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Thermochemical equations A typical chemical equation is S + O 2 SO 2 It is called a thermochemical equation when we add information about H … S + O 2 SO 2 H = – 296.9 kJ If we change the equation, then the H also changes … SO 2 S + O 2 H = + 296.9 kJ If the reaction is reversed the sign is reversed Also, if numbers in the equation change, so will the amount of energy produced/absorbed: 2S + 2O 2 2SO 2 H = – 593.8 kJ Read 1st part of handout. Do Q 1-3

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Answers C + O 2 CO 2 393.5 kJ Q2H 2 + I 2 2HI H = + 53.2 kJ Q1C + O 2 CO 2 H = – 393.5 kJ Q31/2H 2 + 1/2I 2 HI H = + 26.6 kJ 2HI 53.2 kJ H 2 + I 2

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Answers pg. 175, Q 5.45-5.50 5.45 - At STP (25°C and 1 atm) 5.46 - The H° value 5.47 - Moles. (You cant have 1/2 an atom) 5.48 - answer in back of book 4Al(s) + Fe 2 O 3 (s) 2Al 2 O 3 (s) + 4Fe(s) H° = -1708 kJ

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Answers pg. 175, Q 5.45-5.50 5.49 - If 2 = 6542 kJ then 1.5 = 4907 kJ (6542/2 x 1.5) Actually, it should be -4907 kJ since it is exothermic 5.50 - 10CaO(s) + 10H 2 O(l) 10Ca(OH) 2 (s) H° = -653 kJ Equation is reversed, thus H° sign changes Equation is multiplied by 10, thus so is H°

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Standard heats of reaction 5.45° indicates 1 atm (and is usually associated with a temperature of 25°C) 5.46An energy term is added (e.g. H or H°) 5.47 Moles 5.48 Al(s) + 1/2Fe 2 O 3 (s) Fe(s) + 1/2Al 2 O 3 (s) H ° = – 426.9 kJ 4Al(s) + 2Fe 2 O 3 (s) 4Fe(s) + 2Al 2 O 3 (s) H ° = – 1708 kJ 5.49 1.5 is 3/4 of 2. 3/4 of 6542 kJ is 4906.5 kJ 5.50 10CaO(s) + 10H 2 O(l) 10Ca(OH) 2 (s) H ° =–653kJ For more lessons, visit www.chalkbored.com www.chalkbored.com

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