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Using the data (all values are in kcal mol-1 at 25oC) given below, calculate the bond energy of C-C and C-H bonds.
∆Hocombustion (ethane) = -372.0
∆Hocombustion (propane) = -530.0
∆HoC(s) → C(g) = 172.0
Bond energy of H-H = 104.0
∆Hof of H2O(l) = -68.0
∆Hof of CO2(g) = -94.0

Shane Macguire , 11 Years ago
Grade upto college level
anser 1 Answers
Deepak Patra

Last Activity: 11 Years ago

Hello Student,
Please find the answer to your question
For C3H8 : 3C + 4H2 → C3H8; ∆H1 = ?
For C2H6 : 2C + 3H2 → C2H; ∆H2 = ?
∴ ∆H1 = -[2 (C - C) + 8(C - H)] + [3Cs→g + 4(H - H)] ….(1)
∴ ∆H2 = -[1 (C - C) + 6(C - H)] + [2Cs→g + 3(H - H)] ….(2)
Let bond energy of C – C be x kcal and bond energy of C – H by y kcal
∴ By eq. (1) ∆H1 = - (2x + 8y) + [3 * 172 + 4 * 104] ….(3)
∆H2 = - (x + 6y) + [2 * 172 + 3 * 104] ….(4)
Also given C + O2 → CO2; ∆H = - 94.0 k cal ….(5)
H2 + ½O2 → H2O; ∆H = - 68.0 k cal …..(6)
C2H6 + (7/2)O2 → 2CO2 + 3H2O; ∆H = - 372 k cal ….(7)
C3H8 + 5O2 → 3CO2 + 4H2O; ∆H = - 530 k cal …..(8)
By inspection method : 2 * (5) + 3 * (6) – (7) gives
2C + 3H2 → C2H6; ∆H2 = - 20 k cal ….(9)
And 3 * (5) + 4 * (6) – (8) gives
3C + 4H2 → C3H8; ∆H1 = - 20 k cal ….(10)
∴ By eq. (3), (4), (9) and (10)
X + 6y = 676
2x + 8y = 956
∴ x = 82 k cal and y = 99 k cal
Bond energy of C – C bond = 82 k cal
and Bond energy of C – h bond = 99 k cal

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jitender
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