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The enthalpy of vaporisation of a liquid is 30 kJ/mol and entropy of vaporisation is 75 $Jmol^{-1}K^{-1}$. What is the boiling point of the liquid at 1 atm?

(a) 250 K
(b) 400 K
(c) 450 K
(d) 600 K

1 Answer

Answer: 400 K
$\Delta H$ = 30 kJ/mol = 30,000 J/mol
$\Delta S$ = 75 $Jmol^{-1}K^{-1}$
At boiling point, the reversible process liquid $\leftrightharpoons$ vapour is in equilibrium at one atmospheric pressure.
$\therefore \Delta G=0$
$\Delta G = \Delta H - T \Delta S$
$\therefore T = \frac{\Delta H}{\Delta S} = \frac{30,000 J mol^{-1}}{75 JK^{-1}mol^{-1}} = 400$ K
answered Feb 28, 2014 by mosymeow_1

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