Given the following bond dissociation energies: $D_{\mathrm{H-H}} = 436\ \text{kJ mol}^{-1}$, $D_{\mathrm{F-F}} = 158\ \text{kJ mol}^{-1}$, $D_{\mathrm{Cl-Cl}} = 243\ \text{kJ mol}^{-1}$, $D_{\mathrm{H-F}} = 565\ \text{kJ mol}^{-1}$, $D_{\mathrm{H-Cl}} = 432\ \text{kJ mol}^{-1}$. Using these values, calculate the standard enthalpy change for the formation reactions $\mathrm{H_2 + F_2 \rightarrow 2HF}$ and $\mathrm{H_2 + Cl_2 \rightarrow 2HCl}$. Based on the calculated enthalpy changes, which acid, HF or HCl, is the stronger acid? Justify your answer.

Question

Accepted Answer

HF is the stronger acid (its formation releases -536 kJ mol⁻¹, much more exothermic than -185 kJ mol⁻¹ for HCl).

Neco Model Questions Vol1 2024 Chemistry Question 22

Answer Options

About This Question

Master the Exam!