ThermochemistryHard

Question

Reactions involving gold have been of particular interest to a chemist. Consider the following reactions.

Au(OH)3 + 4HCl → HAuCl4 + 3H2O: ΔH = −28 kcal

Au(OH)3 + 4HBr → HAuBr4 + 3H2O: ΔH = −36.8 kcal

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?

Options

A.0.5%
B.0.6%
C.5%
D.50%

Solution

$HAuBr_{4} + 4HCl \rightarrow HAuCl_{4} + 4HBr;$

$\Delta H = ( - 28) - ( - 36.8) = 8.8\text{ kcal}$

$\therefore$Percentage reaction = $\frac{0.44}{8.8} \times 100 = 5\%$

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