Given,H_(s)(g)=2H(g)Delta_(H-H)=103kcalmol^(-1) CH_(4)(g)=CH_(3)(g)+H(g)Delta_(CH_(3)-H)=103kcalmol^(-1) The heat of reaction of CH_(4)(g) = CH_(3)(g)+ H(g)

Given,H_(s)(g)=2H(g)Delta_(H-H)=103kcalmol^(-1) CH_(4)(g)=CH_(3)(g)+H(

1.	Given,H_(s)(g)=2H(g)Delta_(H-H)=103kcalmol^(-1) CH_(4)(g)=CH_(3)(g)+H(g)Delta_(CH_(3)-H)=103kcalmol^(-1) The heat of reaction of CH_(4)(g) = CH_(3)(g)+ H(g)
Answer» `103Kcalmol^(-1)` `206Kcalmol^(-1)` 51.5 zero Solution :`CH_(4)(G)=CH_(3)(g)+H(g)` `D_(CH_(3-H))=103"KCALMOL"^(-1)` `H(g)+H(g)=H_(2)(g)` `(D_(H-H)=-103Kcalmol^(-1))/(Ch_(4)(g)+H(g)=CH_(4)(g)+H_(2)(g))` `DeltaH=103-103=0`