Question

Enthalpy of fusion of water is $6.01kJmo{{l}^{-1}}$. The entropy change of 1 mole of ice at its melting point will be:
A) $22kJmo{{l}^{-1}}$
B) $109kJmo{{l}^{-1}}$
C) $44kJmo{{l}^{-1}}$
D) $11kJmo{{l}^{-1}}$

Answer 1

Hint: The concept of the physical chemistry which deals with the thermodynamics chapter where the calculation of entropy is done using the formula for Gibbs free energy change which in this case is for equilibrium condition between solid and liquid phase of water and the formula is $\Delta G = \Delta H-T\Delta S$

Complete Solution :
We have studied the concept of thermodynamics which comes under the physical chemistry part and are familiar with some of the entities like enthalpy, entropy, Gibbs free energy and many other parameters.
We shall now see what entropy is and how it can be calculated.
- Entropy is the thermodynamic quantity representing the unavailability of a system’s thermal energy to convert into mechanical work and in simple words, it can also be defined as a state of disorder in the system.
- In the above data, they have given about the change of state of water from solid form to the liquid form that is,
\[{{H}_{2}}O(s)\to {{H}_{2}}O(l)\]
- Here, enthalpy of fusion or latent heat of fusion is the change in enthalpy that results from providing energy to specific quantities of the substance and here it is the change from solid to the liquid state.
- Since, at melting point, the solid state is in equilibrium with the liquid state that is, $sl$, the Gibbs free energy change will be as per the formula:
$\Delta G = \Delta H-T\Delta S$
where, $\Delta H$ is the change in enthalpy, T is temperature in Kelvin and here it is ${{0}^{0}}C=273K$
$\Delta S$ is the change in entropy of the system.
- Now, because of the equilibrium conditions the Gibbs free energy change will be zero i.e. $\Delta G=0$
Therefore, the above formula becomes,
$\Delta G = \Delta H-T\Delta S = 0$
$\Rightarrow \Delta H = T\Delta S$
\[\Rightarrow \Delta S = \dfrac{\Delta H}{T}\]
By substituting the values,
\[\Delta S =\dfrac{6.01\times 1000J/mol}{273K} = 22kJmo{{l}^{-1}}\]
So, the correct answer is “Option A”.

Note: Do not be confused about the terms enthalpy and entropy. Enthalpy of the system is the measure of total heat present in the thermodynamic system at constant pressure whereas entropy is the measure of disorder in the thermodynamic system.