Chemistry - Thermochemistry Question with Solution | TestHub
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:
Answer:
Solution:
Subtracting the second equation from the first gives:
HAuBr4 + 4HCl tarrow HAuCl4 + 4HBr
The change in enthalpy (Δ H) for this reaction is calculated as:
Δ H = -28 - (-36.8) = +8.8 kcal/mol.
If the conversion were complete, 8.8 kcal of heat would be absorbed. However, the observed heat absorption is 0.44 kcal.
Therefore, the percentage of conversion is:
Conversion = 0.44 kcal/8.8 kcal = 0.05 = 5%
This corresponds to option (c).
