Heat of Vaporization. (loosely based on Chap Heat of Vaporization (loosely based on Chap. 12 Sec 8 of Jespersen 6th Ed) Dr. C. Yau Spring 2014

Molar Heat of Vaporization = amt of heat needed to vaporize 1 mole of liquid p.552

For water, Hvap = +43.9 kJ mol-1 What exactly does this mean? H2O (l)  H2O (g) H = +43.9 kJ Or H2O (l) +43.9 kJ  H2O (g) 3

How is Hvap Determined? Clausius, a German physicist, and Clapeyron, a French engineer, determined that The relationship between ln P and T where P is the vp and T is the temp in K. Mathematically, as an equation, it is… ln P = k (1/T) where k is the proportionality constant and it turns out to be a negative value.

ln P is proportional to 1/T ln P = k (1/T) For P = vapor pressure of liquid T = temp in deg Kelvin where R = 8.314 J mol-1 K-1 This is an eqn for a straight line: y = mx + b

Graphical Determination of Hvap y = m x + b Plot ln P vs. 1/T Slope = Rearrange eqn to calculate Hvap = - (slope)(R)

ln P vs. 1/T Are the slopes positive or negative? Ans. Negative. Slope for water is more negative than that for acetone.

Hvap must be positive, as expected for vaporization (endothermic) slope = Since the slope is negative, Hvap must be positive, as expected for vaporization (endothermic) Slope of water being more negative than acetone means |Hvap| (water) > |Hvap | (acetone) (Why is this so? What does this mean?)

Graphical Determination of Hvap Measure vp at various temp. Plot ln P versus 1/T (kelvin) Calculate the slope from the graph 4) From the slope calculate Hvap Hvap = - (slope)(R)

Two-Point Method to Determine Hvap Instead of measuring a set of P and T values, Hvap can also be determined by measuring only two sets of P and T. You need only vp at two different temp: (P1, T1) and (P2, T2) See lecture notes for derivation of Clausius-Clapeyron Equation:

Example 12.3: Methanol, CH3OH, experiences hydrogen bonding, dipole-dipole interactions, and London forces. At 64.6 oC, it has a vp of 1.00 atm, and at 12.0oC, it has a vp of 0.0992 atm. What is the heat of vaporization for methanol? Given: R = 8.314 J mol-1 K-1 Write out the eqn & see what is known & what is unknown.

See page 554 for calculation. Ans 35.2 kJ/mol T = 64.6 oC, P =1.00 atm, T =12.0oC, P = 0.0992 atm Which is the unknown? R = 8.314 J mol-1K-1 T1 = 64.6+273.15 = 337.8 K T2 = 12.0+273.15 = 285.2 K P1 = 1.00 atm P2 = 0.0992 atm See page 554 for calculation. Ans 35.2 kJ/mol Do Practice Exercises 12.10 & 12.11