Question

The heat of vaporization for benzaldehyde is 48.8 kj/mol, and its normal boiling point is 451.0 k. use this information to determine benzaldehyde's vapor pressure (in torr) at 61.5°c. report your answer to three significant digits.

Answer 1

Answer:

The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.

Explanation:

• To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
• The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
• The second case: P₂ = ??? needed to be calculated and T₂ = 61.5 °C = 334.5 K.
• ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
• Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
• ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)

• ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.

• (760 torr /P₂) = 0.01075
• Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.

So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.