Question:

The value of \( \Delta n_g \) for the oxidation of 4 mole of sulphur dioxide to sulphur trioxide is _______.

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To find \( \Delta n_g \), always start with a balanced chemical equation for the specific quantities mentioned. Then, carefully sum the stoichiometric coefficients of all gaseous products and subtract the sum of the coefficients of all gaseous reactants.
  • \( -2 \)
  • \( 2 \)
  • \( -4 \)
  • \( 4 \)
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The Correct Option is A

Solution and Explanation

Step 1: Write the balanced chemical equation for the reaction. The standard balanced equation for the oxidation of sulphur dioxide (\(\text{SO}_2\)) to sulphur trioxide (\(\text{SO}_3\)) is: \[ 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \] This equation shows the reaction for 2 moles of \(\text{SO}_2\). The question asks for 4 moles of \(\text{SO}_2\), so we multiply the entire equation by 2: \[ 4\text{SO}_2(g) + 2\text{O}_2(g) \rightleftharpoons 4\text{SO}_3(g) \]
Step 2: Calculate \( \Delta n_g \). The change in the number of moles of gas (\(\Delta n_g\)) is the difference between the total moles of gaseous products and the total moles of gaseous reactants. \[ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) \]
From our balanced equation:
- Moles of gaseous products = 4 mol (from 4\(\text{SO}_3\))
- Moles of gaseous reactants = 4 mol (from 4\(\text{SO}_2\)) + 2 mol (from 2\(\text{O}_2\)) = 6 mol

Step 3: Substitute the values to find \( \Delta n_g \). \[ \Delta n_g = 4 - 6 = -2 \]
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