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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;
Answer
In the first box, an apparatus suitable for drying sulfur dioxide gas can be represented by a drying tube containing anhydrous calcium chloride. This absorbs moisture from the gas. The second box should include a gas collection jar inverted over water to collect the sulfur dioxide gas using upward displacement of air.
Step 2
Step 3
Show, by calculation, that sodium sulfite is the limiting reactant.
Answer
To determine the limiting reactant:
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Calculate the moles of sodium sulfite:
ext{Moles of Na}_2 ext{SO}_3 = rac{0.40 ext{ g}}{126.1 ext{ g mol}^{-1}} \ = 0.00317 ext{ mol}.
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Since the reaction requires 2 moles of HCl for every mole of Na2SO3, we need:
ext{Moles of HCl required} = 2 imes 0.00317 ext{ mol} = 0.00634 ext{ mol}.
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Calculate the moles of HCl available:
ext{Moles of HCl} = c imes V = 1 ext{ mol l}^{-1} imes 0.050 ext{ L} = 0.050 ext{ mol}.
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Since we require only 0.00634 mol of HCl and have 0.050 mol available, sodium sulfite (Na2SO3) is the limiting reactant.
Step 4
Calculate the enthalpy change, in kJ mol⁻¹, for this reaction using the following information.
Answer
The overall enthalpy change for the reaction can be calculated using Hess's law:
ext{ΔH} = ( ext{ΔH of products}) - ( ext{ΔH of reactants})
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From the information given:
ext{Reactants: } C_2S_2 ext{ and } 3 O_2 ightarrow ext{Products: } CO_2 ext{ and } 2 SO_2
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For this reaction:
ext{ΔH} = [(-393) + 2(-296)] - (+87.9) \ = -393 - 592 + 87.9 \ = -897 + 87.9 = -809.1 ext{ kJ mol}^{-1}.
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Therefore, the enthalpy change ext{ΔH} = -897.1 ext{ kJ mol}^{-1}.
Step 5
Step 6
Explain fully why carbon dioxide is much less soluble in water than sulfur dioxide is in water.
Answer
The difference in solubility can be attributed to molecular structure and polarity:
- Carbon dioxide (CO2) is a linear molecule with non-polar characteristics, resulting in weak interactions with water molecules.
- In contrast, sulfur dioxide (SO2) has a bent shape and polar characteristics; thus, it is better able to interact with water molecules through dipole-dipole interactions.
- The polarity of SO2 promotes stronger hydrogen bonding with water, leading to its much higher solubility.