Question

The normal boiling point of liquid methyl acetate is 331 K. Assuming that its molar heat of vaporization is constant at 30.6 kJ/mol, the boiling point of CH3COOCH3 when the external pressure is 1.29 atm is

Answer 1

Answer:

T_2=338.9026K

Boiling point of CH3COOCH3 at external pressure is 338.9026K

Explanation:

We are going to use Clausius-Clapeyron Equation:

$ln\frac{P_2}{P_1} =-\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

Where:

P_2 is the external pressure

P_1 is the atmospheric Pressure=1 atm

ΔH is the heat of vaporization

T_2  boiling point of CH3COOCH3 at external pressure

T_1 normal boiling point of liquid methyl acetate

Now:

$\frac{1}{T_2}=-ln\frac{P_2}{P_1}*\frac{R}{\Delta H}+\frac{1}{T_1} \\\frac{1}{T_2}=-ln\frac{1.29}{1}*\frac{8.314}{30.6*10^3}+\frac{1}{331} \\\frac{1}{T_2}=2.9507*10^-^3\\T_2=\frac{1}{2.9507*10^-^3} \\T_2=338.9026K$

Boiling point of CH3COOCH3 at external pressure is 338.9026K