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))`

Options

A. `206.4 kJ mol ^(-1)`
B. `-206.4 kJ mol ^(-1)`
C. `103.2 kJ mol ^(-1)`
D. `144 kJ mol ^(-1)`

Answer

C

Explanation

The formation reaction is

Hence, `Delta _(f) H= 5 xx 718. 4 + 8 xx 218 - 2 xx 615. 1 - 2 xx 247. 7 - 8 xx 413. 4) kJ mol ^(-1) = 103. 2 kJ mol ^(-1)`