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=1−2=−1;ΔH=−0.31KJmol−1p=1.5atm,ΔV=−50mL=−0.050LΔH=ΔE+pΔV−0.31=ΔE−0.0076;ΔE=−0.3024KJ
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