Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 15 °C?

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Identify the known values: heat of vaporization (\(\Delta H_{vap}\)) = 38.56 \text{ kJ/mol}, T_1 = 78.4 \degree C = 351.4 \text{ K}, P_1 = 1 \text{ atm}, T_2 = 15 \degree C = 288.15 \text{ K}.

Use the Clausius-Clapeyron equation: \(\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\), where \(R = 8.314 \text{ J/mol K}\).

Convert \(\Delta H_{vap}\) from kJ/mol to J/mol by multiplying by 1000.

Substitute the known values into the Clausius-Clapeyron equation to solve for \(\ln\left(\frac{P_2}{1 \text{ atm}}\right)\).

Exponentiate both sides to solve for \(P_2\), the vapor pressure at 15 \degree C.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. For ethanol, this value is 38.56 kJ/mol, indicating the energy needed to overcome intermolecular forces during the phase transition from liquid to gas.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. For ethanol, the normal boiling point is 78.4 °C, meaning at this temperature, ethanol will transition from liquid to gas under standard atmospheric pressure.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It reflects the tendency of molecules to escape from the liquid phase into the gas phase. The vapor pressure of ethanol at 15 °C can be estimated using the Clausius-Clapeyron equation, which relates vapor pressure to temperature and heat of vaporization.

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