Question

Consider the decomposition of red mercury(II) oxide under standard state conditions.

\({\bf{2HgO(s, red )}} \to {\bf{2Hg(l) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Short answer

a) This reaction is not spontaneous under standard state conditions.

b) The reaction becomes spontaneous at the temperature of \(839.39\;{\rm{K}}\).

Step-by-step solution

Determining the spontaneous reaction.

• The definition of spontaneous reaction in science is a reaction that occurs in a specified set of conditions without interference. A spontaneous reaction is one that occurs in the absence of intervention in a given set of conditions. A spontaneous reaction is completed without the assistance of anyone else.
• A reaction is spontaneous if the overall entropy, or disorder, increases, according to the Second Law of Thermodynamics.

Calculating the Gibbs free energy change.

a) In order to determine whether the reaction is spontaneous under standard state conditions, we have to calculate the Gibbs energy change. The reaction is spontaneous under standard state conditions if the Gibbs energy change for the reaction is less than zero.

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

\(\Delta G = 2{G_f}(Hg,l) + {G_f}\left( {{O_2},g} \right) - 2{G_f}({\rm{HgO}},{\rm{ red }})\)

\(\Delta G = (2 \cdot 0 + 0 - 2 \cdot ( - 58.5))\frac{{kJ}}{{mol}}\)

\(\Delta G = 117\frac{{kJ}}{{mol}}\)

This reaction is not spontaneous under standard state conditions.

Determining whether the reaction is spontaneous at room temperature.

b) To determine the temperature at which the reaction becomes spontaneous, we can use the formula

\(\Delta G = \Delta H - T\Delta S\)

We can calculate the \(\Delta H\)and the \(\Delta S\)from the standard state entropies from the table in Appendix \(G\)in the book.

\(\Delta H = {H_f}({\rm{ products }}) - {H_f}({\rm{ reactants }})\)

\(\Delta H = 2{H_f}(Hg,l) + {H_f}\left( {{O_2},g} \right) - 2{H_f}(HgO,{\rm{ red }})\)

\(\Delta H = (2 \cdot 0 + 0 - 2 \cdot ( - 90.83))\frac{{kJ}}{{mol}}\)

\(\Delta H = 181.66\frac{{kJ}}{{mol}}\)

If we write in J per mol it is:

\(\Delta H = 181660\frac{J}{{mol}}\)

We will now calculate \(\Delta S\)

\(\Delta S = S({\rm{ products }}) - S({\rm{ reactants }}){\rm{ }}\)

\(\Delta S = 2S(Ha,l) + S(O,q) - 2S(HaO,{\rm{ red }})\)

\(\Delta S = 216.42\frac{J}{{K \cdot mol}}\)

We can express \(T\)from the equation, getting:

\(T = \frac{{\Delta H - \Delta G}}{{\Delta S}}\)

Since the reaction just becomes spontaneous at this temperature, we assume that \(\Delta G = 0\frac{{k \cdot J}}{{mol}}\)

\(T = \frac{{181660{\rm{Jmo}}{{\rm{l}}^{ - 1}} - 0{\rm{Jmo}}{{\rm{l}}^{ - 1}}}}{{216.42{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}{\rm{ }}\)

\(T = 839.39\;{\rm{K}}\)

Therefore reaction becomes spontaneous at the temperature of \(839.39\;{\rm{K}}.\)