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For a given reaction, AH = -19.9 kJ/mol and AS = -55.5 J/K/mol. The reaction will have AG = 0 at K. Assume that AH and AS do not vary with temperature. A) 359 B) 2789 C) 298 D) 2.79 E) 0.359

Respuesta :

Answer:

The correct answer is option A.

Explanation:

The expression of Gibbs's fee energy is given as:

[tex]\Delta G=\Delta H-T\Delta S[/tex]

[tex]\Delta G[/tex] = Change in Gibbs free energy

[tex]\Delta S[/tex] = Change in an entropy

[tex]\Delta H[/tex] = Enthalpy of reaction

T = Temperature at which reaction is going on.

We have:

[tex]\Delta G=0 J/mol[/tex]

[tex]\Delta S=-55.5 J/K mol[/tex]

[tex]\Delta H=-19.9 kJ/mol=-19,900 J/mol[/tex]

T =?

[tex]0 J/mol=-19,900 J/mol-(T\times (-55.5 J/K mol))[/tex]

T = 359 K

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