Question

Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,
$Au(OH{)}_3+4HCl\to HAuC{l}_4+3H_{2}O,△H=-28kcal$
$Au(OH{)}_3+4HBr\to HAuB{r}_4+3H_{2}O,△H=-36.8kcal$
In an experiment, there was an absorption of 0.44 kcal when one mole of $HAuB{r}_4$ was mixed with 4 moles of HCl. Then, the fraction of $HAuB{r}_4$ converted into $HAuC{l}_4$ (percentage conversion)

• A. 5%
• B. 10%
• C. 15%
• D. 8%

Answer 1

The correct option is A 5%

$Au(OH{)}_3+4HCl\to HAuC{l}_4+3H_{2}O,△H=-28kcal$
$HAu(Br{)}_4+3H_{2}O\to Au(OH{)}_3+4HBr,△H=+36.8kcal$
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$HAuB{r}_4+4HCl\to HAuC{l}_4+4HBr,△H=+8.8kcal$

When 100% conversion occurs, heat absorbed = 8.8 kcal
% conversion $=\frac{0.44}{8.8}\times 100=5$