Predict the boiling point of the following compound using Trouton's Rule:
Heat of vaporization/Tb= 80 J/K
Carbon tetrachloride (CCl4) = ?? degrees Celsius
I found the answer for carbon tetrachloride. I just need to find it for n-butane (C4H10) and ethanol and the database is saying the answers are wrong. I don't understand why if the format is the same as the carbon tetrachloride
To predict the boiling point of carbon tetrachloride (CCl4) using Trouton's Rule, we need to know the molar mass of the compound. Carbon tetrachloride's molar mass can be calculated by adding the atomic masses of each element: C (carbon) = 12.01 g/mol and Cl (chlorine) = 35.45 g/mol.
So, the molar mass of CCl4 (carbon tetrachloride) is:
(12.01 g/mol × 1) + (35.45 g/mol × 4) = 153.82 g/mol
Trouton's Rule states that the heat of vaporization per mole of a substance is approximately constant. The constant value used in Trouton's Rule is 80 J/K·mol.
The boiling point (Tb) can be calculated by using the formula:
Tb = Heat of vaporization / (8.314 J/mol·K × (Molar mass))
Substituting the given values:
Tb = (80 J/K) / (8.314 J/mol·K × 153.82 g/mol)
Now, let's solve for Tb:
Tb ≈ 0.065 K
To convert the boiling point from Kelvin (K) to Celsius (°C), you subtract 273.15 from the value:
Tb in °C ≈ 0.065 K - 273.15 ≈ -273.085 °C
Therefore, according to Trouton's Rule, the predicted boiling point of carbon tetrachloride (CCl4) is approximately -273.085 °C.