Assuming ΔH° and ΔS° are temperature independent, calculate the normal boiling point of bromine.

Question

Quizplus · Accepted Answer

The normal boiling point of a substance is the temperature at which its vapor pressure equals one atmosphere (760 mmHg). We can use the Clausius-Clapeyron equation to relate the boiling point of the substance to its enthalpy of vaporization (ΔH°vap), entropy of vaporization (ΔS°vap), and vapor pressure at a given temperature (P):

ln(P2/P1) = (ΔH°vap/R)((1/T1)-(1/T2))

where P1 is the vapor pressure at a known temperature T1 and P2 is the vapor pressure at the boiling point temperature T2.

At the normal boiling point, T2 = Tnbp = constant, and P2 = 760 mmHg. We can use the known values of ΔH°vap and ΔS°vap for bromine to calculate its boiling point:

ΔH°vap = 29.26 kJ/mol
ΔS°vap = 93.5 J/K·mol
R = 8.314 J/K·mol

ln(760/P1) = (29.26 kJ/mol /8.314 J/K·mol) ((1/T1) - (1/Tnbp))
(1/Tnbp) = (1/T1) - (8.314 J/K·mol / 29.26 kJ/mol) ln(760/P1)
(1/Tnbp) = (1/T1) - 0.002098

Assuming that ΔH° and ΔS° are temperature independent, we can pick any temperature T1 as long as the corresponding vapor pressure P1 is known. For convenience, let's choose T1 = 298 K (room temperature) and P1 = 1 atm = 760 mmHg. Plugging in the numbers and solving for Tnbp, we get:

(1/Tnbp) = (1/298 K) - 0.002098
Tnbp = 332 K or 59°C

Therefore, the best choice is D) 332 K, which is the boiling point of bromine at normal atmospheric pressure.

Assuming ΔH° and ΔS° Are Temperature Independent, Calculate the Normal