Reactions involving gold have been of particular interest to a chemist. Consider the following react — Thermodynamics and Thermochemistry Chemistry Question
Question
Reactions involving gold have been of particular interest to a chemist. Consider the following reactions:<br>Au(OH)3 + 4$HCl$ → HAuCl4 + 3$H_2O$; ΔH = -28 kcal<br>Au(OH)3 + 4$HBr$ → HAuBr4 + 3$H_2O$; ΔH = -36.8 kcal<br>In an experiment, there was absorption of 0.44 kcal when one mole of HAuBr4 was mixed with 4 moles of $HCl$. What is the percentage conversion of HAuBr4 into HAuCl4?
💡 Solution & Explanation
We can represent the conversion reaction as:<br>HAuBr4 + 4$HCl$ → HAuCl4 + 4$HBr$.<br>By Hess's Law, the enthalpy change for this conversion reaction is:<br>ΔH = ΔH1 - Δ$H_2$ = -28.0 - (-36.8) = +8.8 kcal/mol.<br>Since this is endothermic, converting 1 mole of HAuBr4 requires the absorption of 8.8 kcal.<br>In the experiment, 0.44 kcal of heat was absorbed. Therefore, the moles of HAuBr4 converted is:<br>n_converted = 0.44 kcal / 8.8 kcal/mol = 0.05 mol.<br>Percentage conversion = (0.05 / 1.0) × 100 = 5%.