Chemistry HP Unit 10 Exam Th
Name __________________________________________________ Period ________ Date ______________________ Chemistry HP Unit 10 Exam Thermochemistry – Version 1
Total Points: _______/100
Refer to the following graph for Problem 1.
1.
What is the boiling point of the compound in the above heating curve?
A
U
90℃ 2.
What is the final temperature when 2000. J of energy are used to heat 150 g of glass at 150 ºC.? ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 = 𝑇𝑓 =
3.
A
U
W
S
A
U
W
S
A
U
W
S
∆𝐻 𝑚𝐶
∆𝐻 2000. 𝐽 + 𝑇𝑖 = + 150 ℃ = 170 ℃ 𝐽 𝑚𝐶 (150 𝑔) (0.840 ) 𝑔℃
Calculate the specific heat capacity of a substance if 390 J of heat energy raises the temperature of 25.0 g of the substance from 100.0 °C to 140.0 °C. 𝐶=
∆𝐻 390 𝐽 𝐽 = = 0.39 𝑚∆𝑇 (25.0 𝑔)(40.0 ℃) 𝑔℃
4. How much energy is required to turn 10.0 g of ice from – 30.0 ᵒC into steam at 140.0 ᵒC? ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.09
∆𝐻 = 𝐻𝑓𝑢𝑠 𝑚 = (334 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (4.18
𝐽 ) (0℃ − −30.0℃) = 627 𝐽 𝑔℃ 𝐽 ) (10.0 𝑔) = 3340 𝐽 𝑔
𝐽 ) (100.0℃ − 0.0℃) = 4180 𝐽 𝑔℃
∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 = (2256 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.00
𝐽 ) (10.0 𝑔) = 22560 𝐽 𝑔
𝐽 ) (140.0℃ − 100.0℃) = 800. 𝐽 𝑔℃
∆𝐻𝑡𝑜𝑡𝑎𝑙 = 627 𝐽 + 3340 𝐽 + 4180 𝐽 + 22560 𝐽 + 800. 𝐽 = 31500 𝐽
Refer to the following graph for Problems 5-7.
5. What is the activation energy for the forward catalyzed reaction? (Use the appropriate sign).
A
U
A
U
25 𝑘𝐽 6. What is the ∆𝐻 for the reverse reaction? (Use the appropriate sign). - 15 kJ A
7. Is the forward reaction endothermic or exothermic? How do you know? Endothermic, delta H is positive
8.
What mass of acetic acid (HC2H3O2) can be condensed with 5.5 x 103 J of heat energy?
A
U
W
S
A
U
W
S
∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 𝑚=
∆𝐻 5.5 × 103 𝐽 = = 14 𝑔 𝐽 𝐻𝑣𝑎𝑝 395 𝑔
9. A 5.40 g sample of ethane (C2H6) is combusted in a calorimeter filled with 950 g of water. The temperature of water in the calorimeter increases from to 10.0 ˚C to 78.0 ˚C. Determine the heat of combustion of ethane in kJ/mol. Use appropriate signs.
∆𝐻𝑤𝑎𝑡𝑒𝑟 = 𝑚𝐶∆𝑇 = (950 𝑔) (4.18
𝐽 ) (68.0 ℃) = 270 𝑘𝐽 𝑔℃
∆𝐻𝑟𝑥𝑛 = −∆𝐻𝑤𝑎𝑡𝑒𝑟 = −270 𝑘𝐽 ? 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑒 = 5.40 𝑔 𝐶2 𝐻6 ×
1 𝑚𝑜𝑙 = 0.180 𝑚𝑜𝑙 30.068 𝑔
𝑘𝐽 −270 𝑘𝐽 𝑘𝐽 = = −1500 𝑚𝑜𝑙 0.180 𝑚𝑜𝑙 𝑚𝑜𝑙
10. A 20.0 g sample of metal is heated from 30.0 °C to 45.0 °C using 72.0 J of heat energy. Determine the specific heat capacity of the metal. 𝐶=
Lewis Structures:
H2O
U
W
S
NH3
O2
NO2
H2O
∆𝐻 72.0 𝐽 𝐽 = = 0.240 𝑚∆𝑇 (20.0 𝑔)(15.0 ℃) 𝑔℃
11. Draw the following Lewis dot structures:
NH3
A
O2
NO2
12. Use the bond energies to calculate ∆H for the following reaction (Use appropriate sign).
A
U
W
S
A
U
W
S
A
U
W
S
A
U
W
S
4 NH3(g) + 7 O2 (g) ---> 4 NO2 (g) + 6H2O (g) 4 [3 (391
𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 )] + 7 [498 + 607 )] ] − 4 [201 ] − 6 [2 (464 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 = −622
𝑘𝐽 𝑚𝑜𝑙
13. Use the steps provided to calculate the overall heat of the reaction (use appropriate sign). 3Fe2O3 (s) + CO (g) 2Fe3O4 (s) + CO2 (g) Steps: Fe2O3 (s) + 3CO (g) 2Fe(s) + 3CO2 (g)
∆𝐻 = −23.44 𝑘𝐽
Fe3O4 + CO (g) 3FeO (s) + CO2 (g)
∆𝐻 = 21.79 𝑘𝐽
Fe (s) + CO2 (g) FeO (s) + CO (g)
∆𝐻 = −10.94 𝑘𝐽
3 Fe2O3 (s) + 9 CO (g) 6 Fe(s) + 9 CO2 (g)
∆𝐻 = 3(−23.44 𝑘𝐽)
6 FeO (s) + 2 CO2 (g) 2 Fe3O4 + 2 CO (g)
∆𝐻 = −2(21.79 𝑘𝐽)
6 Fe(s) + 6 CO2 (g) 6 FeO (s) + 6 CO (g)
∆𝐻 = 6(−10.94 𝑘𝐽)
3Fe2O3 (s) + CO (g) 2 Fe3O4 (s) + CO2 (g)
∆𝐻 = −179.54 𝑘𝐽
14. Use the heats of formation to calculate the heat of the reaction. (Use the appropriate sign) 2 C2H5OH (l) + 7 O2 (g) ---> 4 CO2 (g) + 6 H2O (l) ∆𝐻 = 4 (−393.5
𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 ) + 6 (−285.8 ) − 2 (−235.2 ) − 7(0) = −2818.4 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙
15. In the following reaction, what mass of O2 would produce 10.0 kJ of heat? CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l); H = ‒890.5 kJ/mol ? 𝑔 𝑂2 = −10.0 𝑘𝐽 ×
2 𝑚𝑜𝑙 𝑂2 32.00 𝑔 × = 0.719 𝑔 −890.5 𝑘𝐽 1 𝑚𝑜𝑙
Total Points: _______/100
Refer to the following graph for Problem 1.
1.
What is the boiling point of the compound in the above heating curve?
A
U
90℃ 2.
What is the final temperature when 2000. J of energy are used to heat 150 g of glass at 150 ºC.? ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 = 𝑇𝑓 =
3.
A
U
W
S
A
U
W
S
A
U
W
S
∆𝐻 𝑚𝐶
∆𝐻 2000. 𝐽 + 𝑇𝑖 = + 150 ℃ = 170 ℃ 𝐽 𝑚𝐶 (150 𝑔) (0.840 ) 𝑔℃
Calculate the specific heat capacity of a substance if 390 J of heat energy raises the temperature of 25.0 g of the substance from 100.0 °C to 140.0 °C. 𝐶=
∆𝐻 390 𝐽 𝐽 = = 0.39 𝑚∆𝑇 (25.0 𝑔)(40.0 ℃) 𝑔℃
4. How much energy is required to turn 10.0 g of ice from – 30.0 ᵒC into steam at 140.0 ᵒC? ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.09
∆𝐻 = 𝐻𝑓𝑢𝑠 𝑚 = (334 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (4.18
𝐽 ) (0℃ − −30.0℃) = 627 𝐽 𝑔℃ 𝐽 ) (10.0 𝑔) = 3340 𝐽 𝑔
𝐽 ) (100.0℃ − 0.0℃) = 4180 𝐽 𝑔℃
∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 = (2256 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.00
𝐽 ) (10.0 𝑔) = 22560 𝐽 𝑔
𝐽 ) (140.0℃ − 100.0℃) = 800. 𝐽 𝑔℃
∆𝐻𝑡𝑜𝑡𝑎𝑙 = 627 𝐽 + 3340 𝐽 + 4180 𝐽 + 22560 𝐽 + 800. 𝐽 = 31500 𝐽
Refer to the following graph for Problems 5-7.
5. What is the activation energy for the forward catalyzed reaction? (Use the appropriate sign).
A
U
A
U
25 𝑘𝐽 6. What is the ∆𝐻 for the reverse reaction? (Use the appropriate sign). - 15 kJ A
7. Is the forward reaction endothermic or exothermic? How do you know? Endothermic, delta H is positive
8.
What mass of acetic acid (HC2H3O2) can be condensed with 5.5 x 103 J of heat energy?
A
U
W
S
A
U
W
S
∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 𝑚=
∆𝐻 5.5 × 103 𝐽 = = 14 𝑔 𝐽 𝐻𝑣𝑎𝑝 395 𝑔
9. A 5.40 g sample of ethane (C2H6) is combusted in a calorimeter filled with 950 g of water. The temperature of water in the calorimeter increases from to 10.0 ˚C to 78.0 ˚C. Determine the heat of combustion of ethane in kJ/mol. Use appropriate signs.
∆𝐻𝑤𝑎𝑡𝑒𝑟 = 𝑚𝐶∆𝑇 = (950 𝑔) (4.18
𝐽 ) (68.0 ℃) = 270 𝑘𝐽 𝑔℃
∆𝐻𝑟𝑥𝑛 = −∆𝐻𝑤𝑎𝑡𝑒𝑟 = −270 𝑘𝐽 ? 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑒 = 5.40 𝑔 𝐶2 𝐻6 ×
1 𝑚𝑜𝑙 = 0.180 𝑚𝑜𝑙 30.068 𝑔
𝑘𝐽 −270 𝑘𝐽 𝑘𝐽 = = −1500 𝑚𝑜𝑙 0.180 𝑚𝑜𝑙 𝑚𝑜𝑙
10. A 20.0 g sample of metal is heated from 30.0 °C to 45.0 °C using 72.0 J of heat energy. Determine the specific heat capacity of the metal. 𝐶=
Lewis Structures:
H2O
U
W
S
NH3
O2
NO2
H2O
∆𝐻 72.0 𝐽 𝐽 = = 0.240 𝑚∆𝑇 (20.0 𝑔)(15.0 ℃) 𝑔℃
11. Draw the following Lewis dot structures:
NH3
A
O2
NO2
12. Use the bond energies to calculate ∆H for the following reaction (Use appropriate sign).
A
U
W
S
A
U
W
S
A
U
W
S
A
U
W
S
4 NH3(g) + 7 O2 (g) ---> 4 NO2 (g) + 6H2O (g) 4 [3 (391
𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 )] + 7 [498 + 607 )] ] − 4 [201 ] − 6 [2 (464 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 = −622
𝑘𝐽 𝑚𝑜𝑙
13. Use the steps provided to calculate the overall heat of the reaction (use appropriate sign). 3Fe2O3 (s) + CO (g) 2Fe3O4 (s) + CO2 (g) Steps: Fe2O3 (s) + 3CO (g) 2Fe(s) + 3CO2 (g)
∆𝐻 = −23.44 𝑘𝐽
Fe3O4 + CO (g) 3FeO (s) + CO2 (g)
∆𝐻 = 21.79 𝑘𝐽
Fe (s) + CO2 (g) FeO (s) + CO (g)
∆𝐻 = −10.94 𝑘𝐽
3 Fe2O3 (s) + 9 CO (g) 6 Fe(s) + 9 CO2 (g)
∆𝐻 = 3(−23.44 𝑘𝐽)
6 FeO (s) + 2 CO2 (g) 2 Fe3O4 + 2 CO (g)
∆𝐻 = −2(21.79 𝑘𝐽)
6 Fe(s) + 6 CO2 (g) 6 FeO (s) + 6 CO (g)
∆𝐻 = 6(−10.94 𝑘𝐽)
3Fe2O3 (s) + CO (g) 2 Fe3O4 (s) + CO2 (g)
∆𝐻 = −179.54 𝑘𝐽
14. Use the heats of formation to calculate the heat of the reaction. (Use the appropriate sign) 2 C2H5OH (l) + 7 O2 (g) ---> 4 CO2 (g) + 6 H2O (l) ∆𝐻 = 4 (−393.5
𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 ) + 6 (−285.8 ) − 2 (−235.2 ) − 7(0) = −2818.4 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙
15. In the following reaction, what mass of O2 would produce 10.0 kJ of heat? CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l); H = ‒890.5 kJ/mol ? 𝑔 𝑂2 = −10.0 𝑘𝐽 ×
2 𝑚𝑜𝑙 𝑂2 32.00 𝑔 × = 0.719 𝑔 −890.5 𝑘𝐽 1 𝑚𝑜𝑙
