if heat at constant volume for combustion of CH4 at 127 degree celsius is -20kj, then heat at consta
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Answer:
The correct option is B)
Combustion of benzene:
C
6
H
6
(l)+
2
15
O
2
(g)⟶6CO
2
(g)+3H
2
O(l)
ΔH=ΔU+PΔV; Δn=6−
2
15
=−1.5
ΔH−ΔU=PΔV
ΔH−ΔU=ΔnRT
ΔH−ΔU=−1.5×R×(127+273)
ΔH−ΔU=−600R .
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