Question

Calculate \(|\Delta H^{\circ}|\) for the following reaction (in kJ/mol):
\[ 2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(l) \] Given data: \[ \Delta H_f^{\circ}(H_2S(g)) = -20.6 \, \text{kJ/mol}, \Delta H_f^{\circ}(SO_2(g)) = -296.8 \, \text{kJ/mol}, \Delta H_f^{\circ}(H_2O(l)) = -285.8 \, \text{kJ/mol} \]

Answer 1

Correct Answer: 1124

Step 1: Write the formula for calculating \(\Delta H^{\circ}\).
\[ \Delta H^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) \]
Step 2: Apply the given values for \(\Delta H_f^{\circ}\).
\[ \Delta H^{\circ} = 2 \times \Delta H_f^{\circ}(SO_2(g)) + 2 \times \Delta H_f^{\circ}(H_2O(l)) - 2 \times \Delta H_f^{\circ}(H_2S(g)) \]
Step 3: Substitute the values for each compound.
\[ \Delta H^{\circ} = 2 \times (-296.8) + 2 \times (-285.8) - 2 \times (-20.6) \]
Step 4: Calculate the result.
\[ \Delta H^{\circ} = -1124 \, \text{kJ/mol} \]