Question

The vapor pressure of benzene (l) is given by ln P = 20.767 - 2773.8T - 53.08, where T and P have the units of K and Pa. (a) Calculate the standard boiling point temperature. (b) Calculate ∆vapH° at 298 K and at the standard boiling point temperature, in the unit of kJ mol-1.

Answer 1

The standard boiling point temperature of benzene is approximately -0.012 °C. The equation for ΔvapH° is not provided in the given information.

Step-by-step explanation:

To calculate the standard boiling point temperature of benzene (l) using the given equation ln P = 20.767 - 2773.8T - 53.08, we can set ln P equal to 0 and solve for T:

0 = 20.767 - 2773.8T - 53.08

2773.8T = 20.767 - 53.08

2773.8T = -32.313

T = -32.313 / 2773.8

T = -0.01165 K

Therefore, the standard boiling point temperature of benzene is approximately -0.012 °C.

To calculate ΔvapH° at 298 K and at the standard boiling point temperature, we need the equation for ΔvapH°. Unfortunately, it is not provided in the given information.