Question:
Calculate the work done in following reaction.
\[ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+\mathrm{HCl}_{(\mathrm{g})} \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}_{(\mathrm{g})} \text { at } 27^{\circ} \mathrm{C} . \\ & \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned} \]
Calculate the work done in following reaction.
\[ \begin{aligned} & \mathrm{C}_2 \mathrm{H}_{4(\mathrm{~g})}+\mathrm{HCl}_{(\mathrm{g})} \longrightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}_{(\mathrm{g})} \text { at } 27^{\circ} \mathrm{C} . \\ & \left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned} \]
Show Hint
Use work done on the system equals minus delta n(g) times R times T.
Updated On: Nov 13, 2026
\(4988.4\text{ }\mathrm{J}\)
\(-1247.1\text{ }\mathrm{J}\)
2494.2 J
3741.3 J
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
For reactions involving gases at constant pressure, \(w = -\Delta n_g RT\) (work done on the system when \(\Delta n_g\) is negative).
Step 2: Find change in moles of gas
\(\text{C}_2\text{H}_4 + \text{HCl} \rightarrow \text{C}_2\text{H}_5\text{Cl}\)
\(\Delta n_g = 1 - 2 = -1\).
Step 3: Calculation
\(T = 27 + 273 = 300\ \text{K}\)
\(w = -(-1)(8.314)(300) = +2494.2\ \text{J}\).
Final Answer:
The work involved is 2494.2 J. \[ \boxed{\text{(C) }2494.2\text{ J}} \]
For reactions involving gases at constant pressure, \(w = -\Delta n_g RT\) (work done on the system when \(\Delta n_g\) is negative).
Step 2: Find change in moles of gas
\(\text{C}_2\text{H}_4 + \text{HCl} \rightarrow \text{C}_2\text{H}_5\text{Cl}\)
\(\Delta n_g = 1 - 2 = -1\).
Step 3: Calculation
\(T = 27 + 273 = 300\ \text{K}\)
\(w = -(-1)(8.314)(300) = +2494.2\ \text{J}\).
Final Answer:
The work involved is 2494.2 J. \[ \boxed{\text{(C) }2494.2\text{ J}} \]
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