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Gibbs free energy (G) is a measure of the spontaneity of a chemical reaction. It is the chemical potential for a reaction, and is minimized at equilibrium. It is defined as G=H−TS where H is enthalpy, T is temperature, and S is entropy. The chemical reaction that causes magnesium to corrode in air is given by 2Mg+O2→2MgO in which at 298 K ΔH∘rxn = −1204 kJ ΔS∘rxn = −217.1 J/K Part A What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K. Express your answer as an integer and include the appropriate units. View Available Hint(s) ΔG∘rxnΔ G r x n ∘ = nothing

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dsdrajlin

Answer:

ΔG° = -1123 kJ

Explanation:

Let's consider the following equation.

2 Mg + O₂ → 2 MgO

The standard Gibbs free energy (ΔG°) can be calculated with the following expression:

ΔG° = ΔH° - T.ΔS°

where,

ΔH° is the standard enthalpy of the reaction

T is the absolute temperature

ΔS° is the standard entropy of the reaction

For this reaction,

ΔG° = -1204 kJ - 298 K × (-271.1) × 10⁻³ kJ/K = -1123 kJ

By convention, ΔG° < 0 means that the reaction is spontaneous.

annyksl

The Gibbs free energy for this reaction is -1123 kJ, which indicates that this reaction is spontaneous.

How to calculate Gibbs free energy?

We use the following equation: ΔG° = ΔH° - T.ΔS°

In this equation, we must know that:

  • "ΔH°" refers to the standard enthalpy of the reaction.
  • "T" is the temperature at which the reaction takes place.
  • "ΔS°" is the entropy of the reaction.

With this information, we can replace the symbols with the values presented in the question.

In this way, we can calculate the equation as follows:

[tex]\Delta G^0= (-1204 - 298) * (-271.1)* 10^-^3 = -1123 kJ[/tex]

More information about Gibbs free energy at the link:

https://brainly.com/question/10012881

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