CH131_12C

CH131_12C - Back to chemistry considerations and...

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Back to chemistry considerations and thermodynamics… For chemical reaction: reactants products Define enthalpy of reaction H = H(products) – H(reactants) At const. P, H = q P = heat transferred into (+) or out of (–) system So if H > 0; q P > 0 system absorbs heat from surroundings H(prod) > H(react) H reactants products H products reactants If H < 0; q P < 0 system releases heat to surroundings H(prod) < H(react) Endothermic reaction Exothermic reaction
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Let’s consider the following chemical reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = –802 KJ @25 o C 1 atm H products reactants CH 4 + 2O 2 CO 2 + 2H 2 O The reaction is exo thermic ( H < 0) => can view heat as a product of the reaction (add to RHS). CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + 802 KJ 1 mol CH 4 + 2 mol O 2 react & release 802 KJ If 2 mol CH 4 + 4 mol O 2 react & release 1604 KJ Enthalpy (H) is an extensive property (extensive => proportional to amount you have. Others; q, V, n) In contrast, some properties are intensive => same throughout material, e.g. T, P When reaction reversed, CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g) Sign of H reversed ( H = +802 KJ) because meaning of products and reactants reversed, H = H(products) - H(reactants)
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H will depend on state of species => gas, liquid, solid For example: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = –890 KJ @25 o C 1 atm ( H is T-dep.; assume 25 o C unless otherwise specified) For H 2 O vapor product, remember H = –802 KJ Since H is a state function, i.e. just depends on initial and final state conditions, we can break H calculation for a given reaction into a sum of H’s for a series of steps equal to the overall reaction. e.g. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = –802 KJ 2H 2 O(g) 2H 2 O(l) H = –88 KJ CH 4 (g) + 2O 2 (g) + 2H 2 O(g) CO 2 (g) + 2H 2 O(l) + 2H 2 O(g) H = –890 KJ overall CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = –890 KJ net This is an example of Hess’s Law: If a reaction is carried out in a series of steps, H for the total reaction is equal to sum of enthalpy changes for each step. Also exothermic; it takes energy to vaporize
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Hess’s Law is extremely useful because you can calculate H for complicated reactions from known values for simpler reactions Note in second step => condensation of 2 mols of water; not a chemical change, just a phase change (exothermic) (1) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) H = –802 KJ (2) 2H 2 O(g) 2H 2 O(l) H = –88 KJ net CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H = –890 KJ If we reverse it (process is vaporization) and consider just 1 mol: H 2 O(l) H 2 O(g) H = +40.7 KJ/mol @ 100 o C This value of H H vap molar enthalpy of vaporization H vap > 0 (endothermic); it takes energy to vaporize something (q P > 0; heat flows into system)
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Similar considerations apply to the other phase transformations, e.g.
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This note was uploaded on 04/17/2008 for the course CAS CH131 taught by Professor Zigler during the Spring '08 term at BU.

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CH131_12C - Back to chemistry considerations and...

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