Question

Chloroform, CHCl3, was once used as an anesthetic. In spy movies it is the liquid put in handkerchiefs to render victims unconscious. Its vapor pressure is 197 mmHg at 23 degrees C and 448 mmHg at 45 degrees
c. Estimate its

a. heat of vaporization
b. normal boiling point I calculated the heat of vaporization to be 29.3 kJ/mol. I'm having some trouble figuring out the normal boiling point, however. I know that the normal boiling point is when, at 1 atm, a liquid boils at a temperature at which its vapor pressure is equal to the pressure above its surface. So P1=P2=1 atm, if P1= vapor pressure and P2= atmospheric pressure/pressure above surface. I figured I could plug this into PV=nRT and solve, but I'm not given a lot of information. I considered assigning arbitrary values for n and V, so I would have
T= (1.00 atm)(1.00 L)/(1.00 moles)(.08206) but is that really the best way to do this problem, or would it even work at all?
I would think you could use the Clausius-Clapeyron equation, just as you did for delta H vap, but this time one of the Ps will be 760 mm and calculate T for that P.
Thank you, that makes sense, but how do I account for P2 and T2 in the equation if I don't know those values either?
But you have two vapor pressures at two temperatures. I would pick 23 C (change to Kelvin, of course) and 197 mm for T1 and P1. Then 760 mm and T2 for the others. You have all of the other numbers. Check my thinking.
oh, of course, I had completely forgotten about that. Thank you.

Answer 1

Answer:

a. $\Delta H_{vap} = 29222.56 J/mol$

b. normal boiling point: 61 C

Explanation:

Data

$T_1 = 23\, C = 296 K$

$T_2 = 45\, C = 318 K$

$P_1 = 197\, mm Hg$

$P_2 = 448\,mm Hg$

R = 8.314 J/mol K, universal gas constant

$\Delta H_{vap}$: heat of vaporization  

From Clausius-Clapeyron equation:

$ln\frac{P_1}{P_2} = \frac{-\Delta H_{vap}}{R} (\frac{1}{T_1} - \frac{1}{T_2})$

$ln\frac{197}{448} = \frac{-\Delta H_{vap}}{8.314} (\frac{1}{296} - \frac{1}{318})$

$-0.8215 = \frac{-\Delta H_{vap}}{8.314} (\frac{11}{47064})$

$\Delta H_{vap} = 0.8215 \times 8.314 \times \frac{47064}{11}$

$\Delta H_{vap} = 29222.56 J/mol$

We can use this result to compute the normal boiling point as follows:

$ln\frac{197}{760} = \frac{-29222.56}{8.314} (\frac{1}{296} - \frac{1}{T_2})$

$-1.35 \times \frac{8.314}{-29222.56 = (\frac{1}{296} - \frac{1}{T_2})$

$\frac{1}{T_2} = \frac{1}{296} -3.84\cdot 10^{-4}$

$T_2 = (3 \cdot 10^{-3})^{-1}$

$T_2 = 334\, K = 61\, C$

Answer 2

a. Using the Clausius-Clapeyron equation
ln P1/P2 = (-Hvap / R) (1/T1 - 1/T2)
P1 = 197 mm Hg
P2 = 448 mm Hg
T1 = 23 C = 296 K
T2 = 45 C = 318 K
R = 8.314 J/mol-K
Solving for Hvap
Hvap = 29225.43 J/mol

b. For normal boiling point
P1 = 197 mm Hg
P2 = 760 mm Hg (normal pressure)
T1 = 23 C = 296 K
T2 = ?
R = 8.314 J/mol-K
Hvap = 29225.43 J/mol
ln P1/P2 = (-Hvap / R) (1/T1 - 1/T2)
Use the equation to solve for T2