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NEET-XI-Chemistry
06: Chemical Thermodynamics
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Qstn# 4
Prvs-Qstn
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#4
Δ
U
θ
of combustion of methane is -
X
kJ mol
-1
. The value of Δ
H
θ
is
(i) = Δ
U
θ
(ii) > Δ
U
θ
(iii) < Δ
U
θ
(iv) = 0
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Ans
:
Since Δ
H
θ
= Δ
U
θ
+ Δ
n
g
R
T
and Δ
U
θ
= -
X
kJ mol
-1
,
Δ
H
θ
= (-
X
) + Δ
n
g
R
T.
⇒ Δ
H
θ
< Δ
U
θ
Therefore, alternative (iii) is correct.
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