Answer:
See explanation below
Explanation:
For start, we need some values here to do this exercise.
In general, you can calculate the normal boiling point of any substance by using the Clausius Clapeyron equation which is the following:
ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)
Where:
P1 and P2: pressure of the substance at T1 and T2.
ΔHvap: enthalpy of vaporization of the substance. In the case of bromine is 29.6 kJ/mol
R: constant gas. In this case is 8.3145 J/mol K
T1 and T2: temperature of the substance.
In order to calculate the normal boiling point, we will assign that value to T2, and the pressure would be 1 atm or 1.01x10^5 Pa
T1 and P1 would be temperature and pressure of this substance at any condition. For this example, I will take the fact that Bromine has 22000 Pa at 20 °C (or 293.15 K)
With this data, let's replace in the clausius Clapeyron equation:
ln(1.01x10^5 / 22000) = -29600/8.3145 (1/T2 - 1/293.15)
-1.5241 * 8.3145 / 29600 = (1/T2 - 1/293.15)
-4.281x10^-4 + 1/293.15 = 1/T2
T2 = 1 / 2.9831x10^-3
T2 = 335.22 or 62.07 °C
The real one is 59 °C so, the difference in the result may come with the values of P1 and T1 that may be not accurate.
