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Ch. 17 – Thermochemistry

Ch. 17 – Thermochemistry. IV. Calculating Enthalpy of Reaction (p 578-582). A. Hess’s Law. Sometimes it is hard to measure the enthalpy change for a reaction: The reaction takes place too slowly to measure The reaction is an intermediate step in a series of reactions

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Ch. 17 – Thermochemistry

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  1. Ch. 17 – Thermochemistry IV. Calculating Enthalpy of Reaction (p 578-582)

  2. A. Hess’s Law • Sometimes it is hard to measure the enthalpy change for a reaction: • The reaction takes place too slowly to measure • The reaction is an intermediate step in a series of reactions • You might not want to destroy the material that undergoes the reaction. Cdiamond (s) → Cgraphite (s) in millions of years

  3. A. Hess’s Law Cdiamond (s) → Cgraphite (s) in millions of years • Hess’s law can be used to find the enthalpy change of this reaction by using the following combustion reactions: a. Cgraphite (s) + O2→ CO2ΔH = -393.5 kJ 0.0 0.0 -393.5 b. Cdiamond(s) + O2→ CO2ΔH = -395.4 kJ 1.9 0.0 -393.5 Since CO2 is our product, we need to reverse equation b.

  4. A. Hess’s Law Cdiamond (s) → Cgraphite (s) in millions of years c. CO2 → Cgraphite (s) + O2 ΔH = 393.5 kJ When a reverse reaction is written, the sign of ΔH must also be reversed. Add equations b & c to get the desired equation: Cdiamond(s) + O2→ CO2ΔH = -395.4 kJ CO2 → Cgraphite (s) + O2 ΔH = 393.5 kJ Cdiamond (s) → Cgraphite (s) ΔH = -1.9kJ

  5. IV. Calculating Enthalpy of Reaction CO is produced when carbon is burned with oxygen in a two step reaction. C(s) + O2(g) → CO2(g) CO2(g) + C(s) → 2CO(g) Because these two reactions occur at the same time and we get a mixture of CO and CO2, it is not possible to directly measure the enthalpy of formation of CO(g) from C(s) and O2(g).

  6. IV. Calculating Enthalpy of Reaction Since ΔHfis determined for 1 mole of product, rewrite the two equations as C(s) + 1/2O2(g) → CO(g) ΔHf =? Even though we cannot directly measure the enthalpy of formation for this reaction, we do know the enthalpy of formation for carbon dioxide and the enthalpy of combustion of carbon monoxide. a. C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ b. CO(g) + 1/2O2(g) → CO2(g) ΔHc = -283.0 kJ

  7. IV. Calculating Enthalpy of Reaction The second equation is reversed because we need CO as a product. C(s) + O2(g) → CO2(g) ΔHf = -393.5 kJ CO2(g) → CO(g) + 1/2O2(g) ΔHc = 283.0 kJ C(s) + 1/2O2(g) → CO(g) ΔHf =-110.5 kJ

  8. IV. Calculating Enthalpy of Reaction Calculate ΔH for the following reaction: 2N2(g) +5O2(g)→ 2N2O5(g) Use the following data in your calculation: H2(g) + 1/2O2(g)→ H2O(l) ΔHf = -285.8 kJ N2O5(g) + H2O(l) → 2HNO3(l) ΔH = -76.6 kJ 1/2N2(g) +3/2O2(g) + H2(g) → HNO3(l) ΔH = -174.1 kJ

  9. IV. Calculating Enthalpy of Formation Calculate ΔH for the following reaction: 2N2(g) + 5O2(g)→ 2N2O5(g) (H2O(l) → H2(g) + 1/2O2(g)ΔHf = +285.8 kJ)2 (2HNO3(l) → N2O5(g) + H2O(l) ΔH = +76.6 kJ)2 (1/2N2(g) +3/2O2(g) + H2(g) → HNO3(l) ΔH = -174.1 kJ)4 2H2O(l) → 2H2(g) + O2(g)ΔHf = +571.6 kJ 4HNO3(l) → 2N2O5(g) + 2H2O(l) ΔH = +153.2 kJ 2N2(g) +6O2(g) +4H2(g) → 4HNO3(l) ΔH = -696.4 kJ 2N2(g) + 5O2(g)→ 2N2O5(g) ΔH = 28.4kJ

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