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CGP EDU Academic Team
Published on: August 14, 2026
The enthalpy change for the reaction of 5 liter of ethylene with 5 liter of H 2 gas at 1.5 atm pressure is Δ H = – 0.5 kJ. The value of Δ U will be : (1 atm Lt = 100 J)
Text Solution
Verified by ExpertsThe correct answer is:
C
C 2 H 2 (g) + H 2 (g) ⎯→ C 2 H 4 (g)
Δ H = Δ V + Δ n g RT
= Δ U + P Δ V
– 0.5 = Δ U + 1.5 (–5) × 
Δ U = – 0.5 + 0.75
Δ U = 0.25 kJ
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