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Thermochemisty (Enthalpy) and Hess’s Law

Thermochemisty (Enthalpy) and Hess’s Law. Chapter 10 Sections 10.6-10.7. Enthalpy. Change in enthalpy ( Δ H p = q p ): the amount of heat exchanged when heat exchange occurs under conditions of constant pressure Enthalpy is a state function

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Thermochemisty (Enthalpy) and Hess’s Law

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  1. Thermochemisty (Enthalpy) and Hess’s Law Chapter 10 Sections 10.6-10.7

  2. Enthalpy • Change in enthalpy (ΔHp = qp): the amount of heat exchanged when heat exchange occurs under conditions of constant pressure • Enthalpy is a state function • State function is defined as a property that changes independent of pathway • ΔH is independent of the path taken

  3. At constant pressure, 890kJ of energy per mole of CH4 is produced as heat Qp = ΔHp = -890 kJ exothermic If 5.8 g of methane is burned, how much heat will be released? 5.8g CH4 x 1mol CH4 = 0.36 mol CH4 16.0 g CH4 0.36 mol CH4 x -890 kJ = -320 kJ mol CH4

  4. Calorimeter • Calorimetry the process of measuring heat flow in the form of temperature changes in a system • Calorimeter an insulated device used to measure temperature changes

  5. Hess’s Law • Hess’s Law • ΔHfor a multi-step process is the sum of ΔH for the individual steps • Hess’s law allows us to calculate ΔHfor new reactions from previously determined values Germain Henri Hess

  6. What is the change in altitude (Δ Alt) between Salt Lake City and Denver? Altitude Denver, CO 5,280 ft Flight 2 ? 1014 ft Salt Lake City, UT 4,266 ft 5280 ft 4266 ft Flight 1 San Francisco, CA 0 ft Flight 1: ΔAlt = -4266 Flight 2: ΔAlt = +5280 SL City-DenverΔ Alt = +1014

  7. Hess’s Law • Applying Hess’s Law • When reversing a reaction, change the sign of ΔH. • When multiplying a reaction by a new coefficient, multiply ΔH

  8. Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

  9. Hess’s Law Calculate the enthalpy change for a partial combustion of 1 mol of graphite to diamond C(s, graphite)  C(s, diamond) ΔH = ? Given: C(s, graphite) + O2(g)  CO2(g) ΔH = -393.5 KJ C(s, diamond) + O2(g)  CO2(g) ΔH = -395.4 KJ Answer: C(s, graphite) + O2(g)  CO2(g) ΔH = -393.5 kJ CO2(g)  C(s, diamond) + O2(g) ΔH = +395.4 kJ C(s, graphite)  C(s, diamond) ΔH = +1.9 kJ

  10. Hess’s Law Example Problem CH4  C + 2H2 +74.80 kJ Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

  11. CH4  C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

  12. CH4  C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ Step #3: Multiply reaction #2 by 2

  13. CH4 C + 2H2 +74.80 kJ C + O2 CO2 -393.50 kJ 2H2 + O2 2 H2O -571.66 kJ CH4 + 2O2 CO2 + 2H2O -890.36 kJ Step #4: Sum up reaction and H

  14. Calculation of Heat of Reaction Hrxn = Hf(products) - Hf(reactants) Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ] Hrxn = -890.36 kJ

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