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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

Use the information in Appendix G to estimate the boiling point of CS2.

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

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?

What is a spontaneous reaction?

What happens to \({\bf{\Delta G}}_{{\bf{298}}}^{\bf{^\circ }}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?

(a) \({\bf{S(s) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{2}}}{\bf{(g)}}\)

(b) \({\bf{2S}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + }}{{\bf{O}}_{\bf{2}}}{\bf{(g)}} \to {\bf{S}}{{\bf{O}}_{\bf{3}}}{\bf{(g)}}\)

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