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Miloflippin7339
20.03.2020 •
Chemistry
A certain reaction is thermodynamically favored at temperatures below 400. K, but it is not favored at temperatures above 400. K. The value of ΔHo for the reaction is –20 kJ/mol. What is the value of ΔSo for the reaction
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Ответ:
-0.050 kJ/mol.K
Explanation:
A certain reaction is thermodynamically favored at temperatures below 400. K, that is, ΔG° < 0 below 400. KThe reaction is not favored at temperatures above 400. K, that is. ΔG° > 0 above 400. KAll in all, ΔG° = 0 at 400. K.
We can find ΔS° using the following expression.
ΔG° = ΔH° - T.ΔS°
0 = -20 kJ/mol - 400. K .ΔS°
ΔS° = -0.050 kJ/mol.K
Ответ:
Explanation:
Hello,
In this case, one considers the Gibbs free energy for the reaction as:
Which should be negative if is spontaneous (thermodynamically favored). Thus, for the given information about the 400K, the entropy of reaction turns out:
Such value is obtained assuming a temperature below 400K and a negative value for the Gibbs free energy of such reaction, for which the entropy must be positive for all the possibilities, accomplishing the thermodynamic favorability.
Best regards,
Ответ:
D. P1/V1 = P2/V2
Explanation:
The ideal gas law combines the three different gas laws. This relates with a constant temperature.