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Chapter 16: Q16.8 CYL (page 898)

Calculate 螖G掳 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 掳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

Calculate the free energy change for the same reaction at 875 掳C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):

\({\text{G}}6{\text{P}} \rightleftharpoons {\text{F}}6{\text{P}}\;\;\;\Delta G_{298}^\circ = 1.7\;{\text{kJ}}\)

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