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 Del

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» Solution :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.