Unit 10 Exam v2 (KEY)
Name __________________________________________________ Period ________ Date ______________________ Chemistry HP Unit 10 Exam Thermochemistry – Version 2 1.
What is the final temperature when 1800 J of energy are used to heat 150 g of glass at 150 ºC.? ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 = 𝑇𝑓 =
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∆𝐻 𝑚𝐶
∆𝐻 1800 𝐽 + 𝑇𝑖 = + 150 ℃ = 160 ℃ 𝐽 𝑚𝐶 (150 𝑔) (0.840 ) 𝑔℃
What mass of acetic acid (HC2H3O2) can be condensed with 5.5 x 103 J of heat energy? ∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 𝑚=
∆𝐻 5.5 × 103 𝐽 = = 14 𝑔 𝐽 𝐻𝑣𝑎𝑝 395 𝑔
3. A 2.30 g sample of ethane (C2H6) is combusted in a calorimeter filled with 1000. 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 = (1000. 𝑔) (4.18
𝐽 ) (68.0 ℃) = 284000 𝐽 𝑔℃
∆𝐻𝑟𝑥𝑛 = −∆𝐻𝑤𝑎𝑡𝑒𝑟 = −284000 𝐽 = −284 𝑘𝐽 ? 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑒 = 2.30 𝑔 𝐶2 𝐻6 ×
1 𝑚𝑜𝑙 = 0.0765 𝑚𝑜𝑙 30.068 𝑔
𝑘𝐽 −284 𝑘𝐽 𝑘𝐽 = = −3710 𝑚𝑜𝑙 0.0765 𝑚𝑜𝑙 𝑚𝑜𝑙
4.
Calculate the specific heat capacity of a substance if 390J 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 (25.0 𝑚∆𝑇 𝑔)(40.0 ℃) 𝑔℃
5. How much energy is required to turn 10.0 g of ice from – 20.0 ᵒC into steam at 150.0 ᵒC? ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.09
∆𝐻 = 𝐻𝑓𝑢𝑠 𝑚 = (334 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (4.18
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𝐽 ) (0℃ − −20.0℃) = 418 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
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𝐽 ) (10.0 𝑔) = 22560 𝐽 𝑔
𝐽 ) (150.0℃ − 100.0℃) = 1.00 × 103 𝐽 𝑔℃
∆𝐻𝑡𝑜𝑡𝑎𝑙 = 418 𝐽 + 3340 𝐽 + 4180 𝐽 + 22560 𝐽 + 1.00 × 103 𝐽 = 31500 𝐽
Refer to the following graph for Problem 6.
6.
What is the boiling point of the compound in the above heating curve? 90 ℃
7.
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=
∆𝐻 72.0 𝐽 𝐽 = = 0.240 (20.0 𝑚∆𝑇 𝑔)(15.0 ℃) 𝑔℃
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8. Use the heats of formation to calculate the heat of the reaction. (Use the appropriate sign)
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NH3
O2
NO2
H2O
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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
9. Draw the following Lewis dot structures: Lewis Structures: NH3
O2
NO2
H2O
10. Use the bond energies to calculate ∆H for the following reaction (Use appropriate sign). 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
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13. What is the activation energy for the reverse catalyzed reaction? (Use the appropriate sign). 10 𝑘𝐽
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14. What is the ∆𝐻 for the forward reaction? (Use the appropriate sign). 15 kJ
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11. 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
12. In the following reaction, what mass of O2 would produce 25.0 kJ of heat? CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l); H = ‒890.5 kJ/mol ? 𝑔 𝑂2 = −25.0 𝑘𝐽 ×
2 𝑚𝑜𝑙 𝑂2 32.00 𝑔 × = 1.80 𝑔 −890.5 𝑘𝐽 1 𝑚𝑜𝑙
Refer to the following graph for Problems 13-15.
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15. Is the reverse reaction endothermic or exothermic? How do you know? Exothermic, delta H is negative.
What is the final temperature when 1800 J of energy are used to heat 150 g of glass at 150 ºC.? ∆𝑇 = 𝑇𝑓 − 𝑇𝑖 = 𝑇𝑓 =
2.
Total Points: _______/100 A
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∆𝐻 𝑚𝐶
∆𝐻 1800 𝐽 + 𝑇𝑖 = + 150 ℃ = 160 ℃ 𝐽 𝑚𝐶 (150 𝑔) (0.840 ) 𝑔℃
What mass of acetic acid (HC2H3O2) can be condensed with 5.5 x 103 J of heat energy? ∆𝐻 = 𝐻𝑣𝑎𝑝 𝑚 𝑚=
∆𝐻 5.5 × 103 𝐽 = = 14 𝑔 𝐽 𝐻𝑣𝑎𝑝 395 𝑔
3. A 2.30 g sample of ethane (C2H6) is combusted in a calorimeter filled with 1000. 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 = (1000. 𝑔) (4.18
𝐽 ) (68.0 ℃) = 284000 𝐽 𝑔℃
∆𝐻𝑟𝑥𝑛 = −∆𝐻𝑤𝑎𝑡𝑒𝑟 = −284000 𝐽 = −284 𝑘𝐽 ? 𝑚𝑜𝑙 𝑒𝑡ℎ𝑎𝑛𝑒 = 2.30 𝑔 𝐶2 𝐻6 ×
1 𝑚𝑜𝑙 = 0.0765 𝑚𝑜𝑙 30.068 𝑔
𝑘𝐽 −284 𝑘𝐽 𝑘𝐽 = = −3710 𝑚𝑜𝑙 0.0765 𝑚𝑜𝑙 𝑚𝑜𝑙
4.
Calculate the specific heat capacity of a substance if 390J 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 (25.0 𝑚∆𝑇 𝑔)(40.0 ℃) 𝑔℃
5. How much energy is required to turn 10.0 g of ice from – 20.0 ᵒC into steam at 150.0 ᵒC? ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (2.09
∆𝐻 = 𝐻𝑓𝑢𝑠 𝑚 = (334 ∆𝐻 = 𝑚𝐶∆𝑇 = (10.0 𝑔) (4.18
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𝐽 ) (0℃ − −20.0℃) = 418 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
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𝐽 ) (10.0 𝑔) = 22560 𝐽 𝑔
𝐽 ) (150.0℃ − 100.0℃) = 1.00 × 103 𝐽 𝑔℃
∆𝐻𝑡𝑜𝑡𝑎𝑙 = 418 𝐽 + 3340 𝐽 + 4180 𝐽 + 22560 𝐽 + 1.00 × 103 𝐽 = 31500 𝐽
Refer to the following graph for Problem 6.
6.
What is the boiling point of the compound in the above heating curve? 90 ℃
7.
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=
∆𝐻 72.0 𝐽 𝐽 = = 0.240 (20.0 𝑚∆𝑇 𝑔)(15.0 ℃) 𝑔℃
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8. Use the heats of formation to calculate the heat of the reaction. (Use the appropriate sign)
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NH3
O2
NO2
H2O
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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
9. Draw the following Lewis dot structures: Lewis Structures: NH3
O2
NO2
H2O
10. Use the bond energies to calculate ∆H for the following reaction (Use appropriate sign). 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
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13. What is the activation energy for the reverse catalyzed reaction? (Use the appropriate sign). 10 𝑘𝐽
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14. What is the ∆𝐻 for the forward reaction? (Use the appropriate sign). 15 kJ
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11. 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
12. In the following reaction, what mass of O2 would produce 25.0 kJ of heat? CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l); H = ‒890.5 kJ/mol ? 𝑔 𝑂2 = −25.0 𝑘𝐽 ×
2 𝑚𝑜𝑙 𝑂2 32.00 𝑔 × = 1.80 𝑔 −890.5 𝑘𝐽 1 𝑚𝑜𝑙
Refer to the following graph for Problems 13-15.
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15. Is the reverse reaction endothermic or exothermic? How do you know? Exothermic, delta H is negative.
