Question

Rhombic and monoclinic Sulphur are at equilibrium at the transition temperature ($95^\circ C$). The enthalpy of transition at $95.5^\circ C$ is 353.8 J $mo{l^{ - 1}}$. The entropy of transition is:
A. 0.064 J ${K^{ - 1}}mo{l^{ - 1}}$.
B. 0.96 J ${K^{ - 1}}mo{l^{ - 1}}$.
C. 0.016 J ${K^{ - 1}}mo{l^{ - 1}}$.
D. 0.008 J ${K^{ - 1}}mo{l^{ - 1}}$.

Answer 1

The entropy of the reaction and enthalpy of the reaction are not directly related with each other. However, with the help of free energy, both the entities can be regrouped in terms of G = H –TS.

Complete step by step answer:
Given,
The transition temperature is $95.5^\circ C$= 368.5 K
Enthalpy of transition is 353.8 J $mo{l^{ - 1}}$.
The amount of energy released or absorbed during the reaction is termed as enthalpy. The enthalpy is denoted by H.
The degree of randomness in a system is termed as the entropy. The entropy is denoted by S.
The free energy change gives the relation between entropy and enthalpy.
At constant temperature, the change in free energy is given as shown below.
$\Delta G = \Delta H - T\Delta S$
For the system at equilibrium the free energy change is zero $\Delta G = 0$.
It is given that the rhombic and monoclinic Sulphur are at equilibrium at the transition temperature.
Then the equation is given as shown below.
$\Delta S = \dfrac{{\Delta H}}{T}$
To calculate the entropy, substitute the values in the above equation.
$\Delta S = \dfrac{{353.8}}{{368.5}}$
$\Rightarrow S = 0.96J{K^{ - 1}}mo{l^{ - 1}}$
Thus, the entropy of transition between monoclinic Sulphur and Rhombic Sulphur is $0.96J{K^{ - 1}}mo{l^{ - 1}}$
Therefore, the correct option is B.

Note:
At a temperature more than $95.5^\circ C$, the entropy value is greater than the enthalpy value. Therefore the free energy change $\Delta G$ is negative that means the reaction will move in a forward direction.