Question

Enthalpy change for the process,
$H_2O$(ice) ${\rightleftharpoons} H_2O$(water)
is $6.01\, kJ\, mol^{-1}$. The entropy change of $1$ mole of ice at its melting point will be

• A. $12 \,J \,K^{-1}\, mol^{-1}$
• B. $22 \,J \,K^{-1}\, mol^{-1}$
• C. $100 \,J \,K^{-1}\, mol^{-1}$
• D. $30 \,J \,K^{-1}\, mol^{-1}$

Answer 1

Answer: (B)

$\Delta S_{fusion} = \frac{\Delta H_{fusion}}{T_{fusion}}$
$= \frac{6.01 \times 1000}{273}$
$= 22 \,J \,K^{-1}\, mol^{-1}$