The normal boiling point of liquid ethyl acetate is 350 K. Assuming that its molar heat of vaporization is constant at 34.4 kJ/mol, the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is K.

337.22 K

Explanation:

Given that:

P₁ = 1 atm

T₁ = 350 K

P₂ = 0.639 atm

T₂ = ??? (unknown)

R(rate constant) =  8.34 J k⁻¹ mol⁻¹

Using Clausius-Clapeyron equation, we can determine the final boiling point of the process.

Clausius-Clapeyron equation can be written as:

Substituting our values given; we have:

∴ the boiling point of CH3COOC2H5 when the external pressure is 0.639 atm is 337.22 K.

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