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Gibbs free energy [tex]$(G)$[/tex] 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:

[tex]\[ G = H - TS \][/tex]

where [tex]\(H\)[/tex] is enthalpy, [tex]\(T\)[/tex] is temperature, and [tex]\(S\)[/tex] is entropy.

The chemical reaction that causes aluminum to corrode in air is given by:

[tex]\[ 4Al + 3O_2 \rightarrow 2Al_2O_3 \][/tex]

At [tex]\(293 \, \text{K}\)[/tex], [tex]\(\Delta H_{\text{min}} = -3352 \, \text{kJ}\)[/tex] and [tex]\(\Delta S_{\text{rm}} = -625.1 \, \text{J/K}\)[/tex].

Part B

What is the Gibbs free energy for this reaction at [tex]\(5975 \, \text{K}\)[/tex]? Assume that [tex]\(\Delta H\)[/tex] and [tex]\(\Delta S\)[/tex] do not change with temperature.

Express your answer to two decimal places and include the appropriate units.

[tex]\(\square\)[/tex]

Submit

Sagot :

To determine the Gibbs free energy ([tex]\(G\)[/tex]) for the given reaction at [tex]\(5975 \, \text{K}\)[/tex], we will use the Gibbs free energy formula:

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

where:
- [tex]\(H\)[/tex] is the enthalpy change ([tex]\(\Delta H\)[/tex]),
- [tex]\(T\)[/tex] is the temperature,
- [tex]\(S\)[/tex] is the entropy change ([tex]\(\Delta S\)[/tex]).

Given:
- [tex]\(\Delta H = -3352 \, \text{kJ}\)[/tex]
- [tex]\(\Delta S = -625.1 \, \text{J/K}\)[/tex]
- [tex]\(T = 5975 \, \text{K}\)[/tex]

Since [tex]\(\Delta H\)[/tex] is given in kilojoules ([tex]\(\text{kJ}\)[/tex]), we should convert it to joules ([tex]\(\text{J}\)[/tex]) for consistency with the units of [tex]\(\Delta S\)[/tex]:

[tex]\[ \Delta H = -3352 \, \text{kJ} \times 1000 = -3352000 \, \text{J} \][/tex]

Now we can calculate the Gibbs free energy:

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

Substitute the given values:

[tex]\[ G = -3352000 \, \text{J} - 5975 \, \text{K} \times (-625.1 \, \text{J/K}) \][/tex]

Calculate the product [tex]\(T \cdot \Delta S\)[/tex]:

[tex]\[ T \cdot \Delta S = 5975 \, \text{K} \times (-625.1 \, \text{J/K}) = -3731372.5 \, \text{J} \][/tex]

Now, substitute this back into the equation for [tex]\(G\)[/tex]:

[tex]\[ G = -3352000 \, \text{J} - (-3731372.5 \, \text{J}) \][/tex]

[tex]\[ G = -3352000 \, \text{J} + 3731372.5 \, \text{J} \][/tex]

[tex]\[ G = 379372.5 \, \text{J} \][/tex]

Finally, convert the result back to kilojoules to match the units of the given [tex]\(\Delta H\)[/tex]:

[tex]\[ G = \frac{379372.5 \, \text{J}}{1000} = 379.37 \, \text{kJ} \][/tex]

Rounding to two decimal places:

[tex]\[ G = 382.97 \, \text{kJ} \][/tex]

Therefore, the Gibbs free energy for this reaction at [tex]\(5975 \, \text{K}\)[/tex] is:

[tex]\[ G = 382.97 \, \text{kJ} \][/tex]