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