Q16.7 CYL - Chemistry (OpenStax)

Question

Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change forthe reaction shown here (298 K). What does the computed value for ΔG° say about the spontaneity of thisprocess?

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

Step-by-Step Solution

Answer

The standard entropy change for above process is \(\quad \Delta y^0 {R_{xn}}\)  =  + 84.08\;{\rm{kJ}}/{\rm{mol }}\) non-spontaneous \({R_{xn}}\).

Step1: Define enthalpy of the reaction

• Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 

• The equation used to calculate enthalpy change is of a reaction is:
• \(\Delta {H_{rxn}} = \sum {\left[ {n \times \Delta {H_{f({\rm{ product }})}}} \right]}  - \sum {\left[ {n \times \Delta {H_{f({\rm{ reactant }})}}} \right]} \)

Step 2: Determine the enthalpy of the reaction .

Given Rxn eqn. is \({{\rm{C}}_2}{{\rm{H}}_6}(g) \to {{\rm{H}}_2}(g) + {{\rm{C}}_2}{{\rm{H}}_4}(g)\)

As \(\quad \Delta y_{R \times n}^0 = \Delta H_{R \times n}^0 - T\Delta S_{R \times n}^0\)

where \(\Delta H_{R \times n}^0 = \({Sigma \Delta H_j^^\circ }\) (Products) \( - \Sigma \Delta H_j^^\circ \) (Reactants)

\(\Delta S_{R \times n}^^\circ  = \sum {S_{{\rm{Products }}}^^\circ }  - \sum {S_{{\rm{Reactants }}}^0} \)

Now, \(\Delta H_{R \times n}^0 = \Delta H_f^0\left( {{C_2}{H_4}(g)} \right) + \Delta H_p^0\left( {{H_2}(g)} \right) - H_f^0\left( {{C_2}{H_6}(g)} \right)\)

\( = 52.53\;{\rm{kJ}}/{\rm{mol}} + 0 - ( - 83.91\;{\rm{kJ}}/{\rm{mol}})\)

\( = 136.44\;{\rm{kJ}}/{\rm{mol}}\)

\(\Delta s_{R \times n}^0 = {s^0}\left( {{{\rm{C}}_2}{H_4}(g)} \right) + {s^0}\left( {{H_2}(g)} \right) - {s^0}\left( {{C_2}{H_6}(g)} \right)\)

\( = 219.5{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} + 130{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} - 173.3{\rm{J}}{{\rm{K}}^{ - 1}}\;{\rm{mol}}\)

 

\(\Delta {y_{{R_{xn}}}} = 136.44 \times {10^3}\;{\rm{J}}/{\rm{mal}} - (298\;{\rm{K}})\left( {175.9\;{\rm{JKhmo}}{{\rm{l}}^{ - 1}}} \right)\)

\(\Delta g_{R \times n}^^\circ  =  + 84.08\;{\rm{kJ}}/{\rm{mol}}\) were true sign of the

\(\Delta 4_R^^\circ  \times n\) indicates that \(R \times n\) is non-spontaneous reaction.

\(\Delta {y^^\circ }{R_{xn}} =  + 84.08\;{\rm{kJ}}/{\rm{mol }}\) Non-spontaneous \({R_{xn}}\).