Consider the Unbalanced Reaction:
Al2O3(s) + CO(g) → Al(s) + CO2(g)
Estimate ∆G at 298 K if 7.1 moles of Al2O3(s) is reacted.
To estimate ∆G at 298 K for the unbalanced reaction Al2O3(s) + CO(g) → Al(s) + CO2(g) when 7.1 moles of Al2O3(s) are reacted, we need to use the concept of Gibbs Free Energy (∆G) and the relationship between ∆G and the moles of reactants and products.
The Gibbs free energy change (∆G) is calculated using the equation:
∆G = ∆G° + RT ln(Q)
Where:
- ∆G is the change in Gibbs free energy
- ∆G° is the standard Gibbs free energy change (at standard conditions)
- R is the gas constant (8.314 J/(mol•K))
- T is the temperature in Kelvin (298 K in this case)
- Q is the reaction quotient (ratio of the product concentrations to the reactant concentrations)
However, to use this equation, we need the standard Gibbs free energy change (∆G°). The standard Gibbs free energy change for the reaction is obtained from tables or databases of thermodynamic values.
Since the reaction is unbalanced, we cannot directly obtain the ∆G° from tables. We first need to balance the reaction:
Al2O3(s) + 3CO(g) → 2Al(s) + 3CO2(g)
Now, we can find the standard Gibbs free energy change (∆G°) from tables or databases. Assuming the value to be -500 kJ/mol.
Once we have ∆G°, we can calculate ∆G using the equation mentioned earlier with the given values:
∆G = ∆G° + RT ln(Q)
Here, Q represents the reaction quotient, which we can calculate using the given number of moles of Al2O3(s) reacted. As per the balanced equation, for the given 7.1 moles of Al2O3(s) reacted, the number of moles of CO(g) used is 3 * 7.1 = 21.3 moles.
Therefore, the ∆G can be estimated by substituting the values into the equation:
∆G = -500000 J/mol + (8.314 J/(mol•K) * 298 K * ln(21.3))
Solving this equation will give the estimated ∆G at 298 K for the given reaction.
To estimate ΔG (change in Gibbs free energy) at 298 K for the given reaction, we need to use the Gibbs free energy change (∆G°) values of the species involved. However, to obtain these values, we need to balance the chemical equation first. The given reaction is:
Al2O3(s) + CO(g) → Al(s) + CO2(g)
To balance this equation, we need to make sure the number of each type of atom is the same on both sides. Balancing the equation, we get:
2 Al2O3(s) + 3 CO(g) → 4 Al(s) + 3 CO2(g)
Now that we have a balanced equation, we can proceed to estimate the Gibbs free energy change (∆G°) for the reaction. The Gibbs free energy change can be calculated using the equation:
∆G° = Σn∆G°f (products) - Σm∆G°f (reactants)
Where ∆G°f represents the standard Gibbs free energy of formation of each species involved and n and m are the stoichiometric coefficients of the products and reactants, respectively.
Let's use the following standard Gibbs free energy of formation values at 298 K (in kJ/mol):
∆G°f (Al2O3) = -1582.3
∆G°f (CO) = -137.2
∆G°f (Al) = 0
∆G°f (CO2) = -394.4
Now we can substitute these values into the equation and calculate the ΔG°:
∆G° = (4 * 0 + 3 * -394.4) - (2 * -1582.3 + 3 * -137.2)
∆G° = -1183.2 kJ/mol
Finally, we can calculate the ΔG for 7.1 moles of Al2O3 (since the reaction is unbalanced):
∆G = ∆G° * moles of Al2O3(s)
∆G = -1183.2 kJ/mol * 7.1 mol
∆G = -8404.72 kJ
Therefore, the estimated ∆G at 298 K for the reaction when 7.1 moles of Al2O3(s) is reacted is -8404.72 kJ.
I will assume you do not have dH or dS available. Look up dGo for the reactant and products. Balance the equation.
Al2O3 + 3CO ==> 2Al + 3CO. Then
dGo rxn - (n*dGo products) - (n*dGo products). That will give you dGo for the reaction for 1 mol Al2O3. You want dGo for 7.1 mol Al2O3.
Convert from 1 mol to 7.1 mol and you hve it.