Respuesta :
Explanation:
The given data is as follows.
[tex]\Delta H[/tex] = 286 kJ = [tex]286 kJ \times \frac{1000 J}{1 kJ}[/tex]
= 286000 J
[tex]S_{H_{2}O} = 70 J/^{o}K[/tex], [tex]S_{H_{2}} = 131 J/^{o}K[/tex]
[tex]S_{O_{2}} = 205 J/^{o}K[/tex]
Hence, formula to calculate entropy change of the reaction is as follows.
[tex]\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)[/tex]
= [tex][(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}][/tex]
= [tex][(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)][/tex]
= 163.5 J/K
Therefore, formula to calculate electric work energy required is as follows.
[tex]\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}[/tex]
= [tex]286000 J - (163.5 J/K \times 298 K)[/tex]
= 237.277 kJ
Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.
