Vapor Pressure of Liquids Equilibria and Thermodynamics
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1 Chemistry 1B-Foothill College Vapor Pressure of Liquids Equilibria and Thermodynamics In this exercise, you will investigate the relationship between the vapor pressure of a liquid and the thermodynamic quantities of H vap, S vap and G vap. When a liquid is added to a closed container, the liquid will evaporate into the air above it in the flask. Eventually, equilibrium is reached between the rate of evaporation and the rate of condensation: liquid vapor At this point, the vapor pressure of the liquid is equal to the partial pressure of its vapor in the flask. The vapor pressure of a liquid (in atm) under equilibrium conditions is numerically equal to the equilibrium constant, K p = K eq, for the process. Vapor pressure, and thus K p, depend strongly on temperature. We can understand this temperature dependence by considering the energetics of evaporation and the sign of H vap. Think about this a bit! Thermodynamics of Vapor Pressure Equilibria: The mathematical relationship between the vapor pressure of a liquid and its temperature is nonlinear as shown below: P = exp ΔH vap T + ΔS vap (1) where P is the vapor pressure in atm, H vap is the heat of vaporization, T is the absolute temperature, is the universal gas constant, and S vap is the entropy change during the vaporization. From thermodynamics, we also know that G is related to H and S : G = H T ( S ) (2) emember that in both of these relationships, (1) and (2), we are assuming that H and S are relatively temperature independent, an assumption that works well over relatively small temperature ranges. We also know that G is related to the vapor pressure of a liquid according to the formula: G = T ln(p) (3) P is the vapor pressure in atm and is equivalent to K eq for this simple system. The sign of G indicates the spontaneous direction of change for the process (forward or reverse) when the vapor pressure is initially set to 1 atm (Q p = 1) at the given T. For example, consider the following system with a pressure of 1 atm water vapor pressure above liquid water: H 2 O(l) H 2 O(g, 1 atm) At temperatures below 100 C, vapor will condense into liquid until equilibrium is reached since we are below the normal boiling point. Hence, for this system below 100 C, G will always be (+), we say that the direction of spontaneous change is to the left. Now, think about the following questions: What is the sign of G at 100 C and what is the sign of G at temperatures above 100 C for this system and why? If we start with nonstandard initial conditions (Q 1) then G can be calculated using one of the following equations: G = T ln(k/q) or G = G + T ln(q) (4) or G = H T S + T ln(q) Again, the sign of G will tell us whether or not we are at equilibrium and, if not, the direction of spontaneous change. Dr. L. J. Larson 1
2 Questions and Problems For numerical problems, please show all your work with units and report answers to the correct precision (correct significant figures). 1) Using Appendix C in your textbook, calculate the literature values for S vap and H vap for methanol. emember, you need to take the difference between the vapor and liquid states! CH 3 OH(l) < > CH 3 OH(g) S vap H vap 2) Using the above values of S vap and H vap, calculate the normal boiling point in C of methanol. You can use either equation (1) or (2). a) Look up the literature (accepted) value for methanol s normal boiling point and report it here: b) Determine the percent error in your calculated value compared to the literature value. c) Does your calculated value agree fairly well with the literature value? Why would they be different? Dr. L. J. Larson 2
3 3) Sketch a graph of vapor pressure (atm) versus temperature ( C) for methanol. Label on the graph the normal boiling point for methanol, indicating the temperature and pressure at this point. (Use the literature value for the boiling point.) a) At what temperature in C is the vapor pressure of methanol equal to 1.25 atm? Mark and label this temperature and pressure on the above graph. 4) Calculate G for the vaporization of methanol at 25 C. Does the sign of G make sense? Explain why or why not. 5) Calculate G for the vaporization of methanol at 100 C. Does the sign of G make sense? Explain why or why not. Dr. L. J. Larson 3
4 6) The Clausius-Clapeyron equation describes the linear relationship between ln (vapor pressure) and absolute temperature: ln(p) = ΔH vap 1 T + ΔS vap Show, step by step, the derivation of this equation starting from equation (1). a) Sketch a graph of ln(p) versus 1/T (P in atm and T in Kelvin) for the vaporization of methanol. Use the axis system given below. ln(p) 0 1/T(K) b) Calculate the values of the slope and y-intercept for the above graph. Include units in your calculations! slope = y-intercept = Dr. L. J. Larson 4
5 7) Sketch a graph of G versus T (T in Kelvin) for the vaporization of methanol. Use the axis system given below. G (kj/mol) 0 T(K) a) eport the values of the slope and y-intercept for the above graph. Include units! slope = y-intercept = b) Calculate the temperature where G will become zero. Label this point on the G vs. T graph. What is the significance of this temperature? 8) At 25 C, the non-equilibrium vapor pressure of methanol is measured to be 56 torr. Find G for this condition for methanol. a) Based upon your calculated G, should there be a net evaporation or a net condensation of methanol under these conditions? 9) Now let s remember what we have learned about intermolecular forces! Please no calculations here, just reasoning! On your previous graph of Vapor Pressure vs. Temperature for methanol drawn in 3), sketch the curve for ethylene glycol (HOCH 2 CH 2 OH) showing this second curve correct ELATIVE TO the methanol curve. Clearly identify the two curves with labels! Dr. L. J. Larson 5
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