Calculate `Delta_(r)H^(@)` for the reaction : Given `Delta_(f)H^(@)` values : `2H_(2)S(g) + 3O_2to 2H_(2)O(l) + 2SO_(2)(g)`
Given `Delta_(f)H^(@)` values : `H_(2)S(g) = -20.60`,
`H_(2)O(l) = -285.83, SO_(2)(g) = -296.83 kJ "mol"^(-1)`

Explanation

To calculate the standard enthalpy change (Δ_rH°) for the reaction:

\[ 2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g) \]

we will use the standard enthalpy of formation (Δ_fH°) values provided:

- Δ_fH°(H₂S(g)) = -20.60 kJ/mol
- Δ_fH°(H₂O(l)) = -285.83 kJ/mol
- Δ_fH°(SO₂(g)) = -296.83 kJ/mol
- Δ_fH°(O₂(g)) = 0 kJ/mol (since it is in its standard state)

### Step-by-step Calculation:

1. **Write the equation for Δ_rH°**:
\[ \Delta_rH° = \sum (\Delta_fH° \text{ of products}) - \sum (\Delta_fH° \text{ of reactants}) \]

2. **Calculate the enthalpy of the products**:
- For \(2H_2O(l)\):
\[ 2 \times \Delta_fH°(H_2O) = 2 \times (-285.83) = -571.66 \text{ kJ} \]
- For \(2SO_2(g)\):
\[ 2 \times \Delta_fH°(SO_2) = 2 \times (-296.83) = -593.66 \text{ kJ} \]
- Total enthalpy of products:
\[ \text{Total products} = -571.66 + (-593.66) = -1165.32 \text{ kJ} \]

3. **Calculate the enthalpy of the reactants**:
- For \(2H_2S(g)\):
\[ 2 \times \Delta_fH°(H_2S) = 2 \times (-20.60) = -41.20 \text{ kJ} \]
- For \(3O_2(g)\):
\[ 3 \times \Delta_fH°(O_2) = 3 \times 0 = 0 \text{ kJ} \]
- Total enthalpy of reactants:
\[ \text{Total reactants} = -41.20 + 0 = -41.20 \text{ kJ} \]

4. **Substitute into the Δ_rH° equation**:
\[ \Delta_rH° = \text{Total products} - \text{Total reactants} \]
\[ \Delta_rH° = -1165.32 - (-41.20) = -1165.32 + 41.20 = -1124.12 \text{ kJ} \]

5. **Final Result**:
The standard enthalpy change for the reaction is:
\[ \Delta_rH° = -1124.12 \text{ kJ/mol} \]