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]\[ G = H - T S \][/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]\[ 4 \text{Al} + 3 \text{O}_2 \rightarrow 2 \text{Al}_2\text{O}_3 \][/tex]

At [tex]\( 293 \text{ K} \)[/tex],

[tex]\[ \Delta H_{\text{min}} = -3352 \text{ kJ} \][/tex]
[tex]\[ \Delta S_{\text{rm}} = -625.1 \text{ J/K} \][/tex]

At what temperature [tex]\( T_{\text{eq}} \)[/tex] do the forward and reverse corrosion reactions occur in equilibrium?

Express your answer as an integer and include the appropriate units.

[tex]\[ T_{\text{eq}} = \begin{array}{|l|l|}
\hline \text{Value} & \text{Units} \\
\hline
\end{array} \][/tex]

Value:

Units:



Answer :

To determine the temperature [tex]\( T_{eq} \)[/tex] at which the forward and reverse corrosion reactions occur in equilibrium, we need to use the Gibbs free energy equation at equilibrium. The Gibbs free energy [tex]\( G \)[/tex] is given by:

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

At equilibrium, the Gibbs free energy change [tex]\( \Delta G \)[/tex] for the reaction is zero:

[tex]\[ \Delta G = \Delta H - T_{eq} \Delta S = 0 \][/tex]

We can rearrange this equation to solve for the equilibrium temperature [tex]\( T_{eq} \)[/tex]:

[tex]\[ T_{eq} = \frac{\Delta H}{\Delta S} \][/tex]

Let's use the given values for the enthalpy change [tex]\( \Delta H \)[/tex] and entropy change [tex]\( \Delta S \)[/tex] to find [tex]\( T_{eq} \)[/tex]. We need to ensure the units are consistent, so we first convert the enthalpy change from kJ to J (since [tex]\( 1 \, \text{kJ} = 1000 \, \text{J} \)[/tex]):

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

Now, we can substitute the values into the equation:

[tex]\[ T_{eq} = \frac{\Delta H}{\Delta S} \][/tex]
[tex]\[ T_{eq} = \frac{-3352000 \, \text{J}}{-625.1 \, \text{J/K}} \][/tex]

Evaluating this expression gives the equilibrium temperature:

[tex]\[ T_{eq} = 5362.342025275956 \, \text{K} \][/tex]

Therefore, the temperature [tex]\( T_{eq} \)[/tex] at which the forward and reverse corrosion reactions occur in equilibrium is approximately:

[tex]\[ \begin{array}{|c|c|} \hline 5362 & \text{K} \\ \hline \end{array} \][/tex]

This can be expressed as:

[tex]\[ \boxed{5362 \, \text{K}} \][/tex]