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Ch. 16: Energy and Chemical Change

Ch. 16: Energy and Chemical Change. Sec. 16.4: Calculating Enthalpy Change. Objectives. Use Hess’s Law of summation to calculate the enthalpy change for a reaction. Explain the basis for the table of standard enthalpies of formation Calculate ΔH rxn using thermochemical equations.

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Ch. 16: Energy and Chemical Change

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  1. Ch. 16: Energy and Chemical Change Sec. 16.4: Calculating Enthalpy Change

  2. Objectives • Use Hess’s Law of summation to calculate the enthalpy change for a reaction. • Explain the basis for the table of standard enthalpies of formation • Calculate ΔHrxn using thermochemical equations. • Determine the enthalpy change for a reaction using enthalpies of formation data.

  3. Calculating Enthalpy Change • C(s, diamond) C(s, graphite) • How can ΔH be determined?

  4. Hess’s Law • Chemists use Hess’s law of heat summation to determine a theoretical value for ΔH. • Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction.

  5. Hess’s Law A + 2B --> D A + B --> C _________________________________________ 3) A + 2B --> D ΔHo = ??? C + B --> D

  6. Hess’s Law • If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

  7. Hess’s Law • The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out

  8. Hess’s Law • Thus we can use information tabulated for a relatively small number of reactions to calculate DH for a large number of different reactions.

  9. Hess’s Law • Consider the combustion reaction of methane to form CO2 and liquid H2O CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) DH = ?

  10. Hess’s Law • This reaction can be thought of as occurring in two steps: • In the first step methane is combusted to produce water vapor: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) DH = -802 kJ

  11. Hess’s Law • In the second step water vapor condenses from the gas phase to the liquid phase: 2H2O(g) -> 2H2O(l) DH = -88 kJ

  12. Hess’s Law • Hess’s Law says we can combine these equations: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) DH = -802 kJ 2H2O(g) -> 2H2O(l) DH = -88 kJ CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l) DH = - 890 kJ

  13. Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • First, look over the equations that contain the substances found in the desired equation and have known enthalpy changes. • S(s) + O2(g) --> SO2(g) DH = -297 kJ • 2SO3(g)--> 2SO2(g) + O2(g) DH = 198 kJ

  14. Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Next, rewrite the found equations so that they are in agreement with the original equation in terms of coefficients and position of the reactants/products. • Equation #1 must be doubled because the original equation has 2 moles of S. S(s) + O2(g) --> SO2(g) DH = -297 kJ becomes 2S(s) + 2O2(g) --> 2SO2(g) DH = -594 kJ

  15. Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Equation #2 must be reversed because you want SO3 to be a product. 2SO3(g)--> 2SO2(g) + O2(g) DH = 198 kJ becomes 2SO2(g) + O2(g --> 2SO3(g) DH = -198 kJ

  16. Calculate DH for the following reaction: 2S(s) + 3O2(g) --> 2SO3(g) • Now add the rewritten equations to obtain the equation for the desired reaction. • 2S(s) + 2O2(g) --> 2SO2(g) DH = -594 kJ • 2SO2(g) + O2(g --> 2SO3(g) DH = -198 kJ • 2S(s) + 3O2(g) --> 2SO3(g) DH = -792 kJ

  17. Hess’s Law • Please note that sometimes thermochemical equations are written with fractional coefficients to determine DH for one mole of product. • 2S(s) + 3O2(g) --> 2SO3(g) DH = -792 kJ can be written as • S(s) + 3/2 O2(g) --> SO3(g) DH = -396 kJ

  18. Practice Problems • Use the equations below to determineDH for the decomposition of H2O2: 2H2O2 --> 2H2O + O2 a) 2H2 + O2 --> 2H2O DH = -572 kJ b) H2 + O2 --> H2O2DH = -188 kJ • Complete practice problems #28 & 29 on pg. 508.

  19. Standard Enthalpy (Heat ) of Formation • Hess’s Law is useful in calculating unknown ΔH values. However, recording ΔH values for all known reactions would be extremely hard. • Scientists have chosen to record ΔH values for one type of reaction - the reaction in which a compound is formed from its elements under standard conditions.

  20. Standard Enthalpy (Heat ) of Formation • The ΔH value for such a reaction is called ΔHfo. • It is defined as the change in enthalpy that accompanies the formation of one mole of a compound from its constituent elements under standard conditions.

  21. Standard Enthalpy (Heat ) of Formation • For example: 2S + 3O2 --> 2SO3 ΔH = -792 kJ becomes S + 3/2O2 --> SO3 ΔHfo = -396 kJ • Standard enthalpies of formation are based on the arbitrary standard that every free element in its standard state has a ΔHfo = 0 kJ.

  22. Standard Enthalpy (Heat ) of Formation • S + 3/2O2 --> SO3 ΔHfo = -396 kJ • In this reaction, ΔHfo of S = 0 kJ & ΔHfo of O2 = 0 kJ. • The energy content of the product SO3 is 396 kJ less than the energy content of the elements from which it is formed.

  23. Using ΔHfo • Find ΔHrxno for H2S + 4F2 --> 2HF + SF6 • To apply Hess’s Law using ΔHfo, you must have one equation for the formation of each compound in the given equation. • Locate these equations in given reference tables: Table 16-7, pg. 510 in your text or in Appendix C, Table C-13, pg. 921.

  24. Using ΔHfo H2S + 4F2 --> 2HF + SF6 • H2 + S --> H2S ΔHfo = -21 kJ • 1/2H2 + 1/2F2 --> HF ΔHfo = -273 kJ • S + 3F2 --> SF6 ΔHfo = -1220 kJ These equations must now be manipulated so they can be “added” according to Hess’s law. • Equation 1 must be reversed. • Equation 2 must be doubled.

  25. Using ΔHfo H2S + 4F2 --> 2HF + SF6 H2S --> H2 + S ΔHfo = 21 kJ H2 + F2 --> 2HF ΔHfo = -546 kJ S + 3F2 --> SF6 ΔHfo = -1220 kJ H2S + 4F2 --> 2HF + SF6 ΔHrxno = -1745 kJ Guess what? There’s an easier way!

  26. Using ΔHfo • Use the following formula: ΔHrxno = ∑ ΔHfo(products) - ∑ ΔHfo(reactants) • The formula says to subtract the sum of heats of formation of the reactants from the sum of the heats of formation of the products.

  27. Using ΔHfo • Find ΔHrxno for H2S + 4F2 --> 2HF + SF6 ΔHrxno = ∑ ΔHfo(products) - ∑ ΔHfo(reactants) = [(2)ΔHfo(HF) + ΔHfo(SF6)] - [ΔHfo(H2S) + (4)ΔHfo(F2)] = [(2)(-273 kJ) + (-1220 kJ)] - [-21 kJ + (4)(0 kJ)] = -1745 kJ

  28. Practice Problems • Use standard enthalpies of formation to calculate ΔHrxno for the following: • CH4 + 2O2 --> CO2 + 2H2O • CaCO3 --> CaO + O2 • 2H2O2 --> 2H2O + O2

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