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>$\text{Au(OH)}_3 + 4\text{HCl} \to \text{HAuCl}_4 + 3\text{H}_2\text{O}; \Delta H = -28\text{ kcal}$<br>$\text{Au(OH)}_3 + 4\text{HBr} \to \text{HAuBr}_4 + 3\text{H}_2\text{O}; \Delta H = -36.8\text{ kcal}$<br>In an experiment, there was absorption of 0.44 kcal when one mole of $\text{HAuBr}_4$ was mixed with 4 moles of HCl. What is the percentage conversion of $\text{HAuBr}_4$ into $\text{HAuCl}_4$?
💡 Solution & Explanation
To get $\text{HAuBr}_4 + 4\text{HCl} \to \text{HAuCl}_4 + 4\text{HBr}$, reverse the second reaction and add to the first: $\Delta H_{\text{rxn}} = -28 - (-36.8) = +8.8\text{ kcal/mol}$. The actual heat absorbed is 0.44 kcal, meaning $0.44 / 8.8 = 0.05$ moles reacted. The percentage conversion is $0.05 \times 100\% = 5\%$.