Lecture_4

# T t p t2 469 since g rt ln k substituting for

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Unformatted text preview: um towards products, removing reactants or adding products drives the equilibrium towards reactants. Temperature Dependence of Equilibrium Constants From Chapter 4, we have é ¹ æ ,G  ö ù = − ,H  . ë ¹T è T ø û P T2 (4.69) Since ,G  = −RT ln K  ,substituting for ∆G°, we get P æ ¹ ln K  ö j d ln K  = ,H  . Pø P è ¹T RT 2 dT P (4.71) The approximate equality holds in most cases because K°P typically has a very weak dependence on P and a very strong dependence on T. On rearranging and integrating, we get  ln K  = − ,H + const. P RT (4.75) which implies that when ln(K°P) is plotted as a function of 1/ T, the slope will be equal to –∆H°/R. However, we also know that ∆G° = ∆H° – T∆S°. Therefore, we get   S ln K  = − ,G = − ,H + ,R . P RT RT (4.76) Comparing Eqs. (4.75) and (4.76), we see that if ∆H° is independent of temperature, the intercept of the line will be equal to the entropy change for the process. It is also possible to show that æ ¹ ln K  ö j d ln K  = ,U  . Cø C è ¹T RT 2 dT P...
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## This document was uploaded on 02/28/2014 for the course CHEM 311 at LA Tech.

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