Question

The enthalpy of vapourisation of liquid diethyl ether, $(C_{2}H_{5})_{2}O$, is 26.0 kJ $mol^{-1}$ at its boiling point (35.0°C). Calculate ΔS for the conversion of liquid to vapour and vapour to liquid at 35.0°C.

Answer 1

T = ∆H/∆S

So,

∆S = ∆H/T

26000/308

= 84.4 Joule per Kelvin.

Answer 2

"Standard enthalpy of vaporization:

Enthalpy change when one mole of liquid completely changes to vapour at its boiling point under standard pressure of 1 bar.

$\Delta { S }_{ vap }\quad =\quad \frac { \Delta { H }_{ vap } }{ T }$

From the given,

Enthalpy of vaporization of Diethyl ether = 26 KJ

Temperatures = 35 + 273 = 308

$\Delta { S }_{ vap }\quad =\quad \frac { 26\quad \times \quad { 10 }^{ 3 } }{ 308 } \quad =\quad +84.41\quad J{ K }^{ -1 }{ mol }^{ -1 }$

The phase changes from vapour to liquid called condensation.

$\Delta S_{ cond. }\quad =\quad \frac { \Delta { H }_{ cond } }{ T } \quad =\quad \frac { 26\quad \times \quad { 10 }^{ 3 } }{ 308 } \quad =\quad +84.41\quad J{ K }^{ -1 }{ mol }^{ -1 }$

${ H }_{ cond }\quad =\quad -26\quad K\quad =\quad -84.41\quad J{ K }^{ -1 }{ \quad mol } ^{ -1 }$"