l05 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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20 .110/5.60 Fall 2005 Lecture #5 page 1 Thermochemistry Much of thermochemistry is based on finding “easy” paths to calculate changes in enthalpy, i.e. understanding how to work with thermodynamic cycles. Goal: To predict H for every reaction, even if it cannot be carried out in the laboratory The heat of reaction rx is the for the isothermal reaction at constant pressure (the complete transfer from reactants to products, not to some equilibrium state). e.g. Fe 2 O 3 (s, T,p ) + 3H 2 (g, ) = 2Fe(s, ) + 3H 2 O( l , ) () ( ) ( ) ( ) 22 2 3 ,2 ,3 ,3 , , [p r o d u c t s r e a c t a n t s ] Fe HO FeO Tp HH =+ −− =− We cannot know values because enthalpy, like energy, is not an absolute scale. We can only measure differences in enthalpy. Define a reference scale for enthalpy (298.15K, 1 bar) 0 For every element in its most stable form at 1 bar and 298.15K e.g. 2 (g) (graphite) 298.15 0 298.15 0 C K ° ° = = The “ ° ” means 1 bar 20.110J / 2.772J / 5.601J Thermodynamics of Biomolecular Systems Instructors: Linda G. Griffith, Kimberly Hamad-Schifferli, Moungi G. Bawendi, Robert W. Field
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20 .110/5.60 Fall 2005 Lecture #5 page 2 : The heat of formation is the heat of reaction to create 1 mole of that compound from its constituent elements in their most stable forms. ( ° 298.15 f K H ) Example ( T = 298.15 K) ½ H 2 (g, ,1 bar) + ½ Br 2 ( l , ,1 bar) = HBr (g, ,1 bar) () °° ° ° = − ∆= ±²²²²²³²²²²²´ 22 , 0 - elements in most stable forms 11 (, 1 ) g , g , , rx HBr Br fHB r TT b a HH l ) We can tabulate values for all known compounds. ( ° 298.15 We can calculate for any reaction. ( ° 298.15 ) e.g. (T=298.15K) CH 4 (g, ,1 bar) + 2O 2 (g, ,1 bar) = CO 2 (g, ,1 bar) + 2H 2 O( l , T, 1 bar) First decompose reactants into elements Second put elements together to form products
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This note was uploaded on 11/11/2011 for the course BIO 20.010j taught by Professor Lindagriffith during the Spring '06 term at MIT.

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l05 - 20.110J / 2.772J / 5.601J Thermodynamics of...

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