Calculate the work done in the oxidation of one mole $\mathrm{HCl}_{(\mathrm{g})}$ at $27^{\circ}…

Calculate the work done in the oxidation of one mole $\mathrm{HCl}_{(\mathrm{g})}$ at $27^{\circ} \mathrm{C}$, according to reaction. $4 \mathrm{HCl}_{(\mathrm{g})}+\mathrm{O}_{2_{(\mathrm{g})}} \rightarrow 2 \mathrm{Cl}_{2_{(\mathrm{g})}}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \quad\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$
  1. 2494.2 J
  2. 623.6 J
  3. $1247 \cdot 1 \mathrm{~J}$
  4. 1870.7 J

Solution

The work done during gaseous reactions under constant temperature and pressure is given by $W = -\Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gas.

For $4\mathrm{HCl}_{(\mathrm{g})} + \mathrm{O}_{2_{(\mathrm{g})}} \rightarrow 2\mathrm{Cl}_{2_{(\mathrm{g})}} + 2\mathrm{H}_2\mathrm{O}_{(\mathrm{g})}$, the gaseous moles are: reactants = 5, products = 4, so $\Delta n_g = -1$.

Per mole of $\mathrm{HCl}_{(\mathrm{g})}$, $\Delta n_g = -1/4 = -0.25$.

At $T = 300\ \mathrm{K}$ and $R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$, $W = -(-0.25) \times 8.314 \times 300 = 623.55\ \mathrm{J}$.

This value corresponds to option B.

Asked in: MHT CET 2025 (05 May Shift 2)

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