Question

Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. The chemical reaction is H2001) H2(g) + 0.502(g) Data (at 298°K and 1 atm): AH = 286 kJ for this reaction, Suzo = 70 JK, SH2 = 131 JIK, and Soz = 205 J/ºK.

Answer 1

Explanation:

The given data is as follows.

          $\Delta H$ = 286 kJ = $286 kJ \times \frac{1000 J}{1 kJ}$

                            = 286000 J

 $S_{H_{2}O} = 70 J/^{o}K$,      $S_{H_{2}} = 131 J/^{o}K$

 $S_{O_{2}} = 205 J/^{o}K$

Hence, formula to calculate entropy change of the reaction is as follows.

          $\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)$

                     = $[(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}]$

                    = $[(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)]$

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             $\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}$

                            = $286000 J - (163.5 J/K \times 298 K)$

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.