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Hess’s Law SECTION 5.3
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Hess’s Law The enthalpy change of a physical or chemical process depends only on the initial and final conditions of the process. The enthalpy change of a multistep process is the sum of the enthalpy changes of its individual steps.
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Problem Iron can be obtained from the following reaction: Fe 2 O 3(s) + 3CO (g) → 3CO 2 + 2Fe (s) Determine the enthalpy change of reaction, given the following: 1. CO (g) + ½ O 2 → CO 2 ∆H°=-283.0kJ 2. 2Fe (s) + 3/2 O 2 → Fe 2 O 3(s) ∆H°=- 824.3kJ
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Solution Manipulate equations to make them match the overall equation: #1 equation proceeds in the correct direction but must be multiplied by 3. So: CO (g) + ½ O 2 → CO 2 ∆H°=-283.0kJ X 3 3CO (g) + 3/2 O 2 → 3CO 2 ∆H°=-849.0 kJ #2 equation needs to be reversed but the coefficients are ok. So: 2Fe (s) + 3/2 O 2 → Fe 2 O 3(s) ∆H°= - 824.3kJ becomes: Fe 2 O 3(s) → 2Fe (s) + 3/2 O 2 ∆H°= 824.3kJ (note the sign is reversed)
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To get the overall equation: Since oxygen is the same on both sides of the equation, cross it out and add the individual enthalpy changes: Fe 2 O 3(s) + 3CO (g) → 3CO 2 + 2Fe (s) ∆H°= -24.7 kJ
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Practice Problems – p. 316 – all questions
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Standard Molar Enthalpies of Formation Change in enthalpy when 1 mole of a compound is formed directly from its elements in their most stable state at SATP (25°C & 100kPa). It’s a synthesis reaction when the compound is formed directly from its elements and not from any other compound. Eg. C (s) + O 2(g) → CO 2 (g) Coefficients are often fractions because there must be only 1 mole of product formed. See Table 5.5 and Appendix B
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Thermal Stability The ability of a substance to resist decomposition when heated. The greater the enthalpy change of a decomposition reaction, the greater its thermal stability.
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Enthalpies of Formation and Hess’s Law ∆H° r =∑(n∆H° f products) - ∑(n ∆H° f reactants) n= coefficient in equation ∑ = “sum of” ∆H° r = enthalpy change of reaction Use standard molar enthalpies of formation on Appendix B
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Determine the enthalpy of formation for the following reaction : CH 4(g) + 2O 2 → CO 2 + 2H 2 O ∆H° r =∑(n∆H° f products) - ∑(n ∆H° f reactants) **Use Appendix B ∆H° r =((1)(-393.2) + (2)(-241.9)) - ((1)(-74.6)(2)(0)) ** since oxygen is an element in its most stable state, standard molar enthalpy is 0. = (-877.1kJ)-(-74.6kJ) = -802.5 kJ
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Videos/Resources to study at home Hess’s Law: http://www.ausetute.com.au/hesslaw.html http://www.ausetute.com.au/hesslaw.html https://www.youtube.com/watch?v=u7aTBxA7sL8 https://www.youtube.com/watch?v=u7aTBxA7sL8 Hess’s Law and Enthalpies of Formation http://www.chemguide.co.uk/physical/energetics/sums.html http://www.chemguide.co.uk/physical/energetics/sums.html https://www.youtube.com/watch?v=iQuy2mgbV9o https://www.youtube.com/watch?v=iQuy2mgbV9o
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Practice Problems – p. 323 – #51, 52, 54, 55, 56, 58. Research Application Questions Choose from: P. 291: #4, 10, 12 P. 311: #8, 12 P. 335: #2, 5, 6, 7, 8 P. 349: #62, 63, 65, 66, 67, 69, 70 P. 351: #23 P. 408-409: #68, 69
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