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

Given H_(2)(g)=2H(g)Delta_(H-H)=103Kcal mol^(-1) The heat of reaction

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