Question 10.6: CALCULATING A HEAT OF VAPORIZATION USING THE CLAUSIUS–CLAPEY...
CALCULATING A HEAT OF VAPORIZATION USING THE CLAUSIUS–CLAPEYRON EQUATION
Ether has P_{vap} = 400 mm Hg at 17.9 °C and a normal boiling point of 34.6 °C. What is the heat of vaporization, ΔH_{vap}, for ether in kJ/mol?
STRATEGY
The heat of vaporization, ΔH_{vap}, of a liquid can be obtained either graphically from the slope of a plot of In P_{vap} versus 1/T or algebraically from the Clausius–Clapeyron equation. As derived in Worked Example 10.5,
which can be solved for ΔH_{vap}:
ΔH_{vap} = \frac{(ln P_{2} – ln P_{1})(R)}{\left(\frac{1}{T_{1}} – \frac{1}{T_{2}}\right)}
where P_{1} = 400 mm Hg and , ln P_{1} = 5.991, P_{2} = 760 mm Hg at the normal boiling point and ln P_{2} = 6.633, R = 8.3145 J/(K · mol), T_{1} = 291.1 K (17.9 °C), and T_{2} = 307.8 K (34.6 °C).
ΔH_{vap} = \frac{(6.633 – 5.991)\left(8.3145 \frac{J}{K · mol}\right)}{\frac{1}{291.1 K} – \frac{1}{307.8 K}} = 28,600 J/mol = 28.6 kJ/mol