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:
The work done in a chemical reaction involving gases 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 equation for 1 mole of \(SO_2\) is: \(SO_{2(g)} + 0.5 O_{2(g)} \rightarrow SO_{3(g)}\).
For 2 moles of \(SO_2\), the balanced equation is:
\[ 2SO_{2(g)} + 1O_{2(g)} \rightarrow 2SO_{3(g)} \]
\(\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1 \text{ mole}\).
Now, substitute into the work formula:
\[ W = - (-1) \times 8.314 \times 298 \]
\[ W = 1 \times 8.314 \times 298 \]
\[ W = 2477.57 \text{ J} \]
Since \(W\) is positive, work is done ON the system (compression).
Step 4: Final Answer:
The work done is \(2477.57 \text{ J}\).