The enthalpy change for the reaction, is –57 kJ. Predict the value of the enthalpy change in the fol — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

The target reaction involves the neutralization of 2 moles of $\text{H}^+$ and $\text{OH}^-$, providing $2 \times (-57) = -114\text{ kJ}$. Additionally, the precipitation of the insoluble $\text{BaSO}_4\text{(s)}$ releases further lattice/precipitation energy. Thus, the total energy released must be significantly more exothermic than -114 kJ, matching strictly with -200 kJ.