What is DG° at 298 K for the following reaction?
Pb(s) + PbO2(s) -> 2PbO(s)
Substance
DG°f (kJ/mol) at 298 K
PbO(s)
–187.9
PbO2(s)
–217.3
Pb(s)
0
Therefore, DG° at 298 K for the reaction is -29.4 kJ/mol.
To find the standard free energy change (ΔG°) at 298 K for the reaction:
Pb(s) + PbO2(s) -> 2PbO(s)
You need to use the thermodynamic data given for the substances involved.
The standard free energy change (ΔG°) for a reaction can be calculated using the following formula:
ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)
Where:
Σn is the stoichiometric coefficient of each species in the balanced equation.
ΔG°f is the standard free energy of formation for each species at 298 K.
Substituting the values given:
ΔG° = (2 * -187.9 kJ/mol) - (1 * -217.3 kJ/mol)
Simplifying the equation:
ΔG° = -375.8 kJ/mol + 217.3 kJ/mol
ΔG° = -158.5 kJ/mol
Therefore, the standard free energy change (ΔG°) at 298 K for the given reaction is -158.5 kJ/mol.
To find the standard free energy change, ΔG°, for the given reaction at 298 K, you can use the formula:
ΔG° = ∑ΔG°f(products) - ∑ΔG°f(reactants)
Start by identifying the products and reactants in the reaction:
Reactants:
Pb(s)
PbO2(s)
Products:
2PbO(s)
Next, you need to look up the standard free energy of formation, ΔG°f, for each substance involved at 298 K:
ΔG°f (PbO(s)) = -187.9 kJ/mol
ΔG°f (PbO2(s)) = -217.3 kJ/mol
Since the reaction involves the formation of 2 moles of PbO(s), you need to multiply the ΔG°f value of PbO(s) by 2 to account for the stoichiometry:
2 × ΔG°f (PbO(s)) = 2 × (-187.9 kJ/mol) = -375.8 kJ/mol
Now, you can substitute the values into the formula for ΔG°:
ΔG° = ∑ΔG°f(products) - ∑ΔG°f(reactants)
ΔG° = [2 × ΔG°f (PbO(s))] - [ΔG°f (Pb(s)) + ΔG°f (PbO2(s))]
ΔG° = [-375.8 kJ/mol] - [0 kJ/mol + (-217.3 kJ/mol)]
Simplifying the equation:
ΔG° = -375.8 kJ/mol + 217.3 kJ/mol
ΔG° = -158.5 kJ/mol
Therefore, the standard free energy change, ΔG°, for the given reaction at 298 K is -158.5 kJ/mol.