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)`A. `103Kcalmol^(-1)`B. `206Kcalmol^(-1)`C. 51.5D. zero

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

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)`A. `103Kcalmol^(-1)`B. `206Kcalmol^(-1)`C. 51.5D. zero
Answer» Correct Answer - C `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`

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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)`A. `103Kcalmol^(-1)`B. `206Kcalmol^(-1)`C. 51.5D. zero

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