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Chemistry 201 Lecture 10 Temperature and pressure dependence of K NC State University Factors that affect K • How does the free energy relate to the standard state? • How does temperature change affect energetics and equilibrium? • What is the pressure dependence of K? Text : Sections 4.7, 5.1 Relationship of free energy to standard state The free energy change for a reaction, DG, is composed of individual free energies Gn that correspond to each reactant. We can write the free energy of each component as where an is the activity of component n and DGno is the standard free energy. Activity means concentration, but it is used for “real” solutions. See the next slide for comparison. Definition of activity Activity is a concentration and has units of molarity. We use activity to account for the fact that there may an effective or non-ideal concentration due to solvent effects. You May also think of the free energy as follows: where [n] is the effective concentration (activity). Combining free energies to make a free energy change The molar free energy change of a reaction is given by the difference between the free energy of the products minus the reactants. For the hypothetical reaction we have Combining free energies to make a free energy change (cont’d) Each free energy is related to the standard free energy for that species so we have We can rearrange this to and finally The standard state Thus, the standard free energy change is the difference in free energy of each species at its standard state, And the reaction quotient is the ratio of each concentration to the standard concentration, which is 1 molar (so it does not appear). Defining the reaction quotient The reaction quotient is unitless. The units of the individual activities (concentrations) are cancelled by the units of the standard state. Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K At what temp. does DGo = 0? (K = 1) Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K At what temp. does DGo = 0? (K = 1) Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K At what temp. does DGo = 0? (K = 1) Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K At what temp. does DGo = 0? (K = 1) Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K At what temp. does DGo = 0? (K = 1) Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K What is K at 425 K? Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K What is K at 425 K? Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K What is K at 425 K? Temperature dependence of K N2O4(g) 2 NO2(g) DHo = +58.20 kJ DSo = +176.6 J/K What is K at 425 K? Remember to convert DHo to units of J (multiply by 1000) Temperature dependence of K H2(g) + 1/2 O2 (g) H2O (g) DHo = -241.82 kJ DSo = -44.38 J/K At what T, if any, does K = 1? Temperature dependence of K H2(g) + 1/2 O2 (g) H2O (g) DHo = -241.82 kJ DSo = -44.38 J/K At what T, if any, does K = 1? The temperature corresponds to DGo = 0 Temperature dependence of K H2(g) + 1/2 O2 (g) H2O (g) DHo = -241.82 kJ DSo = -44.38 J/K Evaluate K at 1500 K. Temperature dependence of K H2(g) + 1/2 O2 (g) H2O (g) DHo = -241.82 kJ DSo = -44.38 J/K Evaluate K at 1500 K. The temperature dependence of DGo As we have shown previously, DG, will decrease until it reaches 0. Then we have reached equilibrium. The equilibrium condition is DGo = -RT ln K Next we consider the fact that we can use the temperature dependence of the free energy to obtain information about the enthalpy. DHo - TDSo = -RT ln K If we assume that DHo and DSo are independent of temperature, then we can obtain the values of K at two temperatures as follows, DHo – T1DSo = -RT1 ln K1 DHo – T2DSo = -RT2 ln K2 The temperature dependence of DGo Then we can divide each equation by its respective temperature to obtain, DHo /T1 – DSo = - R ln K1 DHo /T2 – DSo = - R ln K2 We subtract temperature T2 from T1. DHo (1/T1 – 1/T2) = - R ln(K1/K2) ln(K2/K1) = -DHo /R(1/T2 – 1/T1) This equation says that if we plot ln(K) vs 1/T, we obtain a line, and the slope of that line is -DHo /R. Using equilibrium data to obtain DHo and DSo 2 NO2 (g) 2 NO (g) + O2 (g) T1 at 190 K T2 at 200 K K1 = 18.4 K2 = 681 find DHo and DSo Using equilibrium data to obtain DHo and DSo Solution: starting with the equation Solve for DHo Substitute in the given values Van’t Hoff plots Slope = -DHo/R Note: DHo > 0 The standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/T Van’t Hoff plot for drug binding A practical example of the application of the van’t Hoff equation can be found in drug binding. The equilibrium constant for drug binding to an active site can be measured by fluorescence, NMR, etc. at various temperatures. Then one may plot ln K vs. 1/T and fit the result to a line. In most cases the binding will be exothermic to that DHo < 0 and then slope of the line will be positive rather than negative as shown in the previous slide. Van’t Hoff plot for drug binding Slope = -DHo/R Note: DHo < 0 In this example, the slope is positive because the enthalpy of binding is negative (i.e. binding is exothermic). Example: preventing inflammation by binding to prostaglandin synthase 2 COX-2 crystal structure DHo and DSo can be measured using ln(K) as a function 1/T. Example: preventing depression serotonin transport inhibitors DHo and DSo can be measured using ln(K) as a function 1/T. This article shows specific differences in the enthalpy of binding of drugs based on analysis of so-called van’t Hoff plots (i.e. plots of ln(K) vs. 1/T). Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp. 991-1000 Basic conclusion: fluvoxamine binds exclusively based on entropic driving force Entropically driven binding is relatively rare. Usually the entropy of binding is unfavorable since a flexible drug molecule (large W) will be forced to adopt a fixed conformation (small W or W = 1) upon binding to a protein. Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp. 991-1000 Ligand binding in myoglobin Ligand binding to myoglobin is described as a chemical equilibrium between a bound state MbCO, a dissociated state Mb:CO and a solvent state Mb + CO. Mb + CO Mb:CO MbCO If we ignore the intermediate state the binding can be described by An overall equilibrium process. Mb + CO with equilibrium constant K. We have: MbCO [MbCO] K= [Mb][CO] The fraction bound is: [MbCO] f= [Mb] + [MbCO] We use CO in Many studies. The same equations Hold for O2. Ligand binding curve The fraction bound f can be related to the binding constant K: [MbCO] K= [Mb][CO] [MbCO] K[Mb][CO] K[CO] f= = = [Mb]+[MbCO] [Mb] + K[Mb][CO] 1 + K[CO] This type of binding curve is plotted below. Pressure dependence of species We can see from the gas phase form of the equilibrium constant that pressure of species depend on pressure. For the general gas phase reaction, we can write the equilibrium constant as And the free energy is From Dalton’s law Pressure dependence of species If we substitute these mole fractions and total pressure into the equilibrium constant we have Which depends on the total pressure unless z – c – d = 0. This expression shows that, in general, the free energy depends on the total pressure. This means that for the fixed pressure may affect the proportion of products to reactants. Equilibrium of smog formation Why is the dissociation greater at low pressure? N2O4(g) 2 NO2(g) Can the trend be explained in simple terms? Why is the dissociation greater at low pressure? N2O4(g) 2 NO2(g) Can the trend be explained in simple terms? Yes, this is an example of Le Chatelier’s principle. Why is the dissociation greater at low pressure? N2O4(g) 2 NO2(g) Can the trend be explained in simple terms? Yes, this is an example of Le Chatelier’s principle. Can the trend be explained quantitatively? Why is the dissociation greater at low pressure? N2O4(g) 2 NO2(g) Can the trend be explained in simple terms? Yes, this is an example of Le Chatelier’s principle. Can the trend be explained quantitatively? The Haber-Bosch Process The Haber-Bosch process N2 (g) + 3 H2 (g) 2 NH3(g) Involves both high temperature and pressure. The Haber-Bosch process N2 (g) + 3 H2 (g) 2 NH3(g) Involves both high temperature and pressure. Why high pressure? The Haber-Bosch process N2 (g) + 3 H2 (g) 2 NH3(g) Involves both high temperature and pressure. Why high pressure? Forces the equilibrium to the right. This is an example of Le Chatelier’s principle. The Haber-Bosch process N2 (g) + 3 H2 (g) 2 NH3(g) Involves both high temperature and pressure. Why high pressure? Forces the equilibrium to the right. This is an example of Le Chatelier’s principle. Why high temperature? The Haber-Bosch process N2 (g) + 3 H2 (g) 2 NH3(g) Involves both high temperature and pressure. Why high pressure? Forces the equilibrium to the right. This is an example of Le Chatelier’s principle. Why high temperature? Although this is an exothermic reaction, it also has a large barrier. The process uses a catalyst. But, what does the temperature do to the equilibrium? The Haber-Bosch process 1/2 N2 (g) + 3/2 H2 (g) NH3(g) The enthalpy change of the reaction is equal to the enthalpy of formation of NH3. DfHo = -45.9 kJ/mol The entropy can be calculated from tabulated absolute entropies. Mol NH3 N2 H2 So 192.77 153.3 114.7 The Haber-Bosch Process Under standard conditions The Haber-Bosch Process 1/2 N2 (g) + 3/2 H2 (g) NH3(g) At high pressure there is a shift in the free energy for the process is given by: The entropy DrxnSo is obtained from the tables as well. We note that the standard (tabulated) conditions correspond to Q = 1 and therefore RTlnQ = 0. Suppose we increase the pressure of N2 and H2 to 500 atm, while NH3 is maintained at 1 atm. What happens to the free energy of reaction? The Haber-Bosch Process 1/2 N2 (g) + 3/2 H2 (g) NH3(g) The shift to high pressure will shift the reaction further towards products. This is clear intuitively from Le Chatelier’s principle, but it is quantified using the equilibrium expressions we have learned. Since the free energy change is negative, this might seem unnecessary. However, the Haber-Bosch Process is challenging since the reaction must occur at high temperature as well. We will see how all of the factors combine in future lectures. Hydrogenation: use of pressure Skills • Calculate how DGo changes with temperature and pressure • Calculate how K changes with temperature and pressure • Calculate DHo from the temperature dependence of K