`DeltaU^(@)` of combustion of methane is `-X kJmol^(-1)`. The value of `DeltaH^(@)` is `:` (i) `= DeltaU^(@)` (ii) ` gt DeltaU^(@)` (iii) `lt DeltaU^(@)` (iv)` = 0`

`DeltaU^(@)` of combustion of methane is `-X kJmol^(-1)`. The value of

1.	`DeltaU^(@)` of combustion of methane is `-X kJmol^(-1)`. The value of `DeltaH^(@)` is `:` (i) `= DeltaU^(@)` (ii) ` gt DeltaU^(@)` (iii) `lt DeltaU^(@)` (iv)` = 0`
Answer» The balanced equation for combustion of methane will be `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l)` Thus, `Deltan_(g)= (n_(p) -n_(r))_(g)= 1 -3= -2` `DeltaH^(@) lt DeltaU^(@) + Deltan_(g) RT = - X - 2RT` Thus, `DeltaH^(@) lt DeltaU^(@)` . Hence, (iii) is the correct answer.

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`DeltaU^(@)` of combustion of methane is `-X kJmol^(-1)`. The value of `DeltaH^(@)` is `:` (i) `= DeltaU^(@)` (ii) ` gt DeltaU^(@)` (iii) `lt DeltaU^(@)` (iv)` = 0`

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