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Question
The enthalpy change for the reaction of 50.00 ml  of ethylene with 50.00 ml of H2 at 1.5 atm pressure is ΔH = -0.31 kJ. The value of ΔE will be
Options:
A .  -0.3024 kJ
B .  0.3024 kJ
C .  -2.567 kJ
D .  -0.0076 kJ
Answer: Option A
:
A
C2H4(g)+H2(g)C2H6(g)Δng=12=1;ΔH=0.31KJmol1p=1.5atm,ΔV=50mL=0.050LΔH=ΔE+pΔV0.31=ΔE0.0076;ΔE=0.3024KJ

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