Step 1: Understanding the Concept:
Pressure-volume work done during a chemical reaction at constant temperature and pressure is given by \(W = -P\Delta V\).
Using the ideal gas law, this can be approximated as \(W = -\Delta n_g RT\).
By convention in IUPAC thermodynamics, work done BY the system (expansion) is negative, while work done ON the system (compression) is positive.
Step 2: Key Formula or Approach:
Work \(W = -\Delta n_g RT\).
\(\Delta n_g = \text{(sum of moles of gaseous products)} - \text{(sum of moles of gaseous reactants)}\).
Negative work done (\(W<0\)) occurs when \(\Delta n_g>0\), i.e., an expansion takes place.
Step 3: Detailed Explanation:
Analyze \(\Delta n_g\) for each reaction:
(A) 2H\(_2\)O\(_2(\ell) \rightarrow\) 2H\(_2\)O\((\ell) +\) O\(_2(g)\): \(\Delta n_g = 1 - 0 = 1\).
Since \(\Delta n_g>0\), \(W = -(1)RT\), which is negative. This reaction exhibits negative work (work done by the system).
(B) NH\(_{3(g)} +\) HCl\(_{(g)} \rightarrow\) NH\(_4\)Cl\(_{(s)}\): \(\Delta n_g = 0 - (1 + 1) = -2\).
Since \(\Delta n_g<0\), \(W = -(-2)RT = +2RT\), which is positive.
(C) H\(_{2(g)} +\) Cl\(_{2(g)} \rightarrow\) HCl\(_{(g)}\): As written in the option, \(\Delta n_g = 1 - (1 + 1) = -1\).
Since \(\Delta n_g<0\), \(W = -(-1)RT = +RT\), which is positive.
(D) N\(_{2(g)} + 3\)H\(_{2(g)} \rightarrow\) 2NH\(_{3(g)}\): \(\Delta n_g = 2 - (1 + 3) = -2\).
Since \(\Delta n_g<0\), \(W = -(-2)RT = +2RT\), which is positive.
Step 4: Final Answer:
Reaction (A) exhibits negative work done as it results in a net increase in the number of moles of gas (\(\Delta n_g>0\)).