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Chapter 16: Q16.4-60E (page 912)

Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed.

Short Answer

Expert verified
  • The change of enthalpy is \(\Delta H = - 1.89\frac{{kJ}}{{mol}}\)
  • The change of entropy is \(\Delta S = 3.36\frac{J}{{K \cdot mol}}\)
  • The change of free energy is \(\Delta G = - 2.90\frac{{kJ}}{{mol}}\)
  • The reaction is spontaneous. It is normally observed because the rate of the reaction is very slow.

Step by step solution

01

Definition of Gibbs free energy

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the topmost quantum of work done in a thermodynamic system when temperature and pressure remain constant.

02

Determine the Change of entropy and enthalpy.

The reaction in question is:

\(C({\rm{ diamond }}) \to C(graphite)\)

We can calculate the enthalpy change using the formula

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

We can look up the standard formation enthalpy changes in the table in AppendixG in the book.

\(\Delta H = {H_f}(C,{\rm{ graphite }}) - {H_f}(C,{\rm{ diamond }})\)

\(\begin{array}{l}\Delta H = (0 - 1.89)\frac{{k \cdot J}}{{mol}}\\\Delta H = - 1.89\frac{{kJ}}{{mol}}\end{array}\)

Hence, the change of enthalpy is \(\Delta H = - 1.89\frac{{kJ}}{{mol}}\)

We can calculate the entropy change using the formula

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

We can look up the standard entropies in the table in AppendixG in the book.

\(\begin{array}{l}\Delta S = S(C,{\rm{ graphite }}) - S(C,{\rm{ diamond }})\\\Delta S = (5.740 - 2.38)\frac{J}{{K \cdot mol}}\\\Delta S = 3.36\frac{J}{{K \cdot mol}}\end{array}\)

Hence, the change of entropy is \(\Delta S = 3.36\frac{J}{{K \cdot mol}}\)

03

Determine the Change of free energy.

If we look at the formula for \(\Delta G\) :

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

We can see that, if the enthalpy change is negative and the entropy change is positive, such as the case is with this reaction, \(\Delta G\) will surely be negative and the reaction therefore spontaneous. We can calculate the free energy change using the formula

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

We can look up the standard formation free energy changes for each species in AppendixG in the book.

\(\begin{array}{l}\Delta G = {G_f}(C,{\rm{ graphite }}) - {G_f}(C,{\rm{ diamond }})\\\Delta G = (0 - 2.90)\frac{{k \cdot J}}{{mol}}\\\Delta G = - 2.90\frac{{kJ}}{{mol}}\end{array}\)

Even though the conversion of diamond to graphite is a spontaneous reaction, it is not normally observed because the rate of the reaction is very slow.

Hence, the change of free energy is \(\Delta G = - 2.90\frac{{kJ}}{{mol}}\)

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Most popular questions from this chapter

Although the gas used in the oxyacetylene torch Figure (5.7) is essentially pure acetylene, the heat produced by the combustion of one mole ofacetylene in such a torch is likely, not equal to the enthalpy of combustion of acetylene listed in the table 5.2. Considering the conditions for which the tabulated data are reported. Suggest an explanation.

Under what conditions is \({{\bf{N}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}} \to {\bf{NO(g) + N}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\) spontaneous?

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is \({\bf{F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Al(s)}} \to {\bf{A}}{{\bf{l}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s) + 2Fe(s)}}\). Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb \({\bf{851}}{\bf{.8\;kJ/mol}}\)of heat.

Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at \(850\;{\rm{K}}\).

\({\rm{C}}{{\rm{H}}_4}(g) + 4{\rm{C}}{{\rm{l}}_2}(g) \to {\rm{CC}}{{\rm{l}}_4}(g) + 4{\rm{HCl}}(g)\)

What is the equilibrium constant for the reaction at \(850\;{\rm{K}}\)? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant?

Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

(a) Ammonia\({\bf{:}}{\rm{ }}{\bf{2N}}{{\bf{H}}_{\bf{3}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(b) Diborane\({\bf{:}}{\rm{ }}{{\bf{B}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}} \to {\bf{2\;B(g) + 3}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(c) Hydrazine: \({{\bf{N}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{N}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{(g)}}\)

(d) Hydrogen peroxide: \({{\bf{H}}_{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(l)}} \to {{\bf{H}}_{\bf{2}}}{\bf{O(g) + }}\frac{{\bf{1}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)
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