Question

Enthalpy change for the process,
$H_{2}O\left(ice\right)\rightleftharpoons H_{2}O\left(water\right)$ is $6.01KJ{mol}^{-1}$. The entropy change of $1$ mole of ice at its melting point will be:

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

Answer 1

The correct option is B $22J{K}^{-1}$ ${mol}^{-1}$

Solution:- (B) $22J/mol-K$

Given:-

$\Delta H=6.01kJ/mol=6010J/mol$

As the melting point of ice is $0℃$, i.e., $273K$

Now using,

$\Delta S=\frac{\Delta H}{T}$

$\therefore \Delta S=\frac{6010}{273}=22J/mol-K$

Hence the entropy change of $1$ mole of ice at its melting point will be $22J/mol-K$.