Question

The following data, \[{\rm{\Delta H}} = 9.2\,{\rm{KJmo}}{{\rm{l}}^{ - 1}}\], \[{\rm{\Delta S}} = 0.008\,{\rm{KJ}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]is known about the melting point of a compound AB. Its melting point is:
A. \[{\rm{736}}\,{\rm{K}}\]
B. \[1050\,{\rm{K}}\]
C. \[1150\,{\rm{K}}\]
D. \[1250\,{\rm{K}}\]

Answer 1

Hint: The melting point of a compound may be defined as the temperature at which the solid phase changes into the liquid phase. The melting point of a compound is influenced by intermolecular forces. The melting point of a solid is taken to be the same as the freezing point of the liquid.

Formula used Gibb’s free energy equation can be used to find the melting point of the compound. The relationship is as shown below.
\[{\rm{\Delta G = \Delta H}} - {\rm{T\Delta S}}\]
where, \[{\rm{\Delta G}}\]= change in Gibb’s free energy
\[{\rm{\Delta H = }}\]change in enthalpy
\[{\rm{\Delta S = }}\]change in entropy
\[{\rm{T = }}\]temperature in Kelvin

Complete Step by Step Solution:
Enthalpy change (\[{\rm{\Delta H}}\]) is defined as the total heat content of the system at constant pressure.
Entropy is the extent of disorder of randomness in a system. Entropy change (\[{\rm{\Delta S}}\]) of a substance measures the disorder or randomness in a system.
Gibb’s free energy of a system may be defined as the maximum amount of energy available to a system that can be converted into useful work. In simple words, Gibb’s free energy is the capacity of a system to do useful work. It is denoted by the symbol ‘G’.
Gibb’s free energy equation gives the relationship between change in enthalpy, change in entropy and temperature. The relationship is as: \[{\rm{\Delta G = \Delta H}} - {\rm{T\Delta S}}\]
As per the given data,
Change in enthalpy, \[{\rm{\Delta H}} = 9.2\,{\rm{KJmo}}{{\rm{l}}^{ - 1}}\]
Change in entropy, \[{\rm{\Delta S}} = 0.008\,{\rm{KJ}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}\]
Temperature, \[{\rm{T}}\]=?
Gibb’s free energy change (\[{\rm{\Delta G}}\]) is taken to be zero at the melting point. So, \[{\rm{\Delta G = 0}}\]
Find the melting point (\[{\rm{T}}\]) by using Gibb’s energy equation as shown below.
${\rm{\Delta G = \Delta H}} - {\rm{T\Delta S}}\\$
$\Rightarrow {\rm{0 = \Delta H}} - {\rm{T\Delta S}}\\$
$ \Rightarrow {\rm{\Delta H}} = {\rm{T\Delta S}}\\$
$\Rightarrow {\rm{T}} = \frac{{{\rm{\Delta H}}}}{{{\rm{\Delta S}}}}\\ $
$\Rightarrow T = \dfrac{9.2 \ kJ.mol^{-1}}{0.008 kJ.K^{-1}mol^{-1}}\\
\Rightarrow T = 1150 \ K$
Hence, the melting point of the compound AB is found to be as $1150 \ K$.
Therefore, option C is correct.

Note: Gibb’s energy concept is more useful than the entropy concept to know the feasibility of a process because \[{\rm{\Delta G}}\]refers to the system only while \[{\rm{\Delta S}}\]refers to both system and surroundings.