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Question: Reactions involving gold have been of particular interest to a chemist. Consider the following react...

Reactions involving gold have been of particular interest to a chemist. Consider the following reactions, Au(OH)3+4HCl⟶HAuCl4+3H2O,ΔH=−28 kCal mol−1Au(OH)_3 + 4HCl \longrightarrow HAuCl_4 + 3H_2O, \quad \Delta H = -28 \text{ kCal mol}^{-1} Au(OH)3+4HBr⟶HAuBr4+3H2O,ΔH=−36.8 kCal mol−1Au(OH)_3 + 4HBr \longrightarrow HAuBr_4 + 3H_2O, \quad \Delta H = -36.8 \text{ kCal mol}^{-1} In an experiment there was an absorption of 0.44 kCal when one mole of HAuBr4HAuBr_4 was mixed with 4 moles of HClHCl. What is the percentage conversion of HAuBr4HAuBr_4 into HAuCl4HAuCl_4?

A

5%

B

6%

C

7%

D

8%

Answer

5%

Explanation

Solution

The enthalpy change for the conversion of HAuBr4HAuBr_4 to HAuCl4HAuCl_4 is determined using Hess's Law. Given reactions:

  1. Au(OH)3+4HCl⟶HAuCl4+3H2O,ΔH1=−28 kCal mol−1Au(OH)_3 + 4HCl \longrightarrow HAuCl_4 + 3H_2O, \quad \Delta H_1 = -28 \text{ kCal mol}^{-1}
  2. Au(OH)3+4HBr⟶HAuBr4+3H2O,ΔH2=−36.8 kCal mol−1Au(OH)_3 + 4HBr \longrightarrow HAuBr_4 + 3H_2O, \quad \Delta H_2 = -36.8 \text{ kCal mol}^{-1}

To find the enthalpy change for the reaction HAuBr4+4HCl⟶HAuCl4+4HBrHAuBr_4 + 4HCl \longrightarrow HAuCl_4 + 4HBr, we manipulate the given reactions:

  • Keep reaction (1) as is.
  • Reverse reaction (2) and change its enthalpy sign: HAuBr4+3H2O⟶Au(OH)3+4HBr,ΔH−2=−(−36.8)=+36.8 kCal mol−1HAuBr_4 + 3H_2O \longrightarrow Au(OH)_3 + 4HBr, \quad \Delta H_{-2} = -(-36.8) = +36.8 \text{ kCal mol}^{-1}

Adding reaction (1) and the reversed reaction (2): Au(OH)3+4HCl⟶HAuCl4+3H2OAu(OH)_3 + 4HCl \longrightarrow HAuCl_4 + 3H_2O HAuBr4+3H2O⟶Au(OH)3+4HBrHAuBr_4 + 3H_2O \longrightarrow Au(OH)_3 + 4HBr

HAuBr4+4HCl⟶HAuCl4+4HBrHAuBr_4 + 4HCl \longrightarrow HAuCl_4 + 4HBr

The enthalpy change for this conversion reaction is: ΔHconv=ΔH1+ΔH−2=−28 kCal mol−1+36.8 kCal mol−1=+8.8 kCal mol−1\Delta H_{conv} = \Delta H_1 + \Delta H_{-2} = -28 \text{ kCal mol}^{-1} + 36.8 \text{ kCal mol}^{-1} = +8.8 \text{ kCal mol}^{-1}

This means that for the complete conversion of 1 mole of HAuBr4HAuBr_4 to HAuCl4HAuCl_4, 8.8 kCal of heat is absorbed. In the experiment, 0.44 kCal of heat was absorbed when 1 mole of HAuBr4HAuBr_4 was mixed with 4 moles of HClHCl.

The percentage conversion is calculated as: \text{Percentage Conversion} = \frac{\text{Observed heat absorbed}}{\text{Theoretical heat absorbed for 100% conversion}} \times 100 Percentage Conversion=0.44 kCal8.8 kCal×100\text{Percentage Conversion} = \frac{0.44 \text{ kCal}}{8.8 \text{ kCal}} \times 100 Percentage Conversion=120×100=5%\text{Percentage Conversion} = \frac{1}{20} \times 100 = 5\%