Question

Which from following reactions performs zero work?

• A. \(\text{CH}_{4(\text{g})} + \text{Cl}_{2(\text{g})} \longrightarrow \text{CH}_3\text{Cl}_{(\text{g})} + \text{HCl}_{(\text{g})}\)
• B. \(3\text{H}_{2(\text{g})} + \text{N}_{2(\text{g})} \longrightarrow 2\text{NH}_{3(\text{g})}\)
• C. \(\text{C}_2\text{H}_{2(\text{g})} + \frac{5}{2}\text{O}_{2(\text{g})} \longrightarrow 2\text{CO}_{2(\text{g})} + \text{H}_2\text{O}_{(\text{l})}\)
• D. \(2\text{C}_2\text{H}_{6(\text{g})} + 7\text{O}_{2(\text{g})} \longrightarrow 4\text{CO}_{2(\text{g})} + 6\text{H}_2\text{O}_{(\text{l})}\)

Hint:

If \(\Delta n_g = 0\), work done = 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The work done by or on a gas during a chemical reaction at constant temperature and pressure is given by \(W = -P\Delta V = -\Delta n_g RT\). For the work to be zero, the change in the number of moles of gaseous species (\(\Delta n_g\)) must be zero.
Step 2: Key Formula or Approach:
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \] If \(\Delta n_g = 0\), then \(W = 0\).
Step 3: Detailed Explanation:
Let's check \(\Delta n_g\) for each option:
- (A) \(\Delta n_g = (1 + 1) - (1 + 1) = 2 - 2 = 0\). Thus, \(W = 0\).
- (B) \(\Delta n_g = 2 - (3 + 1) = 2 - 4 = -2\). Thus, \(W \neq 0\).
- (C) \(\Delta n_g = 2 - (1 + 2.5) = 2 - 3.5 = -1.5\). Thus, \(W \neq 0\). (Note: \(\text{H}_2\text{O}\) is liquid).
- (D) \(\Delta n_g = 4 - (2 + 7) = 4 - 9 = -5\). Thus, \(W \neq 0\).
Only reaction A has no net change in the number of moles of gas.
Step 4: Final Answer:
Reaction A performs zero work.