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What is the value for Delta*G at 500 K if Delta*H = 27kJ / m * o * l and triangle S=0.09 kJ/(mol* K) ? OC. Delta*G = 26.9kJ / m * o * l OB. Delta*G = - 18kJ / m * o * l O D. Delta*G = 0kJ / m * o * l OA. Delta*G = 72kJ / m * o * l

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Eduard22slyidn

The value of the Gibbs free energy (ΔG) for the reaction given in the question is –18 KJ/mol

Data obtained from the question

  • Temperature (T) = 500 K
  • Enthalpy change (ΔH) = 27 KJ/mol
  • Change in entropy (ΔS) = 0.09 KJ/Kmol
  • Gibbs free energy (ΔG) =?

How to determine the Gibbs free energy

The Gibbs free energy (ΔG) can be obtained by using the following equation as illustrated below:

ΔG = ΔH – TΔS

ΔG = 27 – (500 × 0.09)

ΔG = 27 – 45

ΔG = –18 KJ/mol

Thus, the value of ΔG for the reaction is –18 KJ/mol

Learn more about Gibbs free energy:

https://brainly.com/question/9552459

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