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Calculate enthalpy change of the followi...

Calculate enthalpy change of the following reaction :
`CH_(2) = CH_(2)(g) + H_(2)(g) rarr CH_(3) - CH_(3)(g)`
The bond energy of C - H, C - C, C = C, H - H are 414, 615 and 436 kJ `mol^(-1)` respectively.

Text Solution

Verified by Experts

`sumBE_(R) - sum BE_(P)`
`[1 xx BE_(C-C) + 4 xx BE_(C+H) + 1 xx B.E_(H-H)]-[1 xx B.E_(C-C) + 6 xx BE_(H-H)]`
`=[1 xx BE_(C-C) + 1 xx BE_(H-H) -1 xx BE_(C-C) - 2 xx BE_(C-H)]`
`=615 + 435 - 347 - 2 xx 414`
`=-125 kJ`
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