Acetic acid has a normal boiling point of 118 ∘c and a δhvap of 23.4 kj/mol. you may want to reference (page) section 11.5 while completing this problem. part a what is the vapor pressure (in mmhg) of acetic acid at 40 ∘c

The vapor pressure of acetic acid at 40 °C is 4569.4 mmHg.

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 mmHg and T₁ = 118 °C = 391 K.The second case: P₂ = ??? needed to be calculated and T₂ = 40 °C = 313 K.ΔHvap = 23.4 KJ/mole = 32.4 x 10³ J/mole and R = 8.314 J/mole.K.Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)ln(760 mmHg /P₂) = (23.4 x 10³ J/mole / 8.314 J/mole.K) (1/391 K - 1/313 K)ln(760 mmHg /P₂) = (2814.53) (-6.373 x 10⁻⁴) = - 1.8(760 mmHg / P₂) = 0.1663Then, P₂ = (760 mmHg) / (0.1663) = 4569.4 mmHg.

So, The vapor pressure of acetic acid at 40 °C is 4569.4 mmHg.

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