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If bond enthalpies of (C- H) = 413 kJ mo...

If bond enthalpies of `(C- H) = 413 kJ mol ^(-1) , (C -C) = 347. 7 kJ mol ^(-1)` , and `(C = C) = 615. 1 kJ mol ^(-1), Delta _(sub) H ` (C, graphite) `= 718kJ mol ^(-1) and Delta _(f) H ( H,g) = 218 kJ mol ^(-1),` the enthalpy of formation of gaseous isoprene will be about `(CH _(2) = C ( CH _(3)) CH = CH _(2))`

A

`206. 4 kJ mol ^(-1)`

B

`- 206. 4 kJ mol ^(-1)`

C

`103. 2 kJ mol ^(-1)`

D

`44 kJ mol ^(-1)`

Text Solution

Verified by Experts

The formation reaction is

Hence, `Delta _(f) H = 5 xx 718. 4 + 8 xx 2. 18 - 2 xx 615 .1 - 2 xx 347.7 - 8 xx 413 . 4 kJ mol ^(-1) = 103. 2 kJ mol ^(-1)`
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