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Hess’s Law

Hess’s Law. If a reaction is carried out in a series of steps, Δ H for the reaction will equal the sum of the Δ H’s for the individual steps.

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Hess’s Law

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  1. Hess’s Law If a reaction is carried out in a series of steps, ΔH for the reaction will equal the sum of the ΔH’s for the individual steps. This is possible because the enthalpy is a state function: the enthalpy of reactants or products depends only on their state (P, T, amount) and not on how they got there. Knowing: C(graphite) + O2(g)  CO2(g) ΔH(25°C) = -393.5 kJ C(diamond) + O2(g)  CO2(g) ΔH(25°C) = -395.4 kJ Lets us find: C(graphite)  C(diamond) ΔH(25°C) = - 393.5 + 395.4 = 1.9 kJ

  2. Hess’s Law Hess’s Law allows us to calculate energy changes for reactions that would be difficult to measure directly. Controlling this reaction is tricky: C(s) + ½O2(g)  CO(g) Hess’s Law allows us instead to run these two reactions, which are much easier to control: C(s) + O2(g)  CO2(g) ΔH(25°C) = -393.5 kJ CO(g) + ½O2(g)  CO2(g) ΔH(25°C) = -283.0 kJ C(s) + ½O2(g)  CO(g) ΔH(25°C) = -393.5 + 283.0 = -110.5 kJ

  3. Hess’s Law - Example Calculate ΔH(25°C) for the reaction NO(g) + O(g)  NO2(g) given the following information: NO(g) + O3(g)  NO2(g) + O2(g) ΔH(25°C)= -198.9 kJ O3(g)  3/2 O2(g) ΔH(25°C) = -142.3 kJ O2(g)  2O(g) ΔH(25°C) = 495.0 kJ NO(g) + O3(g)  NO2(g) + O2(g) ΔH(25°C)= -198.9 kJ 3/2 O2(g) O3(g)ΔH(25°C) = +142.3 kJ O(g) 1/2 O2(g)ΔH(25°C) = (-495.0 kJ)/2 NO(g) + O(g)  NO2(g) ΔH(25°C) = -198.9 + 142.3 - 247.5 = -304.1 kJ

  4. Heats of This and That Enthalpy is not restricted to chemical reactions, but can apply to any change where there is a flow of heat. • Heat of fusion, ΔfusH: solid  liquid • Heat of vaporization, ΔvapH: liquid  gas • Heat of combustion, ΔH or ΔrxnH: compound + O2 (generally) CO2 + H2O • Heat of solution, ΔsolnH: AB + H2O  A+(aq) + B-(aq) • Heat of formation, Δf H: A + B  AB

  5. Heats of Formation • The heat of formation (enthalpy of formation) applies to the formation of a compound from its elements. • **The state of each element MUST be specified.** • THE STANDARD HEAT OF FORMATIONis designatedΔf H° and is for the formation of 1 mole of the compound from its elements in their standard states. • The standard state of any substance is its pure form at 1 bar pressure and the temperature of interest (often 25°C). • **If an element exists in more than one form under standard conditions, the most stable form of the element is used.** • Δf H° for an element in its standard state is 0 at all temperatures.

  6. Heats of Formation - Examples 1. The standard heat of formation of acetic acid (a liquid) is -487.0 kJ at 25°C. Write the thermochemical equation for the chemical reaction to which this enthalpy applies. 2C(s) + 2H2(g) + O2(g)  CH3COOH(l) Δf H°(25°C) = -487.0 kJ

  7. Thermochemical Equations C4H10(l) + 6.5 O2(g)  4CO2(g) + 5H2O(l) ΔH(25°C) = -2856 kJ Note the units of ∆H are kJ and not kJ/mol. In a thermochemical equation, the enthalpy change ∆H applies to the reaction shown (1 mol of butane, 6.5 mol oxygen, 4 mol carbon dioxide, and 5 mol water).

  8. Heats of Formation - Examples The standard heat of formation of O(g) is 247.5 kJ at 25°C, but the heat of formation of O2(g) is 0 kJ. Why? ½ O2(g)  O(g) Δf H°(25°C) = 247.5 kJ The stable state of oxygen at 25°C and 1 atm is O2(g). By definition, Δf H° for the most stable form of an element is 0 at any temperature. Is the formation of O(g) endothermic or exothermic at 25°C? Is the enthalpy of O(g) higher or lower than ½ O2(g) at 25°C?

  9. Heats of Formation Δf H° are available in tables and can be used to calculate heats of reaction, solution, fusion, vaporization, etc. (Get good at using Appendix C)

  10. Using ΔfH° Data to Find Heats of Reaction ΔrxnH° Calculate ΔrxnH° for the reaction NO(g) + O(g)  NO2(g) using the Δf H° data given in Appendix C. ΔrxnH° = ΣνΔf H° (products) – ΣνΔf H°( reactants) (νrepresents the stoichiometric coefficients) ΔrxnH° = Δf H°(NO2) – Δf H°(NO) – Δf H°(O) = 33.84 – 90.37 – 247.5 (all in kJ) = -304.0 kJ

  11. Using ΔfH° Data to Find Heats of Reaction ΔrxnH° Calculate ΔrxnH°(25°C) for the combustion of benzeneusing the Δf H° data given in Appendix C. 2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(l) ΔrxnH° = 12Δf H°(CO2) + 6Δf H°(H2O) - 2Δf H°(C6H6) - 15ΔfH°(O2) = 12(-393.5) + 6(-285.83) – (2(49.0) + 15(0)) = -4722 – 1714.98 – 98.0 = -6535 kJ ΔH°(1 mol of benzene, 25°C) = -6534.98/2 = -3267 kJ ΔfH° for the most stable form of an element is 0. H2O(l) not H2O(g)

  12. Thermochemical Equations and Stoichiometry 2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(l) ΔH°(25°C)= -6535 kJ A thermochemical equation for the combustion of benzene. Treat ΔH° for the reaction like it is one of the products. How much heat could be liberated from the combustion of 28.35 g of benzene at 25°C? 28.35 g C6H6(l) x 1 mol C6H6 x -6535 kJ = -1186 kJ 78.11 g C6H6 2 mol C6H6

  13. Thermochemical Equations and Stoichiometry 2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(l) ΔH°(25°C)= -6535 kJ We want to design a calorimetry experiment in which we ignite benzene (bomb…or coffee-cup?). Our calorimeter has a heat capacity of 30.04 kJ/°C and we would like a temperature change of at least 5.0°C. How many grams of benzene should we tell the students to use? qrxn = -Cbomb calorimeter ΔT = -30.04 kJ/°C x 5.0°C = -150.2 kJ -150.2 kJ x 2 mol C6H6 x 78.11 g C6H6 = 3.6 g -6535 kJ 1 mol C6H6

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