Question

Calculate the amount of work done in the oxidation of 1 mole of HCl(g) at \(127^{\circ }\)C if
\(4\text{HCl(g)}+\text{O}_{2}\text{(g)}\rightarrow 2\text{Cl}_{2}\text{(g)}+2\text{H}_{2}\text{O(g)}\) ; [ R = 8.314 JK\({}^{-1}\) mol\({}^{-1}\)]

• A. \(831.4\) J
• B. \(491.6\) kJ
• C. \(1474.6\) J
• D. \(1662.8\) J

Hint:

Find the change in moles of gas for 1 mole of HCl, then use w equals minus delta n g times RT.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For reactions of gases at constant pressure, \(w=-\Delta n_g RT\).

Step 2: Find \(\Delta n_g\)
For the equation: reactant gas moles \(=4+1=5\), product gas moles \(=2+2=4\), so \(\Delta n_g=-1\) for 4 mol HCl.
For 1 mol HCl, \(\Delta n_g=-\dfrac{1}{4}\).

Step 3: Calculation
\(T=127+273=400\text{ K}\)
\(w=-\left(-\dfrac{1}{4}\right)\times8.314\times400=+831.4\text{ J}\).
The work is positive because the volume decreases and work is done on the system.

Final Answer:
The work done is 831.4 J. \[ \boxed{\text{(A) }831.4\text{ J}} \]