Question

Calculate the work done in the oxidation of one mole $HCl_{(g)}$ at 27$^\circ$C, according to reaction.
$4HCl_{(g)} + O_{2(g)} \rightarrow 2Cl_{2(g)} + 2H_2O_{(g)}$
(R = 8.314 J K$^{-1}$ mol$^{-1}$)

• A. 2494.2 J
• B. 623.6 J
• C. 1247.1 J
• D. 1870.7 J

Hint:

Work is done on the system when $\Delta n_g$ is negative (compression).

Answer 1

The Correct Option is B

Step 1: $\Delta n_g$ Calculation
For the reaction: $4HCl_{(g)} + O_{2(g)} \rightarrow 2Cl_{2(g)} + 2H_2O_{(g)}$
$\Delta n_g = (2+2) - (4+1) = 4 - 5 = -1$ (for 4 moles of $HCl$).

Step 2: $\Delta n_g$ for 1 mole $HCl$
$\Delta n_g (\text{per mole } HCl) = \frac{-1}{4} = -0.25$.

Step 3: Work Done Formula
$W = -\Delta n_g RT$
$W = -(-0.25) \times 8.314 \times 300$
$W = 0.25 \times 2494.2 = 623.55\text{ J}$.

Step 4: Conclusion
The work done is approximately 623.6 J.
Final Answer: (B)