Question

Which of the following reactions shows work of compression?

• A. $\text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)}$
• B. $\text{C}_2\text{H}_{6(g)} \rightarrow 2\text{C}_{(s)} + 3\text{H}_{2(g)}$
• C. $2\text{SO}_{3(g)} \rightarrow 2\text{SO}_{2(g)} + \text{O}_{2(g)}$
• D. $2\text{H}_2\text{O}_{2(l)} \rightarrow 2\text{H}_2\text{O}_{(l)} + \text{O}_{2(g)}$

Hint:

Compression equals a decrease in gas moles ($\Delta n_g < 0$). In option (A), 2 moles of gas combine to make a completely solid powder product, meaning the surrounding air compresses it down completely!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks to find the chemical process that involves the work of compression, which thermodynamically translates to work being done on the system by the surroundings ($W > 0$).

Step 2: Key Formula or Approach:
The pressure-volume ($PV$) work in a chemical reaction is given by: $$ W = -\Delta n_g RT $$ where $\Delta n_g$ is the change in the number of moles of gaseous components: $$ \Delta n_g = \sum n_{g\text{ (products)}} - \sum n_{g\text{ (reactants)}} $$

• If $\Delta n_g > 0$: Volume increases, representing a gas expansion ($W$ is negative, work done by the system).

• If $\Delta n_g < 0$: Volume decreases, representing a gas compression ($W$ is positive, work done on the system).


Step 3: Detailed Explanation:
Let's calculate $\Delta n_g$ for the given options:

• Option (A): $\text{NH}_{3(g)} + \text{HCl}_{(g)} \rightarrow \text{NH}_4\text{Cl}_{(s)}$ Here, $n_{g(\text{products})} = 0$ (since ammonium chloride is a solid) and $n_{g(\text{reactants})} = 1 + 1 = 2$. $$ \Delta n_g = 0 - 2 = -2 $$ Since $\Delta n_g$ is negative, the gas volume shrinks drastically, demonstrating a clear work of compression.

• Option (B): $\Delta n_g = 3 - 1 = +2$ (Expansion)

• Option (C): $\Delta n_g = (2+1) - 2 = +1$ (Expansion)

• Option (D): $\Delta n_g = 1 - 0 = +1$ (Expansion)


Step 4: Final Answer:
The reaction that shows a clear work of compression due to a drop in gaseous mole count is option (A).