Question

What will be melting point $KCl$ if enthalpy change for the reaction is $7.25{\text{ J}}mo{l^{ - 1}}$ and entropy is $0.007{\text{ J }}{K^{ - 1}}mo{l^{ - 1}}?$
(A) 1835.2 K
(B) 173 K
(C) 1035.7 K
(D) 1285.2 K

Answer 1

To calculate the melting point, we will use the concept of Gibbs free energy change i.e. $\Delta G = \Delta H - T\Delta S$

Complete step by step answer:
As you know, Gibbs free energy is the quantity that used to measure the max. amount of work done in a thermodynamics system when temperature and pressure are kept constant.
It is denoted by symbol $'G'$
Units: Joules or Kilojoules (J or KJ)
Moreover, it can also be defined as the max. amount of work that can be extracted from a closed system.
Its equations is $G = H - TS$
G = Gibbs free energy
H = enthalpy
T = temperature
S = entropy
It is a state function so it doesn’t depend on the path. So, change in Gibbs free energy is equal to change in enthalpy minus the product of temperature and entropy change.
For the reaction, given in question.
$KCl\left( s \right) \rightleftharpoons KCl\left( l \right)$
At equilibrium, $\Delta G = 0$
So, from Gibbs free energy equation,
$\Delta G = \Delta H - T\Delta S$
$\Delta H = T\Delta S$
$T = \dfrac{{\Delta H}}{{\Delta S}}$
$T = \dfrac{{7.25}}{{0.007}} = 1035.7K$

So, the correct answer is Option C .

Additional Information:
$\Delta G = \Delta H - T\Delta S$
This equation is known as the Gibbs Helmholtz equation.
If, $\Delta G > 0,$ reaction is non- spontaneous
$\Delta G = 0$, reaction is at equilibrium
$\Delta G < 0,$ reaction is spontaneous and exergonic

Note:
The free energy change is the only criteria that explains the spontaneity of a process and also explains the dependence of reactions on temperature to become spontaneous. Hence, the temperature or the melting point required in the question is calculated through this equation.