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3 years ago

13

Select the correct answer.

1 answer:

Sholpan [36]3 years ago

5 0

Answer:

B. Keq = 6.0 x 10-2.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to remember that any equilibrium constant is computed by dividing the concentration of products by that of reactants:

$Keq=\frac{[Prod]}{[Reac]}$

Thus, a reaction that is reactant-favored will have a Keq>1 because the concentration of reactants prevail over that of products at equilibrium, and thus, the correct answer is B. Keq = 6.0 x 10-2.

Regards!

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3 years ago

Determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298°K and 1 atm. T

Ostrovityanka [42]

Explanation:

The given data is as follows.

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

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$S_{O_{2}} = 205 J/^{o}K$

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

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

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

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

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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.

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4 years ago