Question

Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,

Au(OH)3 + 4 HCl$\rightarrow$HAuCl4 + 3 H2O, $\Delta$H = -28 kcal

Au(OH)3 + 4 HBr$\rightarrow$HAuBr4 + 3 H2O, $\Delta$H = -36.8 kcal

In an experiment, there was an absorption of 0.44 kcal when one mole of HAuBr4 was mixed with 4 moles of HCl. Then, the fraction of HAuBr4 converted into HAuCl4 (percentage conversion)

• A. 5%
• B. 6 %
• C. 7%
• D. 8%

Answer 1

Answer: (A)

5%

Answer 2

Au(OH)3 + 4 HCl ¾® HAuCl4 + 3 H2O,

$\Delta$H = - 28 kcal

HAuBr4 + 3 H2O ¾® Au(OH)3 + 4 HBr ,

$\Delta$H = 36.8 kcal

____________________________________

HAuBr4 + 4 HCl ¾® HAuCl4 + 3 H2O ,

$\Delta$H = 8.8 kcal

When 100% conversion, heat absorbed = 8.8 Kcal

% conversion = $\frac{0.44}{8.8}$ ´ 100 = 5%