Multiple Choice
Using the Clausius-Clapeyron Equation, determine the normal boiling point of a substance whose vapor pressure is 55.1 mm Hg at 35°C and has a ΔHvap of 32.1 kJ/mol.
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13. Liquids, Solids & Intermolecular Forces
Clausius-Clapeyron Equation
Multiple Choice
The vapor pressure of SiCl4 is 100 mmHg at 5.4 °C, and the normal boiling point is 57.7 °C. What is the enthalpy of vaporization (ΔHvap) for SiCl4 in kJ/mol?
A
35.7 kJ/mol
B
45.3 kJ/mol
C
31.4 kJ/mol
D
40.1 kJ/mol
Verified step by step guidance1
To find the enthalpy of vaporization (ΔHvap) for SiCl4, we can use the Clausius-Clapeyron equation, which relates the vapor pressure and temperature to the enthalpy of vaporization. The equation is: , where R is the gas constant (8.314 J/mol·K).
First, convert the temperatures from Celsius to Kelvin. The conversion is done by adding 273.15 to the Celsius temperature. So, T1 = 5.4 °C + 273.15 = 278.55 K and T2 = 57.7 °C + 273.15 = 330.85 K.
Next, identify the vapor pressures at these temperatures. At T1 = 278.55 K, the vapor pressure P1 is given as 100 mmHg. At T2 = 330.85 K, the vapor pressure P2 is the normal boiling point pressure, which is 760 mmHg (since it's the atmospheric pressure at boiling point).
Now, substitute the values into the Clausius-Clapeyron equation: .
Finally, solve for ΔHvap. Rearrange the equation to isolate ΔHvap and calculate its value in J/mol, then convert it to kJ/mol by dividing by 1000. This will give you the enthalpy of vaporization for SiCl4.
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