Reactions involving gold have been of particular interest to alchemists. Consider the following reactions,
`Au(OH)_(3) + 4 HCl ""^(3)//""_(4)®HAuCl_(4) + 3 H_(2)O, DH = - 28 kcal Au(OH)_(3) + 4 HBr""^(3)//""_(4)®HAuBr_(4) + 3 H_(2)O`,
`DH = -36.8 kcal`
In an experiment there was an absorption of 0.44 kcal when one mole of `HAuBr_(4)` was mixed with 4 moles of HCl. Then the fraction of `HAuBr_(4)` converted into `HAuCl_(4)` (percentage conversion)

Options

A. 0.05
B. 0.06
C. 0.07
D. 0.08

Answer

A

Explanation

Reaction for given condition is as follows. `HAuBr_(4) + 4HCl rarr HAuCl_(4) + 4HBr, Delta H` For this
Reaction is `= -28 - (-36.8) = 8.8` K.cal.
So one mole of `HAuBr_(4)` can be converted in to `HAuCl_(4)` by absorption of 8.8K. Cal
So Absorption of 0.44 K.Cal can give `(1)/(8.8) xx 0.44` moles of `HAuCl_(4)`.
So no. of mole of `HAuCl_(4)` obtained = 0.05
So % `=(0.05)/(1) xx 100= 5%`