Photo AI

The diagram shows a sequence of steps in the removal of grease from a surface - HSC - SSCE Chemistry - Question 32 - 2013 - Paper 1

Question icon

Question 32

The-diagram-shows-a-sequence-of-steps-in-the-removal-of-grease-from-a-surface-HSC-SSCE Chemistry-Question 32-2013-Paper 1.png

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 w... 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.

96%

114 rated

Answer

The steps depicted in the diagram illustrate the process of removing grease from a surface using a surfactant.

  1. Step 1: The surfactant molecules are introduced to the greasy surface. These molecules have a hydrophilic (water-attracting) head and a hydrophobic (water-repelling) tail. The hydrophobic tails attach to the grease, while the hydrophilic heads remain in the water, reducing the surface tension and allowing the grease to become emulsified.

  2. Step 2: As the surfactant molecules continue to interact with the grease, they surround and encapsulate the grease particles, forming structures known as micelles. The hydrophilic heads remain in the water, allowing the greasy particles to be lifted away from the surface to which they were attached.

  3. Step 3: Finally, the micelles, now containing the grease, can be rinsed away with water, effectively cleaning the surface. The surfactant's ability to encapsulate grease allows for effective cleaning even in the presence of water.

Step 2

Calculate the equilibrium constant for this reaction.

99%

104 rated

Answer

To calculate the equilibrium constant (K) for the reaction:

2HI(g)H2(g)+I2(g)2HI(g) ⇌ H2(g) + I2(g)

  1. Initial Moles:

    • HI: 0.60 moles in 1.0 L (0.60 M)
    • H2: 0 moles
    • I2: 0 moles
  2. At Equilibrium: 0.25 moles of iodine gas (I2) are present, which implies that 0.25 moles of hydrogen gas (H2) are also formed (stoichiometry of 1:1). Therefore,

  • HI at equilibrium = Initial HI - 2(moles of I2 formed) = 0.60 - 0.25 = 0.10 moles
  1. Equilibrium Concentrations:

    • [HI] = 0.10 M (0.10 moles/liter)
    • [H2] = 0.25 M (0.25 moles/liter)
    • [I2] = 0.25 M (0.25 moles/liter)
  2. Equilibrium Constant Expression:

    K=[H2][I2][HI]2K = \frac{[H2][I2]}{[HI]^2}

    K=(0.25)(0.25)(0.10)2=0.06250.01=6.25K = \frac{(0.25)(0.25)}{(0.10)^2} = \frac{0.0625}{0.01} = 6.25

    Therefore, the equilibrium constant for this reaction is 6.25.

Step 3

Explain the change in the appearance of its contents.

96%

101 rated

Answer

Since the reaction is endothermic, cooling the system will drive the equilibrium to the left, according to Le Chatelier's Principle. Therefore, as the container is cooled, the concentration of the purple iodine gas (I2) will decrease as it reacts to form hydrogen iodide (HI) and the colour will fade. The reduced concentration of I2 will result in a lighter shade of purple, hence affecting the appearance of the contents in the container.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;