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: Understanding the Question:
We need to find the work done for the oxidation of 1 mole of $HCl$ gas based on the stoichiometry of the given chemical reaction.
Step 2: Key Formula or Approach:
Pressure-Volume work for a gaseous chemical reaction at constant temperature is given by:
\[ w = -\Delta n_g RT \]
where $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
Step 3: Detailed Explanation:
The standard reaction is:
$4HCl_{(g)} + O_{2(g)} \rightarrow 2Cl_{2(g)} + 2H_2O_{(g)}$
Total moles of gaseous reactants for this stoichiometric equation = $4 + 1 = 5$ moles.
Total moles of gaseous products = $2 + 2 = 4$ moles.
$\Delta n_g$ for 4 moles of $HCl$ reacted = $4 - 5 = -1$.
The question asks for work done in the oxidation of one mole of $HCl$.
So, for 1 mole of $HCl$, $\Delta n_g = -1/4 = -0.25$.
Given:
$T = 27 + 273 = 300$ K
$R = 8.314$ J/(mol$\cdot$K)
Calculating work:
\[ w = -(-0.25) \times 8.314 \times 300 \]
\[ w = 0.25 \times 8.314 \times 300 = 75 \times 8.314 \]
\[ w = 623.55\text{ J} \approx 623.6\text{ J} \]
Step 4: Final Answer:
The work done is 623.6 J.