Given the data below, determine the normal boiling point of COCl2.
P1= 100 mmHg
T1= -35.6 (degrees)C
(delta) Hvap= 27.4 kj/mol
The Clausius-Clapeyron equation again.
To determine the normal boiling point of COCl2, we can use the Clausius-Clapeyron equation. This equation relates the vapor pressure of a substance at one temperature to its vapor pressure at another temperature.
The Clausius-Clapeyron equation is given as:
ln(P2/P1) = (-delta Hvap / R) * (1/T2 - 1/T1)
where:
P1 is the vapor pressure at temperature T1,
T1 is the initial temperature,
delta Hvap is the enthalpy of vaporization,
R is the ideal gas constant (8.314 J/(mol·K)),
P2 is the vapor pressure at temperature T2,
and T2 is the boiling point we want to determine.
In this case, we are given:
P1 = 100 mmHg
T1 = -35.6 °C = -308.75 K
delta Hvap = 27.4 kJ/mol
We can convert P1 to atm (since R is given in atm) by dividing it by 760 mmHg = 1 atm:
P1 = 100 mmHg / 760 mmHg/atm = 0.1316 atm
Now we have all the required values to solve for T2. Rearranging the equation, we get:
ln(P2/P1) = (-delta Hvap / R) * (1/T2 - 1/T1)
Let's plug in the values:
ln(P2/0.1316) = (-27,400 J/mol / (8.314 J/(mol·K))) * (1/T2 - 1/308.75)
Now we solve for T2.