Use the standard enthalpies of formation values shown in the table below to calculate the enthalpy of the reaction shown below:
2SO2(g)+2H2O(ℓ) ---> 3O2(g)+2H2S(g)
Substance
ΔH°f (kJ/mol)
SO2(g)
84 kJ/mol
H2O(ℓ)
83 kJ/mol
O2(g)
0 kJ/mol
H2S(g)
50 kJ/mol
Reactants:
2SO2(g) + 2H2O(ℓ)
ΔH°f = (2 * 84 kJ/mol) + (2 * 83 kJ/mol)
ΔH°f = 334 kJ/mol
Products:
3O2(g) + 2H2S(g)
ΔH°f = (3 * 0 kJ/mol) + (2 * 50 kJ/mol)
ΔH°f = 100 kJ/mol
Enthalpy of reaction:
ΔH°rxn = ΣΔH°f (products) - ΣΔH°f (reactants)
ΔH°rxn = (3 * 0 kJ/mol) + (2 * 50 kJ/mol) - (2 * 84 kJ/mol) - (2 * 83 kJ/mol)
ΔH°rxn = -469 kJ/mol
Therefore, the enthalpy of the reaction 2SO2(g) + 2H2O(ℓ) ---> 3O2(g) + 2H2S(g) is -469 kJ/mol.
To calculate the enthalpy of the given reaction, we need to use Hess's law, which states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes of its individual steps.
Step 1: Reverse the enthalpy values for the reactants.
HS2(g): -50 kJ/mol
O2(g): 0 kJ/mol
Step 2: Multiply the enthalpy values by the stoichiometric coefficients of each substance in the balanced equation.
2SO2(g): 2 x 84 kJ/mol = 168 kJ/mol
2H2O(ℓ): 2 x 83 kJ/mol = 166 kJ/mol
3O2(g): 3 x 0 kJ/mol = 0 kJ/mol
2H2S(g): 2 x -50 kJ/mol = -100 kJ/mol
Step 3: Sum up the individual enthalpy changes.
∆H° = (168 kJ/mol) + (166 kJ/mol) + (0 kJ/mol) + (-100 kJ/mol)
∆H° = 234 kJ/mol
Therefore, the enthalpy of the given reaction is 234 kJ/mol.