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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 - 2023
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 - 2023
Step 1
Complete the diagram by drawing: in the first box, apparatus suitable for drying the sulfur dioxide gas;
Answer
To dry sulfur dioxide gas, a drying tube can be used, which is filled with anhydrous calcium chloride (or another drying agent). The tube should be connected to the outlet of the gas collection apparatus to absorb moisture from the gas as it passes through.
Step 2
Step 3
Show, by calculation, that sodium sulfite is the limiting reactant.
Answer
First, calculate the moles of sodium sulfite:
Moles of Na2SO3 = [ \text{Mass} = 0.40 \text{g} ] [ \text{Molar mass of Na2SO3} = 126.1 \text{g/mol} ] [ \text{Moles of Na2SO3} = \frac{0.40}{126.1} \approx 0.00317 \text{ moles} ]
Now calculate the moles of HCl present:
Molarity of HCl = 1 mol/l Volume = 50 cm³ = 0.050 l [ \text{Moles of HCl} = 1 \times 0.050 = 0.050 \text{ moles} ]
From the balanced equation, 1 mole of Na2SO3 reacts with 2 moles of HCl. Therefore, 0.00317 moles of Na2SO3 would require: [ 0.00317 \times 2 = 0.00634 \text{ moles of HCl} ]
Since we have more than enough HCl (0.050 moles), sodium sulfite is indeed the limiting reactant.
Step 4
Calculate the enthalpy change, in kJ mol⁻¹, for this reaction.
Answer
To calculate the enthalpy change, use Hess's Law:
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Write down the given reactions.
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Combine the equations in accordance with the reaction:
- C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ mol⁻¹
- 2S(s) + O2(g) → 2SO2(g) ΔH = −2 × 296.8 kJ mol⁻¹ = −593.6 kJ mol⁻¹
- C(s) + 2S(s) → C2S2(l) ΔH = +87.9 kJ mol⁻¹
Thus, the overall reaction enthalpy change is:
[ ΔH = -393.5 + (-593.6) + 87.9 ] [ ΔH = -899.2 kJ mol⁻¹ ]
Therefore, the enthalpy change for the combustion of carbon disulfide is approximately -1075 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
Carbon dioxide (CO2) is linear and non-polar, which means that the dipoles of the CO2 molecule cancel out. This results in less interaction with polar water molecules, significantly decreasing its solubility in water. In contrast, sulfur dioxide (SO2) has a bent shape, which results in a net dipole moment. This polarity increases the interaction between sulfur dioxide and water molecules, leading to much higher solubility.