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The diagram shows a sequence of steps in the removal of grease from a surface - HSC - SSCE Chemistry - Question 32 - 2013 - Paper 1
Question 32
The diagram shows a sequence of steps in the removal of grease from a surface. Explain the process shown in these steps. Hydrogen iodide is a colourless gas that wi... show full transcript
Worked Solution & Example Answer:The diagram shows a sequence of steps in the removal of grease from a surface - HSC - SSCE Chemistry - Question 32 - 2013 - Paper 1
Step 1
Explain the process shown in these steps.
Answer
The removal of grease from a surface involves a series of steps utilizing a surfactant:
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In Step 1, the surfactant is introduced to the grease. The surfactant has a hydrophilic (water-attracting) head and a hydrophobic (water-repelling) tail. The hydrophobic tail interacts with the grease molecules, allowing them to disperse into the water.
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In Step 2, as the surfactant surrounds the grease particles, it forms a stable emulsion. This process is crucial as it helps to break the grease away from the surface, allowing it to be rinsed away.
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Finally, in Step 3, the grease, now in the emulsion form, is washed away with water, leaving the surface clean.
Step 2
Calculate the equilibrium constant for this reaction.
Answer
To calculate the equilibrium constant (K) for the reaction, we first determine the concentrations of the reactants and products at equilibrium.
From the problem statement:
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Initial amount of HI = 0.60 moles in a 1.0 L container gives a concentration of:
[HI] = rac{0.60}{1.0} = 0.60 ext{ M}
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At equilibrium, the concentration of I2 is given as 0.25 moles in 1.0 L:
[I2] = rac{0.25}{1.0} = 0.25 ext{ M}
Using the stoichiometry of the reaction:
- Since 2 moles of HI produce 1 mole of I2, at equilibrium, the amount of HI that has been used can be calculated:
Considering that 0.25 moles of I2 were produced, 0.50 moles of HI must have been consumed. Thus, the equilibrium concentration of HI is:
[HI] =
Now, we have:
- [H2] = [I2] = 0.25 M and [HI] = 0.10 M
The equilibrium constant expression for the reaction is:
K = rac{[H2][I2]}{[HI]^2}
Substituting the values:
K = rac{(0.25)(0.25)}{(0.10)^2} = rac{0.0625}{0.01} = 6.25
Step 3
Explain the change in the appearance of its contents.
Answer
When the container is cooled, the reaction's equilibrium shifts according to Le Chatelier's principle. Since the decomposition of hydrogen iodide is an endothermic process, cooling the system will push the equilibrium to the left to produce more reactants (HI).
As a result, the concentration of purple iodine gas (I2) will decrease, leading the reaction's visual appearance to change. Specifically, the solution will turn to a lighter shade of purple or appear colourless as the amount of iodine decreases.