6.4
∆U°of combustion of methane is - X kJ mol-l. The value of AH is
(i) = ΔU9
(ii) > AU
(iii) < AU
(iv) = 0
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1
Answer:
CH
4
(g)+2O
2
(g)⟶CO
2
(g)+2H
2
O(l)
Now,
Δn
g
=(n
p
−n
r
)=1−3=−2
Again,
ΔH
o
=ΔU
o
+Δn
g
RT
ΔH
o
=−X−2RT
∴ΔH
o
<ΔU
o
Explanation:
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