armoniatrion armoniatrion
  • 23-04-2021
  • Chemistry
contestada

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.
Calculate Δ for this reaction at 298 K.
Δ= ?

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sebassandin
sebassandin sebassandin
  • 28-04-2021

Answer:

[tex]\Delta G =-103.95kJ[/tex]

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

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

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

[tex]\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ[/tex]

Best regards!

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