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**Unformatted text preview: **e equilibrium shifts in the direction of the products so that the volume shrinks. When pressure is decreased, the equilibrium shifts in the direction of the reactants so that volume is increased. (c) change in temperature: If a reaction is exothermic in the forward direction, a decrease in temperature drives the reaction forward to counteract the decrease in temperature. An increase in temperature will drive the reaction in the reverse direction since the reverse reaction will be endothermic. (d) change in concentration or partial pressure: Adding reactants or removing products drives the equilibrium towards products, removing reactants or adding products drives the equilibrium towards reactants.
4 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 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 On rearranging and integrating, we get ln K = − ,H + const. P RT
(4.69) (4.71) (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
ln K = − ,G = − ,H + ,S . P RT RT R (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
Problems 4.32, 4.34 5...

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