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

Step-by-step explanation:

The given data is as follows.

`\Delta H` = 286 kJ =
`286 kJ * (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 \\u_(i)S_(i)_(products) - \sum \\u_(i)S_(i)_(reactants)`

=
`[((1)/(2) * S_{O_(2)}) - (1 * S_{H_(2)})] - [1 * S_{H_(2)O}]`

=
`[((1)/(2) * 205) + (1 * 131)] - [(1 * 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 * 298 K)`

= 237.277 kJ

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