Question

Calculate the work done in oxidation of 2 moles of \(SO_2\) at 298 K, if \(SO_{2(g) + \frac{1}{2}O_{2(g)} \rightarrow SO_{3(g)}\) [Given : R = 8.314 J/K/mol].}

Hint:

Always check if the question asks for work done "by" or "on" the system. A positive value in \(W = -\Delta n RT\) means work is done ON the system (compression).

Answer 1

Step 1: Understanding the Concept:
For a chemical reaction involving gases, the work of expansion/compression is given by the change in the number of gaseous moles.
Step 2: Key Formula or Approach:
\[ W = -\Delta n_g RT \]
Where \(\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})\).
Step 3: Detailed Explanation:
The given reaction is: \(SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow SO_{3(g)}\).
For 1 mole of \(SO_2\), \(\Delta n_g = 1 - (1 + 0.5) = -0.5\).
For 2 moles of \(SO_2\), the reaction is: \(2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}\).
New \(\Delta n_g = 2 - (2 + 1) = -1\).
Substituting into the work formula:
\[ W = -(-1) \times 8.314 \times 298 \]
\[ W = 1 \times 8.314 \times 298 \]
\[ W = 2477.57 \text{ J} \]
Step 4: Final Answer:
The work done is \( + 2477.57 \) J. (Positive sign indicates work done on the system).