Grade upto college level Physical Chemistry

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

Shane Macguire

12 Years agoGrade upto college level

1 Answer

Deepak Patra

12 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)] + [3C_s→g + 4(H - H)] ….(1)

∴ ∆H₂ = -[1 (C - C) + 6(C - H)] + [2C_s→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

Thanks
jitender
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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) = 172.0Bond energy of H-H = 104.0∆Hof of H2O(l) = -68.0∆Hof of CO2(g) = -94.0

Shane Macguire

1 Answer

Deepak Patra

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