Question

Calculate the work done in terms of dm\({}^{3}\) bar when \(200\) ml of ethylene gas and \(150\) ml of HCl gas were allowed to react at \(1\) bar pressure.

• A. \(0.10\)
• B. \(0.15\)
• C. \(0.20\)
• D. \(0.25\)

Hint:

Find the change in gas volume for the reaction C2H4 + HCl giving C2H5Cl, then use w = -p times the volume change.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The reaction is \(\text{C}_2\text{H}_4(g) + \text{HCl}(g) \rightarrow \text{C}_2\text{H}_5\text{Cl}(g)\). Gases react 1:1 by volume, and work is \(w = -p\Delta V\).

Step 2: Volumes
HCl (150 mL) is limiting, so 150 mL of ethylene reacts and 150 mL of chloroethane forms. 50 mL of ethylene is left over.
Initial volume \(= 200 + 150 = 350\) mL.
Final volume \(= 50 + 150 = 200\) mL.

Step 3: Work done
\(\Delta V = -150\) mL \(= -0.15\) dm\(^{3}\).
\(w = -(1\text{ bar})(-0.15\text{ dm}^{3}) = +0.15\) dm\(^{3}\) bar.

Final Answer:
The work done is 0.15 dm3 bar. \[ \boxed{\text{(B) }0.15} \]