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Q.

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

Au(OH)3+4HClHAuCl4+3H2O , ΔH=28kcal

Au(OH)3+4HBrHAuBr4+3H2OΔH=36.8kcal

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 HAuBr4  converted into HAuCl4 : (percentage conversion)

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a

6%

b

7%

c

8%

d

5%

answer is A.

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Detailed Solution

The general formula is 

HAuBr4+4HClHAuCl4+4HBr

Then

ΔH=28+36.8=8.8

0.448.8×100=5%

 

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