Question

Consider the reaction} \[ 2\text{H}_2\text{S (g)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O (l)} + 2\text{SO}_2\text{(g)} \] The magnitude of enthalpy change for the reaction in kJ mol\(^{-1}\) is _______ (Nearest integer).

Answer 1

Correct Answer: -2294

\textcolor{red}{Step 1: Recall the formula for enthalpy change.}
The enthalpy change for the reaction can be calculated using the formula: \[ \Delta H = \sum \Delta_f H^{\circ} (\text{products}) - \sum \Delta_f H^{\circ} (\text{reactants}) \] \textcolor{red}{Step 2: Use the given enthalpy values.}
From the question, we are given the following enthalpy values: \[ \Delta_f H^{\circ} (\text{H}_2\text{S}) = -20.1 \, \text{kJ mol}^{-1}, \quad \Delta_f H^{\circ} (\text{H}_2\text{O}) = -286.0 \, \text{kJ mol}^{-1}, \quad \Delta_f H^{\circ} (\text{SO}_2) = -297.0 \, \text{kJ mol}^{-1} \] \textcolor{red}{Step 3: Calculate the enthalpy change.}
The enthalpy change for the reaction is: \[ \Delta H = [2 \times \Delta_f H^{\circ} (\text{H}_2\text{O}) + 2 \times \Delta_f H^{\circ} (\text{SO}_2)] - [2 \times \Delta_f H^{\circ} (\text{H}_2\text{S}) + 3 \times \Delta_f H^{\circ} (\text{O}_2)] \] Since the enthalpy of formation of \(O_2\) is zero, the equation simplifies to: \[ \Delta H = [2 \times (-286.0) + 2 \times (-297.0)] - [2 \times (-20.1)] \] \textcolor{red}{Step 4: Perform the calculation.}
\[ \Delta H = [2 \times (-286.0) + 2 \times (-297.0)] - [2 \times (-20.1)] = -1146.0 - 1188.0 + 40.2 = -2293.8 \, \text{kJ mol}^{-1} \] The nearest integer is: \[ \boxed{-2294 \, \text{kJ mol}^{-1}} \]