Question

The ∆G° of vaporization for benzene at 298 K and 1.00 atm is 4.883 kJ/mol. Calculate the pressure, in atm, of benzene vapor in equilibrium with benzene liquid at 298 K.

Answer 1

To calculate the pressure of benzene vapor in equilibrium with benzene liquid at 298K, we can use the Clausius-Clapeyron equation.

Step-by-step explanation:

To calculate the pressure of benzene vapor in equilibrium with benzene liquid at 298K, we can use the Clausius-Clapeyron equation: ln(P2/P1) = (∆Hvap/R)(1/T1 - 1/T2), where P1 is the given pressure (1.00 atm), T1 is the given temperature (298K), T2 is the boiling point temperature of benzene, ∆Hvap is the enthalpy of vaporization (-4.883 kJ/mol), and R is the ideal gas constant (8.314 J/(mol K)).

First, we need to convert the given pressure to Pa: 1 atm = 101.325 kPa = 101325 Pa. Then, we can rearrange the equation to solve for T2: T2 = (∆Hvap/R)(1/T1 - ln(P2/P1)). Plug in the values and solve for T2.

Once we have the boiling point temperature (T2), we can convert it to Celsius if needed.