For a given reaction with δδh = -26. 9 kj/mol, the δδg = 0 at 394 k. the value of δδs must be _______ j/k-mol, assuming that δδh and δδs do not vary with temperature.

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For a given reaction with δδh = -26. 9 kj/mol, the δδg = 0 at 394 k. the value of δδs must be -68.27, using the equation ΔG=ΔH−TΔS.

Enthalpy and entropy are combined into a single quantity known as Gibbs free energy, or G. The product of the system's temperature and entropy, added to the enthalpy, equals the change in free energy, or G. ΔG=ΔH−TΔS is the equation for thermodynamics, where G represents the reaction's free energy. The change in entropy and free energy is what determines whether a reaction is spontaneous. The reaction will be spontaneous if there is an increase in entropy and a net decrease in free energy in the products.

As we know that,

ΔG=ΔH−TΔS

ΔH−TΔS = 0

ΔH = TΔS

-26900 = 394×ΔS

ΔS = -68.27

The change in entropy is negative as the system's unpredictability decreases.

Learn more about ΔS here:

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