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At ${27^\circ }C$, Latent heat of fusion of a compound is $2930J/mol$. Entropy change during fusion is:
A.$9.77J/mol - K$
B.$10.77J/mol - K$
C.$9.07J/mol - K$
D.$0.977J/mol - K$

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Last updated date: 20th Jun 2024
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Answer
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Hint: Entropy could be a function of the state of the system, that the change in entropy of a system is decided by its initial and final states. Within the idealization that a process is reversible, the entropy doesn't change, while irreversible processes always increase the full entropy.
Formula used: Entropy change can be calculated by the simple formula $\Delta S = \dfrac{{\Delta {H_f}}}{T}$, for a system in equilibrium where T is the temperature in kelvin, $\Delta S$is the entropy change in $J/mol - K$ and $\Delta {H_f}$is the enthalpy change in $J/mol$

Complete step by step answer:
Enthalpy $(H)$ is defined because the amount of energy released or absorbed during a chemical action. The enthalpy change of a reaction is roughly such as the number of energy lost or gained during the reaction. A reaction is favoured if the enthalpy of the system decreases over the course of the reaction.
Entropy $(S)$ defines the degree of randomness or disorder during a system. The property that matter possesses in terms of the way energy is dispersed is understood as its entropy. In an exceedingly spontaneous process, the entropy of the energy increases. The entropy increases with mass, melting, vaporization or sublimation, chemical complexity, etc
The free energy $(G)$ gives the relation between both terms as shown below:
$G = H - TS$
Where at constant temperature, the change of free energy is expressed as:
$\Delta G = \Delta H - T\Delta S$
Where $\Delta G$ is the change in gibbs free energy, $\Delta H$is the change in enthalpy, $\Delta S$ is the entropy change and T is the temperature in kelvin
Thus, the free energy changes can provide a relation between enthalpy and entropy.
For a system in equilibrium, $\Delta G = 0$, which gives us the formula below:
$\Delta S = \dfrac{{\Delta H}}{T}$- i)
Given,
$T = {27^\circ }C$
$T(K) = T(^\circ C) + {273^\circ }C$
So, the temperature in kelvin will be $ = 300K$
Given,
$\Delta H = 2930$J/mol
So, substituting the value of $\Delta {\rm H}$in the formula above i), we get
$\Delta S = \dfrac{{2930J/mol}}{{300K}}$
On solving we get,
$\Delta S = 9.77J/mol - K$
So, the correct answer is (A).

Note:
-The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant only if all the processes are reversible.
-It also states that the changes in entropy in the universe can never be negative.
$\Delta {S_{univ}} = \Delta {S_{sys}} + \Delta {S_{surr}} > 0$