consider the reaction at 298K. 2H2S(eq) + SO2-----3S(s) + 2H2O (g). Calculate Grxn under these conditions. PH2S=2atm Pso2 =1.5 atm PH2O + 0.01atm
Are these partial pressures at equilibrium? Perhaps they are not. If not Use
dGrxn = dGo + RT*lnQ
for Q = p^2H2O/pSO2*p^2H2S
To calculate the standard Gibbs free energy change (ΔG°) of the reaction under the given conditions, we need to use the reaction equation and the standard Gibbs free energy of formation for each compound involved.
The reaction equation is: 2H2S(eq) + SO2 → 3S(s) + 2H2O(g)
First, we need to determine the standard Gibbs free energy change of formation (ΔG°f) for each compound involved. These values can be found in tables or online sources. For this calculation, we will use the following values:
ΔG°f(H2S) = -20.6 kJ/mol
ΔG°f(SO2) = -300.2 kJ/mol
ΔG°f(S) = 0 kJ/mol
ΔG°f(H2O) = -237.1 kJ/mol
Next, we can calculate the standard Gibbs free energy change (ΔG°) for the reaction using the following equation:
ΔG° = ∑n ΔG°f(products) - ∑n ΔG°f(reactants)
where ∑n represents the sum of the stoichiometric coefficients multiplied by the standard Gibbs free energy change of formation for each compound.
Let's calculate it step by step:
∑n ΔG°f(products) = (3 mol S x 0 kJ/mol) + (2 mol H2O x -237.1 kJ/mol) = -474.2 kJ/mol
∑n ΔG°f(reactants) = (2 mol H2S x -20.6 kJ/mol) + (1 mol SO2 x -300.2 kJ/mol) = -341.4 kJ/mol
ΔG° = -474.2 kJ/mol - (-341.4 kJ/mol) = -132.8 kJ/mol
Therefore, the standard Gibbs free energy change (ΔG°) of the reaction at 298K under the given conditions is -132.8 kJ/mol.
To calculate the standard Gibbs free energy change (ΔG°) of the reaction at 298K, we can use the equation:
ΔG° = ΣΔG°f(products) - ΣΔG°f(reactants)
where ΔG°f is the standard Gibbs free energy of formation for each individual compound.
First, we need to determine the standard Gibbs free energy of formation for each component of the reaction. The values can be found using thermodynamic databases or tables. For this calculation, we will assume the following values:
ΔG°f(H2S) = -20.9 kJ/mol
ΔG°f(SO2) = -300.4 kJ/mol
ΔG°f(S) = 0 kJ/mol
ΔG°f(H2O) = -237.2 kJ/mol
Now we can calculate the standard Gibbs free energy change (ΔG°) for the reaction:
ΔG° = (3*ΔG°f(S)) + (2*ΔG°f(H2O)) - (2*ΔG°f(H2S)) - ΔG°f(SO2)
ΔG° = (3*0 kJ/mol) + (2*(-237.2 kJ/mol)) - (2*(-20.9 kJ/mol)) - (-300.4 kJ/mol)
ΔG° = 711.6 kJ/mol - 41.8 kJ/mol + 300.4 kJ/mol
ΔG° = 970.2 kJ/mol
Therefore, the standard Gibbs free energy change (ΔG°) for the given reaction at 298K is 970.2 kJ/mol.