141 Thermochemistry worksheet key PDF

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General Chemistry I (141) Thermochemistry worksheet answers....

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Thermochemistry Worksheet Key 1. How much heat, in calories, is released or absorbed if a 15.0 g sample of copper is converted from (a) the solid to liquid state. 𝑞 = 𝑚 × ∆𝐻𝑓𝑢𝑠 = 𝟏𝟓. 𝟎 𝑔 × 206

𝐽 1 𝑐𝑎𝑙 = 𝟕𝟑𝟖. 5 = 739 𝑐𝑎𝑙 × 𝑔 4.184 𝐽

So 739 cal is absorbed (b) the gas to the liquid state. 𝑞 = 𝑚 × ∆𝐻𝑐𝑜𝑛 = 𝟏𝟓. 𝟎 𝑔 × −4.7

𝑘𝐽 1000 𝐽 1 𝑐𝑎𝑙 × × = −𝟏𝟔, 𝟖49 𝑔 1 𝑘𝐽 4.184 𝐽 = −1.68 × 104 𝑐𝑎𝑙 𝑜𝑟 − 16.8 𝑘𝑐𝑎𝑙

So 16.8 kcal is released. For copper m.p. = 1084 C ; b.p. = 2927 C; ∆Hfus = 206 J/g, ∆Hvap = 4.7 kJ/g 2. How much heat is absorbed or released if a 12.5 g sample of aluminum changes temperature from 25.0 to 87.5 C? The specific heat capacity of aluminum is 0.900 J/g.K Tfinal = 87.5 + 273.15 = 360.65 K

Tinitial = 25.0 + 273.15 = 298.15 K

𝐽 × (360. 𝟔5 − 298. 𝟏5)𝐾 𝑔. 𝐾 𝐽 × 62.5 𝐾 = 703.1 = 703 𝐽 = 12.5 𝑔 × 0.900 𝑔. 𝐾

𝑞 = 𝑚 × 𝑐 × ∆𝑇 = 𝟏𝟐. 𝟓 𝑔 × 0.900

3. What was the mass of a sample of water if the addition of 4.28 x 103 joules raised the temperature of the sample from 22.0 0C to 34.5 0C? Tfinal = 34.5 + 273.15 = 307.65 K T increases so endothermic so q > 0

𝑠𝑜 𝑚 =

𝑞 = 𝑐 × ∆𝑇

Tinitial = 22.0 + 273.15 = 295.15 K 𝑞 = 𝑚 × 𝑐 × ∆𝑇

4.28 × 103 𝐽

𝐽 4.184 𝑔. 𝐾 × (307. 𝟔5 − 295. 𝟏5)𝐾

=

𝟒. 𝟐𝟖 × 103 𝐽 = 𝟖𝟏. 𝟖3 = 81.8 𝑔 𝐽 4.184 𝑔. 𝐾 × 12.5 𝐾

4. Calculate the total heat energy (in kilojoules, kJ) needed to convert 25.0 g of ice at 0.0 °C to steam at 130.0 °C. (For H2O: b.p. = 100.0 °C ; m.p. = 0.0 °C ; ∆Hfus = 333 J/g ; ∆Hvap = 2.26 kJ/g ; c (ice) = 2.108 J/g.K ; c (water) = 4.184 J/g.K ; c (steam) = 1.996 J/g.K).

𝑞𝑡𝑜𝑡𝑎𝑙 = 𝑞𝐴→𝐵 + 𝑞𝐵→𝐶 + 𝑞𝐶→𝐷 + 𝑞𝐷→𝐸 = 𝑚∆𝐻𝑓𝑢𝑠 + 𝑚𝑐∆𝑇 + 𝑚∆𝐻𝑣𝑎𝑝 + 𝑚𝑐∆𝑇

E 130 ºC C

0C A

D 100ºC

0.0 C = 273.15 K; 100.0 C = 373.15 K; 130 C = 400.15 K

B

𝑞𝑡𝑜𝑡𝑎𝑙 = (25.0 𝑔 × 333

1 𝑘𝐽 𝐽 × ) 𝑔 1000 𝐽

1 𝑘𝐽 𝐽 × (373.15 𝐾 − 273.15 𝐾) × ) +(25.0 𝑔 × 4.184 𝑔. 𝐾 1000 𝐽 𝑘𝐽 +(25.0 𝑔 × 2.26 ) 𝑔

𝐽 1 𝑘𝐽 +(25.0 𝑔 × 1.996 × (400.15 𝐾 − 373.15 𝐾 ) × ) 𝑔℃ 1000 𝐽

𝑞𝑡𝑜𝑡𝑎𝑙 = 8. 𝟑𝟐5 𝑘𝐽 + 10. 𝟒6 𝑘𝐽 + 56. 𝟓 𝑘𝐽 + 1. 𝟒𝟗7 𝑘𝐽 = 76. 𝟕82 𝑘𝐽 = 76.8 𝑘𝐽 5. Calculate the total heat energy, q, (in kJ) when 50.0 g of steam at 100.0 °C converts to ice at 0.0 °C. (For H2O: b.p. = 100.0 °C ; m.p. = 0.0 °C ; ∆Hfus = 333 J/g ; ∆Hvap = 2.26 kJ/g ; c (ice) = 2.108 J/g.K ; c (water) = 4.184 J/g.K ; c (steam) = 1.996 J/g.K). B

0C D

A 100ºC

C

𝑞𝑡𝑜𝑡𝑎𝑙 = (50.0 𝑔) (−2.26

𝑞𝑡𝑜𝑡𝑎𝑙 = 𝑞𝐴→𝐵 + 𝑞𝐵→𝐶 + 𝑞𝐶→𝐷 = 𝑚∆𝐻𝑐𝑜𝑛 + 𝑚𝑐∆𝑇 + 𝑚∆𝐻𝑓𝑟𝑒 0.0 C = 273.15 K; 100.0 C = 373.15 K 𝑘𝐽 ) 𝑔

𝐽 1 𝑘𝐽 +(50.0 𝑔) (4.184 ) (273.15 − 373.15)𝐾 × 1000𝐽 𝑔. 𝐾 𝐽 1 𝑘𝐽 +(50.0 𝑔) (−333 ) × 1000𝐽 𝑔

𝑞𝑡𝑜𝑡𝑎𝑙 = −113 𝑘𝐽 + (−20. 𝟗2 𝑘𝐽) + (−16. 𝟔5 𝑘𝐽) = −𝟏𝟓𝟎. 57 𝑘𝐽 = −151 𝑘𝐽

6. A 245 gram sample of a metal at 99.5 C was added to a 114 gram sample of water in a perfect calorimeter. The original temperature in the calorimeter was 23.5 C. The final temperature of the metal-water mixture was 35.6 C. Calculate the specific heat of this metal in joules per gram per Celsius degree. For metal Tinitial = 99.5 +273.15 = 372.65 K For water Tinitial = 23.5 + 273.15 = 296.65 K For metal and water, Tfinal = 35.6 + 273.15 = 308.75 K 𝑞𝑚𝑒𝑡𝑎𝑙 = −𝑞𝑤𝑎𝑡𝑒𝑟 𝑚𝑚𝑒𝑡𝑎𝑙 × 𝑐𝑚𝑒𝑡𝑎𝑙 × ∆𝑇𝑚𝑒𝑡𝑎𝑙 = −𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 × ∆𝑇𝑤𝑎𝑡𝑒𝑟 𝑐𝑚𝑒𝑡𝑎𝑙

𝐽 −114 𝑔 × 4.184 × (308.75 − 296.65)𝐾 −𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 × ∆𝑇𝑤𝑎𝑡𝑒𝑟 𝑔. 𝐾 = = 245 𝑔 × (308.75 − 372.65)𝐾 𝑚𝑚𝑒𝑡𝑎𝑙 × ∆𝑇𝑚𝑒𝑡𝑎𝑙 =

𝐽 𝐽 𝑔. 𝐾 × 12.1 𝐾 = 0. 𝟑𝟔𝟖65 = 0.369 245 𝑔 × −63.9 𝐾 𝑔. 𝐾

−𝟏𝟏𝟒 𝑔 × 4.184

7. Calculate the final temperature, in C, that results from mixing 245 grams of cobalt, specific heat 0.446 J/g.K, at 142 C with 106 grams of water at 24.8 C. For metal Tinitial = 142 +273.15 = 415.15 K For metal and water, Tfinal = X

For water Tinitial = 24.8 + 273.15 = 297.95 K

𝑞𝑚𝑒𝑡𝑎𝑙 = −𝑞𝑤𝑎𝑡𝑒𝑟 𝑚𝑚𝑒𝑡𝑎𝑙 × 𝑐𝑚𝑒𝑡𝑎𝑙 × ∆𝑇𝑚𝑒𝑡𝑎𝑙 = −𝑚𝑤𝑎𝑡𝑒𝑟 × 𝑐𝑤𝑎𝑡𝑒𝑟 × ∆𝑇𝑤𝑎𝑡𝑒𝑟 245 𝑔 × 0.446

𝐽 𝐽 (𝑋 − 415. 𝟏5)𝐾 = −106 𝑔 × 4.184 × (𝑋 − 297. 𝟗5)𝐾 𝑔. 𝐾 𝑔. 𝐾

𝟏𝟎𝟗. 27

𝐽 𝐽 (𝑋 − 41𝟓. 15)𝐾 = −𝟒𝟒𝟑. 5 × (𝑋 − 297. 𝟗5)𝐾 𝐾 𝐾

109.27 𝐽 𝑋 − 𝟒𝟓𝟑63.44 𝐽 = −𝟒𝟒𝟑. 5 𝐽 𝑋 + 𝟏𝟑𝟐140.825 𝐾 𝟏𝟎𝟗. 27 𝑋 + 𝟒𝟒𝟑. 5 𝑋 = 𝟏𝟑𝟐140.825 + 𝟒𝟓𝟑63.44 𝟓𝟓𝟐. 77 𝑋 = 𝟏𝟕𝟕504.265 𝑋=

𝟏𝟕𝟕504.265 = 𝟑𝟐𝟏. 11 𝐾 = 𝟑𝟐𝟏. 11 − 273.15 = 47.96 = 48 ℃ 𝟓𝟓𝟐. 77

8. Consider the thermochemical equation below: C6H12O6 (s) + 6 O2 (g)  6 CO2 (g) + 6 H2O (g) ∆rH = -2803 kJ (a) Is this reaction endothermic or exothermic?

Exothermic (∆rH 0) (b) How much heat will be absorbed or released if 10.0 g of NO is reacted with 10.0 g H2O? Limiting reagent problem: Which produces the least energy? 𝟏𝟎. 𝟎 𝑔 𝑁𝑂 ×

1 𝑚𝑜𝑙 𝑁𝑂 1170 𝑘𝐽 = 𝟗𝟕. 𝟒6 = 97.5 𝑘𝐽 × 4 𝑚𝑜𝑙 𝑁𝑂 30.01 𝑔

𝟏𝟎. 𝟎 𝑔 𝐻2 𝑂 ×

1 𝑚𝑜𝑙 𝐻2 𝑂 1170 𝑘𝐽 × = 𝟏𝟎𝟖. 21 = 108 𝑘𝐽 18.02 𝑔 6 𝑚𝑜𝑙 𝐻2 𝑂

So NO is the limiting reagent and the theoretical yield of energy is 97.5 kJ (c) How much heat energy is required to produce 500. g of NH3? 𝟓𝟎𝟎. 𝑔 𝑁𝐻3 ×

1 𝑚𝑜𝑙 𝑁𝐻3 1170 𝑘𝐽 = 𝟖𝟓𝟖7.7 = 8590 𝑘𝐽 × 4 𝑚𝑜𝑙 𝑁𝐻3 17.03 𝑔

10. When 6.50 g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee cup calorimeter, the temperature rises from 21.6 C to 37.8 C. (a) Is the formation of the sodium hydroxide solution exothermic or endothermic? Why? Exothermic. Temperature of water increases so the water absorbed energy which means the sodium hydroxide released energy as it dissolved. (b) Calculate ΔH (in kJ/mol NaOH) for the solution process. Assume that the specific heat of the solution is the same as that of pure water. 𝑞𝑟 + 𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 0 Tfinal = 37.8 + 273.15 = 310.95 K

Tintial = 21.6 + 273.15 = 294.75 K

𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 𝑐𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × ∆𝑇𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝐽 × (310.95 − 294.75) 𝐾 = (100.0 𝑔 + 6.50 𝑔) × 4.184 𝑔. 𝐾 𝐽 1 𝑘𝐽 = 106.5 𝑔 × 4.184 × 𝟏𝟔. 𝟐 𝐾 = 7218.6 × = 𝟕. 𝟐𝟏86 𝑘𝐽 𝑔. 𝐾 1000𝐽

𝑞𝑟 = −𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = −𝟕. 𝟐𝟏86 𝑘𝐽 𝑀𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 = 𝟔. 𝟓𝟎 𝑔 × 𝑆𝑜 ∆𝐻 =

1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 = 0. 𝟏𝟔𝟐4 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 40.01 𝑔 𝑁𝑎𝑂𝐻

−𝟕. 𝟐𝟏86 𝑘𝐽 𝑘𝐽 = −𝟒𝟒. 𝟒33 = −44.4 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0. 𝟏𝟔𝟐4 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

11. In a coffee-cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions were originally at 24.6 C. After the reaction, the temperature is 31.3 C. What is the enthalpy change for the neutralization of one mole of HCl by NaOH? (Hint: assume that the density of the mixture is 1.00 g/mL, and the specific heat of the mixture is the same as water) 𝑞𝑟 + 𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 0 Tfinal = 31.3 + 273.15 = 304.45 K

Tintial = 24.6 + 273.15 = 297.75 K

𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 𝑐𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × ∆𝑇𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 1.00 𝑔 𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = (100.0 𝑚𝐿 + 100.0 𝑚𝐿) × = 200. 𝑔 𝑚𝐿 𝐽 × (304.45 − 297.75) 𝐾 𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 200. 𝑔 × 4.184 𝑔. 𝐾 𝐽 1 𝑘𝐽 = 200. 𝑔 × 4.184 × 𝟔. 𝟕 𝐾 = 𝟓𝟔06.56 𝐽 × = 5.60656 𝑘𝐽 𝑔. 𝐾 1000𝐽 𝑞𝑟 = −𝑞𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = −𝟓. 𝟔0656 𝑘𝐽

𝑚𝑜𝑙 𝐻𝐶𝑙 1𝐿 × 𝟏. 𝟎 = 0. 𝟏𝟎 𝑚𝑜𝑙 𝐻𝐶𝑙 1000 𝐿 1𝐿 𝑘𝐽 −𝟓. 𝟔0656 𝑘𝐽 = −𝟓𝟔. 0656 = −56 𝑆𝑜 ∆𝑟 𝐻 = 0. 𝟏𝟎 𝑚𝑜𝑙 𝐻𝐶𝑙 𝑚𝑜𝑙 𝐻𝐶𝑙 Notice rH < 0 as reaction is exothermic (temperature of solution increases) 𝑀𝑜𝑙𝑒𝑠 𝐻𝐶𝑙 = 100.0 𝑚𝐿 ×

12. In a coffee-cup calorimeter, 100.0 mL of distilled water, originally at 22.8 C, is combined with 13.95 g of ice at 0.00 C. After the ice has melted, the temperature is 12.5 C. What is the heat capacity of the coffee cup? Assume the density of water is 1.00 g/mL. Note: c (ice) = 2.108 J/g.K ; c (water) = 4.184 J/g.K ; c (steam) = 1.996 J/g.K; ∆Hfus = 333 J/g ; ∆Hvap = 2.26 kJ/g Notice water and cup lose energy, ice gains energy (first to melt at 0.0 C then to warm up to 12.5 C) so: 𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑐𝑢𝑝 + 𝑞𝑓𝑢𝑠𝑖𝑜𝑛,𝑖𝑐𝑒 + 𝑞𝑤𝑎𝑟𝑚𝑖𝑛𝑔,𝑖𝑐𝑒 = 0 For ice, water and cup, Tfinal = 12.5 + 273.15 = 285.65 K For ice, Tinitial = 0.00 + 273.15 = 273.15 K For water, and cup, Tinitial = 22.8 + 273.15 = 295.95 K 𝑔 𝐽 × (285.65 − 295.95)𝐾 𝑞𝑤𝑎𝑡𝑒𝑟 = 𝑚 𝑐 ∆𝑇 = (100.0 𝑚𝐿 × 1.00 ) × 4.184 𝑔. 𝐾 𝑚𝐿 𝐽 × −𝟏𝟎. 𝟑 𝐾 = −𝟒𝟑𝟎9.5 𝐽 = 100. 𝑔 × 4.184 𝑔. 𝐾 𝐽 = 𝟒𝟔𝟒5.35 𝐽 𝑞𝑓𝑢𝑠𝑖𝑜𝑛,𝑖𝑐𝑒 = 𝑚 ∆𝐻𝑓𝑢𝑠 = 13.95 𝑔 × 333 𝑔 𝐽 × (285.65 − 273.15) 𝐾 𝑞𝑤𝑎𝑟𝑚𝑖𝑛𝑔,𝑖𝑐𝑒 = 𝑚 𝑐 ∆𝑇 = 13.95 𝑔 × 4.184 𝑔. 𝐾 𝐽 × 𝟏𝟐. 𝟓 𝐾 = 𝟕𝟐𝟗. 585 𝐽 = 13.95 𝑔 × 4.184 𝑔. 𝐾

𝑞𝑐𝑢𝑝 = 𝑚 𝑐 ∆𝑇 = 𝑚𝑐𝑢𝑝 × 𝑐𝑐𝑢𝑝 × (285.65 − 295.95) 𝐾 = 𝑚𝑐𝑢𝑝 × 𝑐𝑐𝑢𝑝 × −𝟏𝟎. 𝟑 𝐾 𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑐𝑢𝑝 + 𝑞𝑓𝑢𝑠𝑖𝑜𝑛,𝑖𝑐𝑒 + 𝑞𝑤𝑎𝑟𝑚𝑖𝑛𝑔,𝑖𝑐𝑒 = 0

−𝟒𝟑𝟎9.5 𝐽 + 4645.35 𝐽 + 𝟕𝟐𝟗. 585 𝐽 + (𝑚𝑐𝑢𝑝 × 𝑐𝑐𝑢𝑝 × −𝟏𝟎. 𝟑 𝐾) = 0 𝑚𝑐𝑢𝑝 × 𝑐𝑐𝑢𝑝 =

𝟒𝟑𝟎9.5 𝐽 − 𝟒𝟔𝟒5.35 𝐽 − 𝟕𝟐𝟗. 585 𝐽 𝐽 = 𝟏𝟎𝟑. 44 = 103 𝐾 −𝟏𝟎. 𝟑 𝐾

13. When a 7.68 g sample of sulfur (S8) is burned in a bomb calorimeter with excess oxygen, the temperature increases from 20.34 C to 36.75 C. The calorimeter contains 815 g of water, and the heat capacity of the bomb is 923 J/K. Calculate ΔU for the reaction in kJ/mol S8. 𝑞𝑟 + 𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑏𝑜𝑚𝑏 = 0 For water and bomb, Tfinal = 36.75 + 273.15 = 309.90 K Tinitial = 20.34 + 273.15 = 293.49 K 𝐽 × (309.90 − 293.49) 𝐾 𝑞𝑤𝑎𝑡𝑒𝑟 = 815 𝑔 × 4.184 𝑔. 𝐾 1 𝑘𝐽 𝐽 × 16.41 𝐾 = 𝟓𝟓𝟗57.4 × = 𝟓𝟓. 𝟗574 𝑘𝐽 = 815 𝑔 × 4.184 𝑔. 𝐾 1000𝐽 𝐽 𝐽 1 𝑘𝐽 = 𝟏𝟓. 𝟏464 𝑘𝐽 𝑞𝑏𝑜𝑚𝑏 = 923 × (309.90 − 293.49) 𝐾 = 923 × 16.41 𝐾 = 𝟏𝟓𝟏46.4 × 1000𝐽 𝐾 𝐾 𝑞𝑟 + 𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑏𝑜𝑚𝑏 = 0

𝑞𝑟 = −(𝑞𝑤𝑎𝑡𝑒𝑟 + 𝑞𝑏𝑜𝑚𝑏 ) = −( 𝟓𝟓. 𝟗574 𝑘𝐽 + 𝟏𝟓. 𝟏464 𝑘𝐽) = −𝟕𝟏. 𝟏038 𝑘𝐽

𝑀𝑜𝑙𝑒𝑠 𝑠𝑢𝑙𝑓𝑢𝑟 = 𝟕. 𝟔𝟖 𝑔 × 𝑆𝑜 ∆𝑈 =

1 𝑚𝑜𝑙 𝑆8 = 0.0𝟐𝟗𝟗4 𝑚𝑜𝑙 𝑆8 256.5 𝑔 𝑆8

−𝟕𝟏. 𝟏038 𝑘𝐽 𝑘𝐽 = −𝟐𝟑𝟕4.8 = −2370 0.0𝟐𝟗𝟗4 𝑚𝑜𝑙 𝑆8 𝑚𝑜𝑙 𝑆8

14. Calculate the ∆rH0 for the following reaction: 2 H2O2 (l)  2 H2O (g)

+ O2 (g)

You are given these two equations:

Start with

2 H2 (g) + O2 (g)  2 H2O (l)

∆rH0 = -572 kJ

H2 (g) + O2 (g)

 H2O2 (l)

∆rH0 = -188 kJ

H2 (g) + O2 (g)

 H2O2 (l)

∆rH0 = -188 kJ

Need H2O2 as reactant so reverse it: H2O2 (l)

 H2 (g) + O2 (g) ∆rH0 = +188 kJ

Then multiply by 2 to get 2 H2O2 2 H2O2 (l)

 2 H2 (g) + 2 O2 (g)

∆rH0 = +(2 x 188) kJ = +376 kJ

Eqn 1

Can now combine Eqn 1 with the 2 H2 + O2 → 2 H2O reaction, cancelling terms that appear on both sides and adding together the ∆rH0: 2 H2O2 (l)

 2 H2 (g) + 2 O2 (g)

∆rH0 = +376 kJ

2 H2 (g) + O2 (g)  2 H2O (l)

∆rH0 = -572 kJ

2 H2O2 (l)

∆rH0 = +376 kJ + (-572 kJ) = -196 kJ

 2 H2O (l) + O2 (g)

15. Calculate the ∆rH0 for the following reaction: NO (g) + O (g)  NO2 (g) You are given these three equations:

Start with

O2 (g)  2 O (g)

∆rH0 = +498.4 kJ

2 O3 (g)  3 O2 (g)

∆rH0 = -285.3 kJ

NO (g) + O3 (g)  NO2 (g) + O2 (g)

∆rH0 = -199.0 kJ

NO (g) + O3 (g)  NO2 (g) + O2 (g)

∆rH0 = -199.0 kJ

Eqn 1

Reverse the second equation: 3 O2 (g)  2 O3 (g)

∆rH0 = +285.3 kJ

Now adjust the coefficients to produce just one O3 by dividing by 2 (remember to divide ∆rH0 as well) 3/2 O2 (g)  O3 (g)

∆rH0 = +142.7 kJ

Eqn 2

Next add Eqn 1 and Eqn 2 cancelling terms that appear on both sides and adding together the ∆rH0. NO (g) + ½ O2 (g)  NO2 (g)

∆rH0 = -199.0 kJ + 142.7 kJ = -56.3 kJ Eqn 3

Now take the O2 → 2 O reaction, reverse it and divide by 2: O (g)  ½ O2 (g)

∆rH0 = -249.2 kJ

Add this to Eqn 3 and cancelling terms gives NO (g) + O (g) → NO2 (g) 16. Write the formation reactions for the following: (a) H2O2 (l) H2 (g) + O2 (g) → H2O2 (l) (b) Potassium sulfate 2 K (s) + S (s) + 2 O2 (g) → K2SO4 (s) (c) Lithium nitrate Li (s) + ½ N2 (g) + 3/2 O2 (g) → LiNO3 (s) (d) Liquid Glycerol (C3H8O3) 3 C (s) + 4 H2 (g) + 3/2 O2 (g) → C3H8O3 (l) (e) Ammonia ½ N2 (g) + 3/2 H2 (g) → NH3 (g) (f) N2O5 (a solid) N2 (g) + 5/2 O2 (g) → N2O5 (s)

∆rH0 = -249.2 kJ + (-56.3) kJ = -305.5 kJ

17. Calculate the ∆rH0 from the ∆fH0 (use Appendix L of textbook) for the following reactions: (a)

PbO (s) + CO (g)

-------> Pb (s)

+ CO2 (g)

check reaction balanced (yes) Look up ∆rH0 PbO (s)

∆rH0 = -219 kJ/mol

CO (g)

∆rH0 = -110.5 kJ/mol

Pb (s)

∆rH0 = 0

CO2 (g)

∆rH0 =-393.5 kJ/mol

∆rH0 = ∑n∆fH0(products) - ∑n∆fH0(reactants)

∆𝑓 𝐻0 = [(

1 𝑚𝑜𝑙 𝑃𝑏 𝑘𝐽 𝑘𝐽 1 𝑚𝑜𝑙 𝐶𝑂2 ×0 × −393.5 )+( )] 1 𝑚𝑜𝑙 𝑟𝑥𝑛 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 𝑚𝑜𝑙 − [(

∆𝑓 𝐻0 = −395.5

1 𝑚𝑜𝑙 𝑃𝑏𝑂 𝑘𝐽 1 𝑚𝑜𝑙 𝐶𝑂 𝑘𝐽 )] × −219 )+( × −110.5 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑟𝑥𝑛

𝑘𝐽 𝑘𝐽 𝑘𝐽 𝑘𝐽 + 219 + 110.5 = −66 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙

(b) SiH4 (g) + 2 O2 (g)

-------> SiO2 (s) + 2 H2O (l)

check reaction balanced (yes) Look up ∆rH0 SiH4 (g)

∆rH0 = 34.41 kJ/mol

O2 (g)

∆rH0 = 0 kJ/mol

SiO2 (s)

∆rH0 = -910.86 kJ/mol

H2O (l)

∆rH0 = -285.83 kJ/mol

∆rH0 = ∑n∆fH0(products) - ∑n∆fH0(reactants)

∆𝑓 𝐻0 = [(

1 𝑚𝑜𝑙 𝑆𝑖𝑂2 𝑘𝐽 𝑘𝐽 2 𝑚𝑜𝑙 𝐻2 𝑂 )+( × −910.86 × −285.83 )] 1 𝑚𝑜𝑙 𝑟𝑥𝑛 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 𝑚𝑜𝑙 − [(

∆𝑓 𝐻0 = (−910.86

1 𝑚𝑜𝑙 𝑆𝑖𝐻4 𝑘𝐽 𝑘𝐽 2 𝑚𝑜𝑙 𝑂2 )] × 34.31 ×0 )+( 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑟𝑥𝑛 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙

𝑘𝐽 𝑘𝐽 𝑘𝐽 ) = −1516.83 𝑘𝐽/𝑚𝑜𝑙 − 571.66 ) − (34.31 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙

(c) 4 FeS2 (s) + 11 O2 (g)

-------> 2 Fe2O3 (s) + 8 SO2 (g)

check reaction balanced (yes) Look up ∆rH0 FeS2 (s)

∆rH0 = -178.2 kJ/mol

O2 (g)

∆rH0 = 0 kJ/mol

Fe2O3 (s)

∆rH0 = -825.5 kJ/mol

SO2 (g)

∆rH0 = -296.84 kJ/mol

∆rH0 = ∑n∆fH0(products) - ∑n∆fH0(reactants)

∆𝑓 𝐻0 = [(

2 𝑚𝑜𝑙 𝐹𝑒2 𝑂3 𝑘𝐽 𝑘𝐽 8 𝑚𝑜𝑙 𝑆𝑂2 × −825.5 × −296.84 )+( )] 1 𝑚𝑜𝑙 𝑟𝑥𝑛 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙 𝑚𝑜𝑙 − [(

∆𝑓 𝐻0 = (−1651

4 𝑚𝑜𝑙 𝐹𝑒𝑆2 𝑘𝐽 𝑘𝐽 11 𝑚𝑜𝑙 𝑂2 )] × −178.2 ×0 )+( 𝑚𝑜𝑙 1 𝑚𝑜𝑙 𝑟𝑥𝑛 1 𝑚𝑜𝑙 𝑟𝑥𝑛 𝑚𝑜𝑙

𝑘𝐽 𝑘𝐽 𝑘𝐽 ) = −331𝟐. 9 = −3313 𝑘𝐽/𝑚𝑜𝑙 − 237𝟒. 7 ) + (712.8 𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙...