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.
To calculate the normal boiling point of chloroform (CHCl3), you can use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at two different temperatures to its enthalpy of vaporization.
The Clausius-Clapeyron equation is given by:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ΔHvap is the heat of vaporization, R is the gas constant (8.314 J/mol·K), and T1 and T2 are the temperatures in Kelvin.
To calculate the normal boiling point, we can use the equation with known values and solve for T2. From the question, we are given the following values:
P1 = 197 mmHg (converted to atm by dividing by 760 mmHg/atm)
P2 = 1 atm (normal atmospheric pressure)
ΔHvap = 29.3 kJ/mol (converted to J/mol by multiplying by 1000)
R = 8.314 J/mol·K
T1 = 23°C (converted to Kelvin by adding 273.15)
Plugging these values into the equation, we have:
ln(1/0.259) = (29300 J/mol / 8.314 J/mol·K) * (1/296.15 K - 1/T2)
Simplifying the equation, we get:
-1.343 = 3519.96 * (0.003383 - 1/T2)
Rearranging the equation to solve for T2, we have:
T2 = 1 / (1 / 296.15 K - (-1.343) / 3519.96)
Solving this equation will give you the value of T2, which represents the normal boiling point of chloroform.