Question

Identify from following reactions that exhibits negative work done.

• A. \(2\mathrm{H}_{2}\mathrm{O}_{2(\ell )}\rightarrow 2\mathrm{H}_{2}\mathrm{O}_{(\ell )}+\mathrm{O}_{2(\text{ }\mathrm{g})}\)
• B. \(\mathrm{NH}_{3(\text{ }\mathrm{g})}+\mathrm{HCl}_{(\mathrm{g})}\rightarrow \mathrm{NH}_{4}\mathrm{Cl}_{(\mathrm{s})}\)
• C. \(\mathrm{H}_{2(\text{ }\mathrm{g})}+\mathrm{Cl}_{2(\text{ }\mathrm{g})}\rightarrow 2\mathrm{HCl}_{(\mathrm{g})}\)
• D. \(\mathrm{N}_{2(\text{ }\mathrm{g})}+3\mathrm{H}_{2(\text{ }\mathrm{g})}\rightarrow 2\mathrm{NH}_{3(\text{ }\mathrm{g})}\)

Hint:

Work done on the system is w = -P times change in volume; it is negative when the gas volume increases.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For reactions involving gases, \(w=-P\Delta V=-\Delta n_g RT\).
Work is negative when \(\Delta n_g\gt0\), that is when the number of gas moles increases (expansion).

Step 2: Find the change in gas moles
(A) \(2\text{H}_2\text{O}_2(\ell)\rightarrow2\text{H}_2\text{O}(\ell)+\text{O}_2(g)\): \(\Delta n_g=+1\).
(B) \(\Delta n_g=0-2=-2\). (C) \(\Delta n_g=2-2=0\). (D) \(\Delta n_g=2-4=-2\).

Step 3: Compare the options
Only reaction (A) produces gas from liquid, so the system expands and \(w=-\Delta n_gRT\) is negative. Option (A) is correct.

Final Answer:
Only the decomposition of hydrogen peroxide forms gas from a liquid, so its work term is negative. \[ \boxed{\text{(A) }2\text{H}_2\text{O}_{2(\ell)}\rightarrow2\text{H}_2\text{O}_{(\ell)}+\text{O}_{2(g)}} \]