From the following bond energies :-
`H-H "bond energy" = 420 KJmol^(-1)`
`C=-c "bond energy" = 601 KJmol^(-1)`
`C-C "bond energy"=340KJmol^(-1)`
`C-H "bond energy" = 425 KJ mol^(-1)`
Enthalpy for the reaction :-
`C_(2)H_(2)+2H_(2)toC_(2)H_(6)`

Options

A. `-599 KJ mol^(-1)`
B. `-580 KJ mol^(-1)`
C. `-625 KJ mol^(-1)`
D. `-325 KJ mol^(-1)`

Answer

A

Explanation

`DeltaH_(rxn)=2B.E(C-H)+B.E(C-=C)+2B.E(H-H)-6B.E(C-H)-B.E(C-C)`
`rArrB.E(C-=C)+2B.E(H-H)-4B.E(C-H)-B.E(C-C)`
`rArr 601+2xx240-4xx425-340`
`DeltaH_(rxn=-599KJ//mol`