Question:
Calculate \(\Delta_r H^0\) for the reaction \(A^+ (aq) + B^- (aq) \rightarrow AB(s)\) at 25°C. Given:
\[
\Delta_f H^0 (A^+, aq) = 200~\text{kJ mol}^{-1}, \quad
\Delta_f H^0 (B^-, aq) = -300~\text{kJ mol}^{-1},
\]
\[
\Delta_f H^0 (A^+, g) = -150~\text{kJ mol}^{-1}, \quad
\Delta_f H^0 (AB, s) = -250~\text{kJ mol}^{-1},
\quad
\Delta_f H^0 (AB, g) = +130~\text{kJ mol}^{-1}
\]
Calculate \(\Delta_r H^0\) for the reaction \(A^+ (aq) + B^- (aq) \rightarrow AB(s)\) at 25°C. Given:
\[ \Delta_f H^0 (A^+, aq) = 200~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (B^-, aq) = -300~\text{kJ mol}^{-1}, \] \[ \Delta_f H^0 (A^+, g) = -150~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, s) = -250~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, g) = +130~\text{kJ mol}^{-1} \]
\[ \Delta_f H^0 (A^+, aq) = 200~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (B^-, aq) = -300~\text{kJ mol}^{-1}, \] \[ \Delta_f H^0 (A^+, g) = -150~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, s) = -250~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, g) = +130~\text{kJ mol}^{-1} \]
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Always remember that reaction enthalpy is calculated as the sum of formation enthalpies of products minus reactants. Pay attention to the correct states (aq, s, g) given in the problem.
Updated On: Jun 19, 2026
- 150 kJ mol\(^{-1}\)
- -150 kJ mol\(^{-1}\)
- 0 kJ mol\(^{-1}\)
- +130 kJ mol\(^{-1}\)
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The Correct Option is B
Solution and Explanation
Step 1: Recall the reaction enthalpy formula.
\[ \Delta_r H^0 = \sum \Delta_f H^0 (\text{products}) - \sum \Delta_f H^0 (\text{reactants}) \]
Step 2: Identify products and reactants.
Products: \(ce{AB(s)}\) Reactants: \(ce{A^+(aq)}\) and \(ce{B^-(aq)}\)
Step 3: Substitute the given enthalpy values.
\[ \Delta_r H^\circ = \Delta_f H^\circ\bigl(\ce{AB(s)}\bigr) - \left[ \Delta_f H^\circ\bigl(\ce{A+ (aq)}\bigr) + \Delta_f H^\circ\bigl(\ce{B- (aq)}\bigr) \right] \] \[ \Delta_r H^0 = -250 - [200 + (-300)] \]
Step 4: Simplify the brackets.
\[ 200 + (-300) = 200 - 300 = -100 \]
Step 5: Compute the reaction enthalpy.
\[ \Delta_r H^0 = -250 - (-100) = -250 + 100 = -150~\text{kJ mol}^{-1} \]
Step 6: Conclusion.
The standard enthalpy change for the reaction is \(-150~\text{kJ mol}^{-1}\).
\[ \Delta_r H^0 = \sum \Delta_f H^0 (\text{products}) - \sum \Delta_f H^0 (\text{reactants}) \]
Step 2: Identify products and reactants.
Products: \(ce{AB(s)}\) Reactants: \(ce{A^+(aq)}\) and \(ce{B^-(aq)}\)
Step 3: Substitute the given enthalpy values.
\[ \Delta_r H^\circ = \Delta_f H^\circ\bigl(\ce{AB(s)}\bigr) - \left[ \Delta_f H^\circ\bigl(\ce{A+ (aq)}\bigr) + \Delta_f H^\circ\bigl(\ce{B- (aq)}\bigr) \right] \] \[ \Delta_r H^0 = -250 - [200 + (-300)] \]
Step 4: Simplify the brackets.
\[ 200 + (-300) = 200 - 300 = -100 \]
Step 5: Compute the reaction enthalpy.
\[ \Delta_r H^0 = -250 - (-100) = -250 + 100 = -150~\text{kJ mol}^{-1} \]
Step 6: Conclusion.
The standard enthalpy change for the reaction is \(-150~\text{kJ mol}^{-1}\).
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