Question

Calculate \(\Delta_r H^0\) for the reaction \(A^+ (aq) + B^- (aq) \rightarrow AB(s)\) at 25°C. Given:
\[ \Delta_f H^0 (A^+, aq) = 200~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (B^-, aq) = -300~\text{kJ mol}^{-1}, \] \[ \Delta_f H^0 (A^+, g) = -150~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, s) = -250~\text{kJ mol}^{-1}, \quad \Delta_f H^0 (AB, g) = +130~\text{kJ mol}^{-1} \]

• A. 150 kJ mol\(^{-1}\)
• B. -150 kJ mol\(^{-1}\)
• C. 0 kJ mol\(^{-1}\)
• D. +130 kJ mol\(^{-1}\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Standard reaction enthalpy formula.
ΔrH° = Σ ΔfH° (products) - Σ ΔfH° (reactants).

Step 2: Recognizing products and reactants.
Product: AB(s). Reactants: A⁺(aq) and B⁻(aq).

Step 3: Inserting the supplied enthalpies.
ΔrH° = ΔfH°(AB(s)) - [ΔfH°(A⁺(aq)) + ΔfH°(B⁻(aq))] = -250 - [200 + (-300)].

Step 4: Simplifying the bracket.
200 + (-300) = 200 - 300 = -100.

Step 5: Computing the final value.
ΔrH° = -250 - (-100) = -250 + 100 = -150 kJ mol⁻¹.

Step 6: Conclusion.
The standard enthalpy change for the reaction is -150 kJ mol⁻¹.