Calculate the normal boiling point of a compound that has a vapor pressure of 19.30kPa at 167 Fahrenheit and a vapor pressure of 1.86kPa at 77 Fahrenheit.

Use the Clausius-Claperyron equation. Using p1, p2, T1 and T2, solve for delta H vaporization. Then plug into a new Clausius-Claperyron equation EITHER p and T pair, and plug in 760 mm and Tboiling for the other pair. Solve for Tboling.

To calculate the normal boiling point of a compound based on its vapor pressure at different temperatures, we can use the Clausius-Clapeyron equation.

The Clausius-Clapeyron equation is given by:

ln(P1/P2) = (-ΔHvap / R)((1/T1) - (1/T2))

Where:
P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively.
ΔHvap is the enthalpy of vaporization.
R is the ideal gas constant.

First, let's convert the temperatures from Fahrenheit to Kelvin:

T1 = (167 - 32) * (5/9) + 273.15 = 341.94 K
T2 = (77 - 32) * (5/9) + 273.15 = 298.15 K

Next, let's plug in the values into the equation:

ln(19.30 / 1.86) = (-ΔHvap / R) * ((1/341.94) - (1/298.15))

Now, we need to calculate the natural logarithm:

ln(10.38) = (-ΔHvap / R) * (0.00292 - 0.00336)

We can simplify the equation:

2.339 - 0.914 = -ΔHvap / R * (-0.00044)

2.339 / 0.914 = ΔHvap / R * 0.00044

ΔHvap / R = (2.339 / 0.914) / 0.00044

Finally, we can calculate the ratio ΔHvap / R:

ΔHvap / R = 6.408 / 0.00044 ≈ 14,563.6

Now, the typical value for R is 8.314 J/(mol·K). Therefore:

ΔHvap = 14,563.6 * 8.314 = 121,056.39 J/mol

Since the boiling point occurs when the vapor pressure reaches atmospheric pressure (1 atm or 101.325 kPa), we can use the Antoine equation or reference tables to find the corresponding temperature.