Question

1. A handbook lists the normal boiling point of an organic liquid as 42.6 Celsius and it’s enthalpy of vaporization as 24.9 kJ/mol. Calculate the vapor pressure in torr of this compound at 25.0 Celsius

2. Hydrazine N2H4 has a normal boiling point of 113.5 Celsius and a critical point at 380.0 Celsius and 145.4 atm. Estimate the vapor pressure of hydrazine at 75.0 Celsius (treat the critical point as last point on the vapor pressure curve)

3. The normal boiling point of a liquid is 214.1 Celsius and it’s enthalpy of vaporization at the normal boiling point is 68.2 kJ/mol. At what temperature will the liquid have a vapor pressure of 475.0 torr?

Answer 1

Answer on Question #72621-Chemistry-Organic Chemistry

1. A handbook lists the normal boiling point of an organic liquid as 42.6 Celsius and it's enthalpy of vaporization as $24.9\ \mathrm{kJ/mol}$. Calculate the vapor pressure in torr of this compound at 25.0 Celsius

Solution

$$\ln \frac{p_1}{p_2} = \frac{H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ $$\ln \frac{760}{p_2} = \frac{24900}{8.31} \left(\frac{1}{214.1 + 25} - \frac{1}{214.1 + 42.6}\right)$$ $$p_2 = 322\ \text{torr}.$$

2. Hydrazine N2H4 has a normal boiling point of 113.5 Celsius and a critical point at 380.0 Celsius and 145.4 atm. Estimate the vapor pressure of hydrazine at 75.0 Celsius (treat the critical point as last point on the vapor pressure curve)

Solution

$$\ln \frac{p_1}{p_2} = \frac{H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ $$\ln \frac{145.4}{1} = \frac{H}{8.31} \left(\frac{1}{113.5 + 273.15} - \frac{1}{380 + 273.15}\right)$$ $$H = 39212\ \frac{J}{mol}$$ $$\ln \frac{145.4}{p_2} = \frac{39212}{8.31} \left(\frac{1}{75 + 273.15} - \frac{1}{380 + 273.15}\right)$$ $$p_2 = 0.259\ \text{atm.}$$

3. The normal boiling point of a liquid is 214.1 Celsius and it's enthalpy of vaporization at the normal boiling point is $68.2\ \mathrm{kJ/mol}$. At what temperature will the liquid have a vapor pressure of 475.0 torr?

Solution

$$\ln \frac{p_1}{p_2} = \frac{H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ $$\ln \frac{760}{475} = \frac{68200}{8.31} \left(\frac{1}{T_2} - \frac{1}{214.1 + 273.15}\right)$$ $$T_2 = 474.02\ K = 474.02 - 273.15 = 200.87\ ^{\circ}\mathrm{C}$$

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