Calculate delta G and Kc.

2Al(s) + 3I2(s) to 2Al3+(aq) + 6I-(aq)

To calculate ΔG (standard Gibbs free energy change) and Kc (equilibrium constant), we need to use thermodynamic data and the equation:

ΔG = -RTlnK

Where:
ΔG represents the change in Gibbs free energy
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
K is the equilibrium constant

Let's begin by balancing the equation:

2Al(s) + 3I2(s) → 2Al3+(aq) + 6I-(aq)

Now, let's gather relevant thermodynamic data for the chemical species involved. We will need the standard Gibbs free energy values (ΔGf°) for each compound at standard conditions (298 K and 1 atm).

ΔGf°(Al(s)) = 0 (standard condition for the element in its standard state)
ΔGf°(I2(s)) = 0 (standard condition for the element in its standard state)
ΔGf°(Al3+(aq)) = -524.2 kJ/mol
ΔGf°(I-(aq)) = 0 (standard condition for the element in its standard state)

Next, we can calculate the ΔG° (standard Gibbs free energy change) for the reaction using the formula:

ΔG° = ΣnΔGf°(products) - ΣmΔGf°(reactants)

Where:
Σn represents the stoichiometric coefficients of the products
Σm represents the stoichiometric coefficients of the reactants

ΔG° = [2ΔGf°(Al3+(aq)) + 6ΔGf°(I-(aq))] - [0 + 0]
ΔG° = 2(-524.2 kJ/mol) + 6(0 kJ/mol)
ΔG° = -1048.4 kJ/mol

Now, we can convert kJ to J by multiplying by 1000:
ΔG° = -1048.4 kJ/mol × 1000 J/kJ
ΔG° = -1,048,400 J/mol

Finally, we can calculate Kc using the relation between ΔG° and K:

ΔG° = -RTlnKc

We can rearrange the equation to solve for Kc:

Kc = e^(-ΔG°/RT)

Substituting the values:
Kc = e^(-(-1,048,400 J/mol) / (8.314 J/(mol·K) × 298 K))

Now, calculate Kc using the above expression.