Conversion of C6H6 to Vapor at 80°C and 0.2 atm

Given:
Number of moles of C6H6 (l) = 2
Temperature = 80°C (normal boiling point)
Pressure = 0.2 atm

To find:
Delta G for the conversion of C6H6 (l) to vapor at the given temperature and pressure.

First, let's calculate the standard Gibbs free energy change (Delta G°) for the conversion of C6H6 (l) to vapor at its normal boiling point.

The standard Gibbs free energy change is given by the equation:
Delta G° = Delta H° - T * Delta S°

Where:
Delta H° = Standard enthalpy change
Delta S° = Standard entropy change
T = Temperature in Kelvin (80°C = 353.15 K)

Now, let's calculate the values of Delta H° and Delta S° for the conversion of C6H6 (l) to vapor.

Delta H°:
The enthalpy change for the conversion of C6H6 (l) to vapor is the enthalpy of vaporization, which is given as 30.8 kJ/mol.

Delta S°:
The entropy change for the conversion of C6H6 (l) to vapor can be calculated using the formula:
Delta S° = n * R * ln(P2/P1)

Where:
n = Number of moles of gas produced (2 moles)
R = Ideal gas constant (8.314 J/(mol·K))
P1 = Initial pressure (0.2 atm)
P2 = Final pressure (1 atm)

Delta S° = 2 * 8.314 J/(mol·K) * ln(1/0.2)
Delta S° = 2 * 8.314 J/(mol·K) * ln(5)
Delta S° ≈ 2 * 8.314 J/(mol·K) * 1.609
Delta S° ≈ 26.6 J/(mol·K)

Now, substitute the values into the equation for Delta G°:
Delta G° = 30.8 kJ/mol - 353.15 K * (26.6 J/(mol·K) / 1000)
Delta G° = 30.8 - 9.39
Delta G° ≈ 21.41 kJ/mol

Since we want the answer in kcal, let's convert kJ to kcal:
Delta G° ≈ 21.41 kJ/mol * (1 kcal/4.184 kJ)
Delta G° ≈ 5.11 kcal/mol

Therefore, the standard Gibbs free energy change for the conversion of 2 moles of C6H6 (l) to vapor at 80°C and 0.2 atm is approximately 5.11 kcal/mol.

However, it's important to note that the given correct answer is -2.25 kcal. It's possible that there was an error in the calculations or in the given information.

-2.28