Question

Organic liquids such as benzene can be used as nonpolar solvents, but some have significant vapor pressures at room temperature. The vapor pressure of benzene at 25°C is 95.0 torr, and benzene's normal boiling point is 80.0°C. Calculate an estimate of the heat of vaporization of benzene.

Answer 1

The heat of vaporization of benzene can be estimated using the Clausius-Clapeyron equation with data on its vapor pressure at room temperature, and its normal boiling point.

Step-by-step explanation:

To estimate the heat of vaporization of benzene, you can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance at different temperatures to its heat of vaporization (∆Hvap). The equation is ln(P1/P2) = (∆Hvap/R)((1/T2) - (1/T1)), where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, R is the gas constant. In this case, we have the vapor pressure at room temperature (P1 = 95.0 torr) and the normal boiling point of benzene (T2 = 80°C or 353.15 K).

The boiling point temperature must be converted from °C to Kelvin (T1 = 25°C or 298.15 K). We also need to convert the vapor pressure from torr to atm, because the gas constant R is commonly given in units of L·atm/mol·K. Using the conversion 1 atm = 760 torr, P1 = 95.0 torr * (1 atm / 760 torr) = 0.125 atm. The normal boiling point corresponds to the vapor pressure equal to standard atmospheric pressure, so P2 = 1 atm.

Plugging in these values into the Clausius-Clapeyron equation and solving for ∆Hvap gives us the heat of vaporization. The gas constant R is typically 0.0821 L·atm/mol·K. Rearranging and solving for ∆Hvap gives us ΔHvap = -R * ln(P1/P2) * ((T1 * T2) / (T2 - T1)). Substituting in the known values, we can calculate the estimate for the heat of vaporization of benzene.