1.

Given that DeltaH_("comb")of C(s) , H_(2) (g) and CH_(4) (g) are, -394 , -294 and -829 kJ//mol respectively. The heat of formation fo CH_(4) is

Answer»

`70 kJ//mol`
` -71. 8 kJ // mol`
`-244 kJ//mol`
`-748 kJ // mol `

Solution :the formation of `CH_(4)` is represented as :
`C(s)+2H_(2)(g)to CH_(4)(g),`
` DeltaH_(F)=(DeltaH_("comb"))_(R)-(DeltaH("comb")_(P)`
` =[-394+2 (-284)-(-892)kJ`
`-962 + 892 =-70 kJ mol^(-1)`


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