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Calculate enthalpy change of the following reaction `:` `H_(2)C=CH_(2(g))+H_(2(g)) rarr H_(3)C-CH_(3(g))` The bond energy of `C-H,C-C,C=C,H-H` are `414,347,615` and `435k J mol^(-1)` respectively.A. `+125kJ`B. `-125kJ`C. `+250kJ`D. `-250kJ` |
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Answer» Correct Answer - B `DeltaH=-[e_(C-C)+6xxe_(C-H)]+[e_(C=C)+4xxe_(C-H)+e_(H-H)]` `=-[347+6xx414]+[615+4xx414+435]` `=-125kJ` |
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