Q16.5 CYL - Chemistry (OpenStax)

Question

Calculate the standard entropy change for the following process:

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

Step-by-Step Solution

Answer

The standard entropy change for above process is \( - 120.66\;{\rm{J}}/{\rm{Kmol}}\)

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 .

\({{\rm{H}}_{2(\;{\rm{g}})}} + {{\rm{C}}_2}{{\rm{H}}_{4(\;{\rm{g}})}} \cdots {{\rm{C}}_2}{{\rm{H}}_{6(\;{\rm{g}})}}\)

\(\Delta {S_{{\rm{rxn }}}} = \Delta {S_{{\rm{Product }}}} - \Delta {S_{{\rm{Reactant }}}}\)

\(\Delta {\rm{S}}\) of \({{\rm{H}}_{2(\;{\rm{g}})}} = 130.7\;{\rm{J}}/{\rm{K}} \cdot \) mole

\(\Delta S\) of \({{\rm{C}}_2}{{\rm{H}}_{4(\;{\rm{g}})}} = 219.56\;{\rm{J}}/{\rm{K}} \cdot \) mole

\(\Delta {\rm{S}}\) of \({{\rm{C}}_2}{{\rm{H}}_{6(\;{\rm{g}})}} = 229.6\;{\rm{J}}/{\rm{K}} \cdot \) mole

\(\Delta {S_{{\rm{rxn }}}} = \Delta {S_{{\rm{Product }}}} - \Delta {S_{{\rm{Reactant }}}}\)

\( = \left\{ 1 \right.\) mole \( \times \Delta S\) of \(\left. {{C_2}{H_{6(g)}}} \right\} - \left\{ 1 \right.\) mole \( \times \Delta S\) of \({H_{2(g)}} + 1\) mole \( \times \Delta S\) of \(\left. {{C_2}{H_{4(g)}}} \right\}\)

\( = \{ 229.6\;{\rm{J}}/{\rm{K}}\}  - \{ 130.7\;{\rm{J}}/{\rm{K}} + 219.56\;{\rm{J}}/{\rm{K}}\} \)

\( =  - 120.66\;{\rm{J}}/{\rm{K}}\)

\(\Delta {S_{{\rm{rxn}}}} =  - 120.66\;{\rm{J}}/{\rm{K}}\)