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

-0.050 kJ/mol.K

Explanation:

All 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

or simply positive.

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.

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c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current

it's c. elements are placed in rows based on how may valencies electrons have.