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Chapter 16: Q16.8 CYL (page 898)

Calculate ΔG° using

(a) free energies of formation and

(b) enthalpies of formation and entropies(Appendix G). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

\({{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}}\)

Short Answer

Expert verified
  1. The Gibbs free energy change using enthalpies of formation and entropies is\(141.55\;{\rm{kJ}}/{\rm{mol}}\).

2. The calculated values are greater than zero, so the Gibbs free energy change is positive. Hence, the reaction is nonspontaneous (not spontaneous).

Step by step solution

01

Define enthalpy of the reaction

Gibbs free energy changeis used to determine the spontaneity of a process. It is expressed in terms of enthalpy and entropy of a system.

02

a) Determine the Gibbs free energy change using free energies of formation

Given:

Temperature\( = {25^^\circ }{\rm{C}}\)

\({{\rm{C}}_2}{{\rm{H}}_4}(\;{\rm{g}}) \to {{\rm{H}}_2}(\;{\rm{g}}) + {{\rm{C}}_2}{{\rm{H}}_2}(\;{\rm{g}})\)

To calculate the Gibbs free energy change using free energies of formation as follows,

\(\begin{array}{l}\Delta G_{rxn}^^\circ = \Delta G_f^^\circ {\rm{ products }} - \Delta G_f^^\circ {\rm{ reactants}}\\ = (0 + 209.2){\rm{kJ}}/{\rm{mol}} - (68.4\;{\rm{kJ}}/{\rm{mol}})\\ = (209.2 - 68.4){\rm{kJ}}/{\rm{mol}}\Delta G_{rxn}^^\circ \\ = 140.8\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

Therefore, the Gibbs free energy change using free energies of formation is\(140.8\;{\rm{kJ}}/{\rm{mol}}\).

03

b) Determine the Gibbs free energy change using free energies of formation and entropies

To calculate the Gibbs free energy change using enthalpies of formation and entropies as follows,

Calculating the enthalpy of formation,

\(\begin{array}{l}\Delta H_{rxn}^^\circ = \Delta H_f^^\circ {\rm{ products }} - \Delta H_f^^\circ {\rm{ reactants }}\\{\rm{ = }}(0 + 227.4){\rm{kJ}}/{\rm{mol}} - (52.4\;{\rm{kJ}}/{\rm{mol}})\\ = (227.4 - 52.4){\rm{kJ}}/{\rm{mol}}\Delta H_{rxn}^^\circ = 175\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

04

Determine the Gibbs free energy change using free energies of formation

Calculating the entropy of reaction,

\(\begin{array}{l}\Delta S_{xxn}^^\circ = \Delta S_f^^\circ {\rm{ products }} - \Delta S_f^^\circ {\rm{ reactants }}\\{\rm{ = }}(130.7 + 200.9){\rm{J}}/{\rm{K}} \cdot {\rm{mol}} - (219.3\;{\rm{J}}/{\rm{K}} \cdot {\rm{mol}})\\ = (331.6 - 219.3){\rm{J}}/{\rm{K}}.{\rm{mol}}\Delta S_{rxn}^^\circ \\ = 112.3\;{\rm{J}}/{\rm{K}}.{\rm{mol}}\end{array}\)

Calculating the Gibbs free energy change,

\(\begin{array}{l}\Delta S_{rxn}^^\circ = 112.3\;{\rm{J}}/{\rm{K}} \cdot {\rm{mol}}\\\Delta H_{rxn}^^\circ = 175\;{\rm{kJ}}/{\rm{molT}}\\ = {25^^\circ }{\rm{C}}\\ = (25 + 273){\rm{K}}\\ = 298\;{\rm{K}}\end{array}\)

Substituting the values in the Gibbs free energy change expression,

\(\begin{array}{l}\Delta {G^^\circ } = \Delta {H^^\circ } - T\Delta {S^^\circ }\\ = (175\;{\rm{kJ}}/{\rm{mol}}) - (298\;{\rm{K}})(112.3\;{\rm{J}}/{\rm{K}}.{\rm{mol}})\left( {\frac{{1\;{\rm{kJ}}}}{{1000\;{\rm{J}}}}} \right)\\ = 175\;{\rm{kJ}}/{\rm{mol}} - 33.45\;{\rm{kJ}}/{\rm{mol}}\Delta {G^^\circ }\\ = 141.55\;{\rm{kJ}}/{\rm{mol}}\end{array}\)

Therefore, the Gibbs free energy change using enthalpies of formation and entropies is\(141.55\;{\rm{kJ}}/{\rm{mol}}\).The calculated values are greater than zero, so the Gibbs free energy change is positive. Hence, the reaction is nonspontaneous (not spontaneous).

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Most popular questions from this chapter

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?

Calculate the standard entropy change for the following process:

\({{\bf{H}}_{\bf{2}}}{\bf{(g) + }}{{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{4}}}{\bf{(g)}} \to {{\bf{C}}_{\bf{2}}}{{\bf{H}}_{\bf{6}}}{\bf{(g)}}\)

Is the formation of ozone \(\left( {{{\bf{O}}_{\bf{3}}}{\bf{(g)}}} \right)\) from oxygen \(\left( {{{\bf{O}}_{\bf{2}}}{\bf{(g)}}} \right)\) spontaneous at room temperature under standard state conditions?

Popular chemical hand warmers generate heat by the air-oxidation of iron:\({\bf{4Fe(s) + 3}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{2F}}{{\bf{e}}_{\bf{2}}}{{\bf{O}}_{\bf{3}}}{\bf{(s)}}\).How does the spontaneity of this process depend upon temperature?

An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) decomposes to form copper and sulfur described by the following equation:

\({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{\;S(s)}} \to {\bf{Cu(s) + S(s)}}\)

(a) Determine \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the decomposition of \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S(\;s)}}\).

(b) The reaction of sulfur with oxygen yields sulfur dioxide as the only product. Write an equation that describes this reaction, and determine\({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\)for the process.

(c) The production of copper from chalcocite is performed by roasting the \({\bf{C}}{{\bf{u}}_{\bf{2}}}{\bf{S}}\) in air to produce the \({\bf{Cu}}\). By combining the equations from Parts (a) and (b), write the equation that describes the roasting of the chalcocite, and explain why coupling these reactions together makes for a more efficient process for the production of the copper.
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