The standard molar enthalpy of fusion and the standard molar enthalpy of vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the standard molar entropy changes for the solid ↔ liquid and liquid ↔ vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
A. ΔS°fus = 22.6 J/K·mol; ΔS°vap = 67.1 J/K·mol
B. ΔS°fus = 39.1 J/K·mol; ΔS°vap = 87.8 J/K·mol
C. ΔS°fus = 19.8 J/K·mol; ΔS°vap = 3.87 J/K·mol
D. ΔS°fus = 46.1 J/K·mol; ΔS°vap = 101. J/K·mol
I got "B". Can anyone verify?
That's right. Good work.
To calculate the standard molar entropy changes for the solid ↔ liquid and liquid ↔ vapor transitions for benzene, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)
Where P1 and P2 are the initial and final pressures, ΔHvap is the standard molar enthalpy of vaporization, R is the ideal gas constant, T1 and T2 are the initial and final temperatures.
For the solid ↔ liquid transition:
T1 = 5.5°C = 278.65 K
T2 = T1 (melting point)
P1 = P2 = 1 atm
ln(1/1) = ΔHfus/R * (1/T1 - 1/T2)
Simplifying:
0 = ΔHfus/R * (1/278.65 - 1/T2)
For the liquid ↔ vapor transition:
T1 = T2 (boiling point)
T2 = 80.1°C = 353.25 K
P1 = P2 = 1 atm
ln(1/1) = ΔHvap/R * (1/T1 - 1/353.25)
Simplifying:
0 = ΔHvap/R * (1/T2 - 1/353.25)
To calculate the standard molar entropy change (ΔS) for each transition, we can use the equations:
ΔSfus = ΔHfus/T
ΔSvap = ΔHvap/T
Let's substitute the values given:
ΔHfus = 10.9 kJ/mol = 10.9 * 10^3 J/mol
ΔHvap = 31.0 kJ/mol = 31.0 * 10^3 J/mol
T = 278.65 K for the solid ↔ liquid transition
T = 353.25 K for the liquid ↔ vapor transition
Calculations:
ΔSfus = ΔHfus/T = (10.9 * 10^3 J/mol) / 278.65 K = 39.1 J/K·mol
ΔSvap = ΔHvap/T = (31.0 * 10^3 J/mol) / 353.25 K = 87.8 J/K·mol
Therefore, the correct answer is:
B. ΔS°fus = 39.1 J/K·mol; ΔS°vap = 87.8 J/K·mol
To calculate the standard molar entropy changes for the solid ↔ liquid and liquid ↔ vapor transitions, we can use the following equations:
ΔS°fus = ΔH°fus / T
ΔS°vap = ΔH°vap / T
where ΔH°fus is the standard molar enthalpy of fusion, ΔH°vap is the standard molar enthalpy of vaporization, and T is the temperature in Kelvin.
First, we need to convert the temperatures from Celsius to Kelvin:
T_melting = 5.5°C + 273.15 = 278.65 K
T_boiling = 80.1°C + 273.15 = 353.25 K
Now we can calculate the standard molar entropy changes:
ΔS°fus = 10.9 kJ/mol / 278.65 K = 0.0391 kJ/(mol·K) = 39.1 J/(mol·K)
ΔS°vap = 31.0 kJ/mol / 353.25 K = 0.0878 kJ/(mol·K) = 87.8 J/(mol·K)
Therefore, the correct answer is:
B. ΔS°fus = 39.1 J/K·mol; ΔS°vap = 87.8 J/K·mol