Calculate the resonance energy of isoprene (C 5 H 8 ) from the data given.
Given that Δ H C=C = 615 kJmole –1 ; Δ H c–C = 348 kJ mole –1 ; Δ H C–H = 413 kJmole –1 ; Δ H H–H = 435 kJmole –1
The standard heat of sublimation of graphite is 718 kJmole –1 and heat of formation of C 5 H 8 (g) is 79 kJmole –1
(Give your answer in kcal mole –1 ; approximate integer).
Text Solution
Verified by Experts5
(5)
2C (s) ⎯→ 5C (g) Δ H 1 = 5 × 718 = 3590
4H 2(g) ⎯→ 8H (g) Δ H 2 = 4 × 435 = 1740
5C (g) + 8H (g) — → C 5 H 8(g) Δ H 3 = –(8 × 913 + 2 × 348 + 2 × 615) = –5230
+ +
5C (s) + 4H 2(g) ⎯→ C 5 H 8(g) Δ H r×n = Δ H f (C 5 H 8 ) = Δ H 1 + Δ H 2 + Δ H 3 = 100
so resonance energy = 79 – 100 = –21 kJ mole –1
Hence resonance energy in Kcalmole –1 = 5
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