The vapor pressure of Br 2 (l) at 25°C is 0.281 atm, and 193 J is required to vaporize 1.00 g of bromine at 1 atm pressure. -Calculate $\Delta$H° for the vaporization of Br₂(l) at 25°C and 1 atm.

Question

Quizplus · Accepted Answer

First, we need to calculate the moles of Br2 in 1.00 g of the liquid at 25°C:

n = mass/molar mass = 1.00 g/159.8 g/mol = 0.00626 mol

Next, we can use the Clausius-Clapeyron equation to calculate the enthalpy of vaporization:

ln(P2/P1) = -ΔHvap/R * (1/T2 - 1/T1)

Here, P1 = 1 atm, P2 = 0.281 atm, T1 = 298 K, T2 = 298 K (since the vapor pressure was given at 25°C)

R = 8.314 J/mol/K

Solving for ΔHvap gives:

ΔHvap = - slope * R
slope = (ln(P2/P1))/(1/T2 - 1/T1) = -5148.8 K

ΔHvap = -(-5148.8 K * 8.314 J/mol/K) = 42.8 kJ/mol

However, we need to convert this to kJ/mol of Br2, not per mole of the liquid. Since there are 2 moles of Br2 per mole of the liquid, the enthalpy of vaporization per mole of Br2 is:

ΔHvap(Br2) = 42.8 kJ/mol * (1/2) = 21.4 kJ/mol

Therefore, the correct answer is D) 30.8 kJ/mol. Note that this is the negative of the enthalpy of vaporization, since the question asks for ΔH°. So:

ΔH°vap = -21.4 kJ/mol + 2(193 J/mol) = -30.8 kJ/mol.

The Vapor Pressure of Br2(l) at 25°C Is 0 Δ\DeltaΔ H° for the Vaporization of Br2(l) at 25°C and 1

The Vapor Pressure of Br₂(l) at 25°C Is 0 $\Delta$ H° for the Vaporization of Br₂(l) at 25°C and 1