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Sulfur dioxide is a colourless, toxic gas that is soluble in water and more dense than air - Scottish Highers Chemistry - Question 5 - 2017
Question 5
Sulfur dioxide is a colourless, toxic gas that is soluble in water and more dense than air. (a) One laboratory method for preparation of sulfur dioxide gas involves... show full transcript
Worked Solution & Example Answer:Sulfur dioxide is a colourless, toxic gas that is soluble in water and more dense than air - Scottish Highers Chemistry - Question 5 - 2017
Step 1
Complete the diagram by drawing: - in the first box, apparatus suitable for drying the sulfur dioxide gas; - in the second box, apparatus suitable for collecting the gas.
Answer
To complete the diagram, the first box should show an apparatus like a drying tube or a purifier that typically includes drying agents such as calcium chloride to remove moisture from the sulfur dioxide gas. The second box should represent a gas collection method, such as an inverted gas jar or a gas collection tube where the sulfur dioxide gas accumulates, using upward displacement of air.
Step 2
Show, by calculation, that sodium sulfite is the limiting reactant.
Answer
First, calculate the moles of sodium sulfite:
Mass of Na2SO3 = 0.40 g Molar mass of Na2SO3 = 126.1 g/mol.
Moles of Na2SO3 = ( \frac{0.40 , g}{126.1 , g/mol} = 0.00317 , mol )
From the reaction equation, 1 mole of Na2SO3 reacts with 2 moles of HCl. Therefore, 0.00317 moles of Na2SO3 would need:
Moles of HCl required = ( 2 \times 0.00317 , mol = 0.00634 , mol )
Now, calculate the moles of HCl provided in 50 cm³ (0.050 L) of 1 mol/L HCl:
Moles of HCl = concentration × volume = ( 1 , mol/L \times 0.050 , L = 0.050 , mol )
Since 0.00317 moles of Na2SO3 require 0.00634 moles of HCl but only 0.050 moles of HCl are available, Na2SO3 is the limiting reactant.
Step 3
Calculate the enthalpy change, in kJ mol⁻¹, for this reaction.
Answer
To calculate the enthalpy change, we need to apply Hess's law. The relevant equations are:
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Forming CO2: C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ mol⁻¹
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Forming SO2: S(s) + O2(g) → SO2(g) ΔH = −296 kJ mol⁻¹
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Formation of C2S2: C(s) + 2S(s) → C2S2(l) ΔH = +87.9 kJ mol⁻¹
Using Hess's law, we can summarize the reaction:
- Start with: ( 2SO2) (which we need to form) and adjust for enthalpy accordingly.
- The total enthalpy change will be derived from the addition and subtraction of the aforementioned enthalpy changes. The final calculation yields an enthalpy change of ΔH = -1075 kJ mol⁻¹.
Step 4
Step 5
Explain fully why carbon dioxide is much less soluble in water than sulfur dioxide is in water.
Answer
Carbon dioxide (CO2) is a linear molecule with low polarity, which limits its interaction with polar water molecules. In contrast, sulfur dioxide (SO2) is a bent molecule, resulting in a greater polarity. This polarity allows sulfur dioxide to interact more effectively with water molecules, thus enhancing its solubility. Additionally, the dipole-dipole interactions in SO2 further increase its solubility compared to CO2.