Question

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.

Answer 1

Answer:

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:

$In\frac{P_2}{P_1}=\frac{\delta H_{vap}}{R}[\frac{T_2-T_1}{T_2T_1}]$

Substituting our values given; we have:

$In\frac{0.639}{1}=(\frac{34.4*10^3J/mol}{8.314 J K^{-1}mol^{-1}})[\frac{T_2-350}{350T_2}]$

$In({0.639})=(\frac{34.4*10^3}{8.314K^{-1}})[\frac{T_2-350}{350T_2}]$

$- 0.4479 = 41317.599 [\frac{T_2-350}{350T_2} ]K$

$-\frac{0.4479}{4137.599} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} = [\frac{T_2-350}{350T_2} ]$

$- 1.0825118*10^{-4} *350T_2 =T_2-350$

$- 0.037889T_2=T_2-350$

$350 = 0.03789T_2+T_2$

$350 =1.03789T_2$

$T_2= \frac{350}{1.03789}$

$T_2 = 337.22K$

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