Q16.8 CYL - Chemistry (OpenStax)

Question

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?

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

Step-by-Step Solution

Answer

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).

Step1: Define enthalpy of the reaction

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

Step 2: 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 \(140.8\;{\rm{kJ}}/{\rm{mol}}\).

Step 3: 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,

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

Step 4: 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).