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For the gaseous reaction of xenon and fluorine to form xenon hexafluoride:(a) Calculate ΔS° at 298 K ( ΔH°=-402kJ/mol and ΔG° = -280kJ/mol ).

Respuesta :

ΔS° at 298 K is ΔS°= -0.409kJ/mol.K

The energy is absorbed by the reactants and thus it is endothermic and  and the number of moles of gaseous substances are increasing as we proceed towards products and thus the entropy increases.

For the reaction:

Xe(g)+3F2(g) --------->  XeF6(g)

ΔH°=-402kJ/mol and  

ΔG° = -280kJ/mol

The fact that

ΔG°=ΔH°-TΔS°

ΔS°=ΔH - ΔG/T

Hence,

=[-402kJ/mol-(-280kJ/mol)]/298K

ΔS°= -0.409kJ/mol.K

Learn more about ΔS here:

https://brainly.com/question/17087340

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