394
Total/Sum -1,029 Total/Sum -1,207
∆H = ∑ Hproducts – ∑ Hreactants
= -1,207 – (-1,029)
= -1,207 + 1,029
= - 178 kJ = Exothermic