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Thermochemical equations

Thermochemical equations

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Thermochemical equations

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  1. Thermochemical equations • A typical chemical equation is S + O2 SO2 • It is called a “thermochemical equation” when we add information about H … S + O2 SO2H = – 296.9 kJ • If we change the equation, then the H also changes … SO2 S + O2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 + 2O2 2SO2H = – 593.8 kJ • Read 1st part of handout. Do Q 1-3

  2. C + O2 393.5 kJ CO2 2HI H2 + I2 53.2 kJ Answers Q1 C + O2 CO2H = – 393.5 kJ Q2 H2 + I2 2HIH = +53.2 kJ Q3 1/2H2 + 1/2I2 HIH = +26.6 kJ

  3. 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 can’t have 1/2 an atom) 5.48 - answer in back of book 4Al(s) + Fe2O3(s)  2Al2O3(s) + 4Fe(s) H° = -1708 kJ

  4. 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) + 10H2O(l)  10Ca(OH)2(s) H° = -653 kJ Equation is reversed, thus H° sign changes Equation is multiplied by 10, thus so is H°

  5. Standard heats of reaction 5.45 ° indicates 1 atm (and is usually associated with a temperature of 25°C) 5.46 An energy term is added (e.g. H or H°) 5.47 Moles 5.48 Al(s) + 1/2Fe2O3 (s)Fe(s) + 1/2Al2O3 (s) H°= – 426.9 kJ 4Al(s) + 2Fe2O3(s)  4Fe(s) + 2Al2O3 (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) + 10H2O(l) 10Ca(OH)2(s) H°=–653kJ For more lessons, visit www.chalkbored.com