We can determine the heat of vaporization by using the Clausius–Clapeyron equation if we know the vapor pressures of a substance at two different temperatures. Determine the heat of vaporization of diethyl ether if the vapor pressure is 1.0 mm Hg at –74.3 °C and 425 mm Hg at 18.7 °C.

-9.2 × 10⁴ J/mol

Explanation:

Step 1: Convert the temperatures to Kelvin

We will use the following expression.

K = °C + 273.15

T₁: K = 74.3°C + 273.15 = 347.5 K

T₂: K = 18.7°C + 273.15 = 291.9 K

Step 2: Convert the pressures to Pascal (SI unit)

We will use the conversion factor 760 mmHg = 101325 Pa.

P₁: 1.0 mmHg × 101325 Pa/760 mmHg = 1.3 × 10² Pa

P₂: 425 mmHg × 101325 Pa/760 mmHg = 5.67 × 10⁴ Pa

Step 3: Determine the heat of vaporization of diethyl ether

We will use Clausius–Clapeyron equation.

ln (P₂/P₁) = -ΔHvap/R × (1/T₂ - 1/T₁)

ln (5.67 × 10⁴ Pa/1.3 × 10² Pa) = -ΔHvap/(8.314 J/mol.K) × (1/291.9 K - 1/347.5 K)

ΔHvap = -9.2 × 10⁴ J/mol

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your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.