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
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
Was this answer helpful ?
:
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
Was this answer helpful ?
Submit Solution