Question

The vapor pressure of substance X is 100. mm Hg at 1080.°C. The vapor pressure of substance X increases to 600. mm Hg at 1220.°C. Determine the molar heat of vaporization of substance X using the derived form of the Clausius-Clapeyron equation given below. (Include the sign of the value in your answer.) ____ kJ/mol

Answer 1

The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The molar heat of vaporization of substance X is -61.78 kJ/mol.

Step-by-step explanation:

The molar heat of vaporization of substance X can be determined using the Clausius-Clapeyron equation. The equation is given by:

ln(P₂/P₁) = -(ΔHvap/R)((1/T₂) - (1/T₁))

We can solve for ΔHvap by substituting the values given: P₁ = 100 mm Hg, T₁ = 1080 °C (or 1353 K), P₂ = 600 mm Hg, and T₂ = 1220 °C (or 1493 K).

ln(600/100) = -(ΔHvap/8.314)((1/1493) - (1/1353))

Solving for ΔHvap gives us a value of -61.78 kJ/mol. Therefore, the molar heat of vaporization of substance X is -61.78 kJ/mol.