Multiple Choice
Using Raoult's Law, calculate the vapor pressure of SiCl4 at 33.5 °C given that its vapor pressure is 100 mm Hg at 5.4 °C and ΔHvap = 30.2 kJ/mol.
14. Solutions
Vapor Pressure Lowering (Raoult's Law)
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While mercury is very useful in barometers, mercury vapor is toxic. Given that mercury has a ΔHvap of 59.11 kJ/mol and its normal boiling point is 356.7°C, calculate the vapor pressure in mmHg at room temperature, 25°C.
A
2 mmHg
B
20 mmHg
C
0.2 mmHg
D
0.002 mmHg
Verified step by step guidance1
Identify the known values: ΔHvap (enthalpy of vaporization) is 59.11 kJ/mol, the normal boiling point is 356.7°C, and the temperature at which we want to find the vapor pressure is 25°C.
Use the Clausius-Clapeyron equation to relate the vapor pressures at two different temperatures: \( \ln \left( \frac{P_2}{P_1} \right) = -\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \), where \( R \) is the ideal gas constant (8.314 J/mol·K).
Convert the temperatures from Celsius to Kelvin by adding 273.15 to each: \( T_1 = 356.7 + 273.15 \) K and \( T_2 = 25 + 273.15 \) K.
Assume the vapor pressure at the normal boiling point (\( T_1 \)) is 1 atm, which is equivalent to 760 mmHg. Substitute \( P_1 = 760 \) mmHg, \( T_1 \), \( T_2 \), and \( \Delta H_{\text{vap}} \) into the Clausius-Clapeyron equation.
Solve the equation for \( P_2 \), the vapor pressure at 25°C, by rearranging the equation to find \( P_2 \) and then exponentiating to remove the natural logarithm: \( P_2 = P_1 \times e^{-\frac{\Delta H_{\text{vap}}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right)} \).
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