Student Worksheet for Thermochemistry

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



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.



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.



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.


C-H 413

N-H 391

C-N 305

H-H 432

CH4 + NH3 CH3NH2 + H2



9.


H-Cl 427

S-Cl 271

H-S 347

Cl-Cl 243

2 HCl + SCl2 H2S + 2 Cl2

10.


N-N 160

H-H 432

N-H 391

N2 + 3 H2 2 NH3

11.


C-Cl 339

O-H 467

H-Cl 427

C-O 358

CHCl3 + H2O HCl + CHCl2OH



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.


NH3 -368

N2O4 27

N2 0

H2O -242

8 NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g)



14.


Fe 0

O2 0

Fe2O3 -824

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)

15.


CaO -635

CO2 -394

CaCO3 -1,207

CaO (s) + CO2 (g) CaCO3 (s)



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.


= 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)



= −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 = 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.



For questions 7-11, use the bond energies provided to classify them as Endothermic or

Exothermic.

7.


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


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.


C-H 413

N-H 391

C-N 305

H-H 432

CH4 + NH3 CH3NH2 + H2

Reactants


CH 1 413 413



NH 1 391 391

Total 804

Products


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


HCl 2 427 854

SCl 2 271 542

Total 1396

Products


HS 2 347 694

Cl2 2 243 486

Total 1180

1,396 kJ – 1,180 kJ = + 216 kJ = Endothermic



10.


N-N 160

H-H 432

N-H 391

N2 + 3 H2 2 NH3

Reactants


N2 1 160 160

H2 3 432 1296

Total 1456

Products


NH 6 391 2346

Total 2346

1456 kJ – 2346 kJ = -890 kJ = Exothermic

11.


C-Cl 339

O-H 467

H-Cl 427

C-O 358

CHCl3 + H2O HCl + CHCl2OH

Reactants


CCl 1 339 339

OH 1 467 467

Total 806

Products


HCl 1 427 427



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.


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



13.


NH3 -368

N2O4 27

N2 0

H2O -242

8 NH3 (g) + 3 N2O4 (g) 7 N2 (g) + 12 H2O (g)

Reactants Products


NH3 -46 * 8 = -368 N2 0 * 7 = 0

N2O4 +9 * 3 = 27 H2O -242 * 12 = -2,904

Total/Sum -341 Total/Sum -2,904


= -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.


Fe 0

O2 0

Fe2O3 -824

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)

Reactants Products


Fe 0 * 4 = 0 Fe2O3 2 * - 824 = -1,648

O2 0 * 3 = 0

Total/Sum 0 Total/Sum -1,648




= -1,648 – 0

= -1,648 kJ = Exothermic

15.


CaO -635

CO2 -394

CaCO3 -1,207

CaO (s) + CO2 (g) CaCO3 (s)

Reactants Products


CaO -635 CaCO3 -1,207

CO2 -394

Total/Sum -1,029 Total/Sum -1,207


= -1,207 – (-1,029)

= -1,207 + 1,029

= - 178 kJ = Exothermic

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Student Worksheet for Thermochemistry