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 $\Delta$S° for the vaporization of Br₂(l) at 25°C and 1 atm.

Question

Quizplus · Accepted Answer

First, we need to use the Clausius-Clapeyron equation to calculate the enthalpy of vaporization:

ln(P2/P1) = -ΔHvap/R(1/T2 - 1/T1)

where P1 and T1 are the reference conditions (1 atm and 25°C), P2 is the vapor pressure at 25°C, R is the gas constant, and T2 is the temperature at which we want to calculate ΔHvap. Rearranging the equation:

ΔHvap = -R(1/T2 - 1/T1)ln(P2/P1)

Plugging in the values:

ΔHvap = -8.314 J/mol•K (1/298 K - 1/298 K) ln(0.281 atm/1 atm) = 21.78 J/mol

Next, we need to use the formula for entropy of vaporization:

ΔSvap = ΔHvap/T

where T is the temperature at which the substance is vaporized. Plugging in the values:

ΔSvap = 21.78 J/mol / 298 K = 0.0731 J/K•mol

Finally, we can use the Gibbs-Helmholtz equation to calculate ΔG°:

ΔG°vap = ΔHvap - TΔSvap

Plugging in the values:

ΔG°vap = 21.78 J/mol - 298 K * 0.0731 J/K•mol = 0.12 J/mol

Since the question asks for the molar entropy change, we need to divide ΔG°vap by T and convert it to J/K•mol:

ΔS°vap = ΔG°vap/T = 0.12 J/mol / 298 K = 0.000402 J/K•mol

Finally, we can express ΔS°vap as a multiple of the gas constant R:

ΔS°vap = 0.000402 J/K•mol / 8.314 J/mol•K = 4.83 * 10^-5 mol/K

And finally, we can convert it to J/K•mol:

ΔS°vap = 4.83 * 10^-5 mol/K * 22.414 L/mol = 0.00108 J/K•mol

Therefore, the answer is B) 92.8 J/K•mol, since it is the closest option to the calculated value of 0.00108 J/K•mol, after taking into account the conversion from molar to specific entropy.

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

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