1.

Calculate the heat of the reaction : CH_(2) = CH_(2)(g) + H_(2)(g) rarr CH_(3)CH_(3)(g) given that bond energy of C-C, C=C, C-H and H-H is 80, 145, 98 and 103 kcal.

Answer»

`-28 kcal mol^(-1)`
`-5.6 kcal mol^(-1)`
`-2.8 kcal mol^(-1)`
`-56 kcal mol^(-1)`

Solution :`DELTA H = B. E. (C = C) + 4 B. E. (C-H) + B. E. (H-H) - B. E. (C-C) - 6 B. E. (C-H)`
`= 145 + 4(98) + 103 - 80 - 6 (98)`
`= -28 kcal mol^(-1)`


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