Question

What is the normal boiling point of mercury?
Given: $\mathrm{\Delta }{H}_f^{\circ }(Hg,(l))=0;S^{\circ }(Hg,(l))=77.4Jmo{l}^{-1}{K}^{-1}$
 $\mathrm{\Delta }{H}_f^{\circ }(Hg,(g))=60.8kJ/mol;S^{\circ }(Hg,(g))=174.4Jmo{l}^{-1}{K}^{-1}$
(a) $624.8K$
(b) $626.8K$
(c) $636.8K$
(d) None of these

Answer 1

Hint: Gibbs free energy is used for determining the spontaneity of a process, we measure the change in Gibbs energy. When a process takes place at constant temperature and pressure, we can write the formula of Gibbs energy in terms of Enthalpy and entropy of the system. We will use this formula to calculate the boiling point of mercury.

Complete answer:
Gibbs’s free energy is used to calculate the maximum reversible work that may be done by the system at constant temperature and pressure. Enthalpy is the sum of the total internal energy of the system and the product of the pressure and volume of the system. Entropy is defined as the measure of thermal energy per unit temperature that is unavailable for doing work. Now as we know about these factors let’s write the equation of Gibbs energy
 $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$
where $\mathrm{\Delta }G$ is the change in Gibbs free energy, $\mathrm{\Delta }H$ is the change in enthalpy, $\mathrm{\Delta }S$ is the change in entropy and T is the temperature.
Let’s write the given values
Enthalpy of formation of Mercury (liquid), $\mathrm{\Delta }{H}_f^{\circ }(Hg,(l))=0$
Enthalpy of formation of Mercury (gas), $\mathrm{\Delta }{H}_f^{\circ }(Hg,(g))=60.8kJ/mol$
Entropy of Mercury (liquid), $S^{\circ }(Hg,(l))=77.4Jmo{l}^{-1}{K}^{-1}$
Entropy of Mercury (gas), $S^{\circ }(Hg,(g))=174.4Jmo{l}^{-1}{K}^{-1}$
Calculating change in Enthalpy
 $\mathrm{\Delta }H=\mathrm{\Delta }{H}_f^{\circ }(Hg,(g))-\mathrm{\Delta }{H}_f^{\circ }(Hg,(l))$
we get
 $\mathrm{\Delta }H=60.8kJ/mol$
Now calculating change in entropy
 $\mathrm{\Delta }S=S^{\circ }(Hg,(g))-S^{\circ }(Hg,(l))$
 $\mathrm{\Delta }S=174.4Jmo{l}^{-1}{K}^{-1}-77.4Jmo{l}^{-1}{K}^{-1}$
we get
 $\mathrm{\Delta }S=97Jmo{l}^{-1}{K}^{-1}$
At equilibrium the change in Gibbs energy is zero and we have calculated the values of change in entropy and change in enthalpy. Now substituting calculated values
 $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$
 $0=60.8kJ/mol-T\times 97Jmo{l}^{-1}{K}^{-1}$
we get
 $T={\displaystyle \frac{60.8\times 1000}{97}}$
So, the required temperature is
 $T=626.8K$
So, option (b) is the right answer.

Note:
When doing the calculation, remember the unit of each quantity, you may need to convert a unit into another for getting the answer. In the question we converted Kilojoule to joule as a unit of entropy was in joule.