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: Understanding the Question:
The problem asks to determine the magnitude of the standard enthalpy of reaction ($|\Delta_rH^\circ|$) using the given standard enthalpies of formation ($\Delta_fH^\circ$) for all reactants and products.
Step 2: Key Formula or Approach:
The standard enthalpy of a reaction is calculated using the formula:
\[ \Delta_rH^\circ = \sum \Delta_fH^\circ(\text{products}) - \sum \Delta_fH^\circ(\text{reactants}) \]
Step 3: Detailed Explanation:
For the reaction: $2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(\ell)$
The enthalpy of reaction is:
\[ \Delta_rH^\circ = [2 \times \Delta_fH^\circ(SO_2,g) + 2 \times \Delta_fH^\circ(H_2O,\ell)] - [2 \times \Delta_fH^\circ(H_2S,g) + 3 \times \Delta_fH^\circ(O_2,g)] \]
Note that the standard enthalpy of formation of an element in its standard state, like $O_2(g)$, is zero.
\[ \Delta_rH^\circ = [2 \times (- 296.8) + 2 \times (- 285.8)] - [2 \times (- 20.6) + 3 \times 0] \]
\[ \Delta_rH^\circ = [- 593.6 - 571.6] - [- 41.2] \]
\[ \Delta_rH^\circ = - 1165.2 + 41.2 = - 1124 \text{ kJ/mol} \]
The magnitude $|\Delta_rH^\circ| = |- 1124| = 1124$ kJ/mol.
Step 4: Final Answer:
The value of $|\Delta_rH^\circ|$ is 1124 kJ/mol.