Find delta G knot for the following reaction:
2Ch3OH(l) + 3O2(g) --> 2CO2(g) + 4H2O(g)
See your other post; i.e., DGo = DGo products - DGo reactants.
Calculate delaG°
To find the standard Gibbs free energy change (ΔG°) for a reaction, we can use the following equation:
ΔG° = ΣnΔG° f (products) - ΣnΔG° f (reactants)
Where:
- n is the stoichiometric coefficient of each species in the balanced chemical equation.
- ΔG° f is the standard Gibbs free energy of formation for each species.
First, we need to find the standard Gibbs free energy of formation for each species involved in the reaction. The values can be found in tables.
The standard Gibbs free energy of formation for CH3OH(l), CO2(g), and H2O(g) are usually taken as zero per definition.
The standard Gibbs free energy of formation for O2(g) is 0 kJ/mol.
Using these values and the stoichiometry of the reaction, we can calculate ΔG°:
ΔG° = [2 × ΔG° f (CO2(g))] + [4 × ΔG° f (H2O(g))] - [2 × ΔG° f (CH3OH(l))] - [3 × ΔG° f (O2(g))]
Since CH3OH(l), CO2(g), and H2O(g) have ΔG° f values of zero, the equation simplifies to:
ΔG° = [2 × 0] + [4 × 0] - [2 × ΔG° f (CH3OH(l))] - [3 × 0]
ΔG° = -2 × ΔG° f (CH3OH(l))
The value of ΔG° f for CH3OH(l) can be found in tables. Assuming we find it to be ΔG° f = -166 kJ/mol, we can substitute it into the equation:
ΔG° = -2 × (-166 kJ/mol) = +332 kJ/mol
Therefore, the standard Gibbs free energy change (ΔG°) for the given reaction is +332 kJ/mol.
To find ΔG° for the given reaction, you need the standard Gibbs free energy change (ΔG°f) values of the reactants and products. The ΔG°f values represent the change in Gibbs free energy when 1 mole of a substance is formed from its constituent elements at standard conditions (1 atm pressure and 25°C temperature).
Here are the standard ΔG°f values for the given substances:
ΔG°f(CH3OH(l)) = -166.3 kJ/mol
ΔG°f(O2(g)) = 0 kJ/mol
ΔG°f(CO2(g)) = -394.4 kJ/mol
ΔG°f(H2O(g)) = -228.6 kJ/mol
Now we can calculate the standard Gibbs free energy change (ΔG°) for the reaction using the formula:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
In this case, the stoichiometric coefficients are:
reactants: 2 CH3OH(l), 3 O2(g)
products: 2 CO2(g), 4 H2O(g)
Substituting the values, we get:
ΔG° = [2ΔG°f(CO2(g)) + 4ΔG°f(H2O(g))] - [2ΔG°f(CH3OH(l)) + 3ΔG°f(O2(g))]
ΔG° = [2(-394.4 kJ/mol) + 4(-228.6 kJ/mol)] - [2(-166.3 kJ/mol) + 3(0 kJ/mol)]
ΔG° = -1054.8 kJ/mol - (-332.6 kJ/mol)
ΔG° = -722.2 kJ/mol
Therefore, the ΔG° for the given reaction is -722.2 kJ/mol.