using the data (all values are in kilocalories per mole at 25C) given below, calculate the bond energy of C-C and C-H bonds
∆H◦combustion(ethane) = -372.0
∆H◦Combustion (propane) =-530
∆H◦ for C(graphite) àC(g) =172
Bond energy of H-H = 104
∆H◦ of H2O (l) = -68
∆H◦ for CO2 (g) = -94

Dear Student,

Please find below the solution to the asked query:

Let the bond energy of C−C be x k cal and that of C−H be y k cal.

Given,
C + O2  CO2 ; H = -94.0 k cal      ...1H2 +12 O2  H2O ; H = -68.0 k cal     ...2C2H6 +72 O2  2 CO2 + 3 H2O ; H = -372.0 k cal   ...3C3H8 + 5 O2  3 CO2 + 4 H2O ; H = -530.0 k cal     ...4

The reaction for the formation of propane is given as:

3 C + 4 H2  C3H8; H1

The reaction for the formation of ethane is given as:

2 C + 3 H2  C2H6; H2

Thus,
H1 = 3×172 + 4×104 - 2x + 8y   ...5H2 = 2×172 + 3×104 - x + 6y    ...6

Now, 2 × Eq. 1 + 3 × Eq. 2 - Eq. 3, gives

2 C + 3 H2  C2H6; H2 = - 20 k cal      ...7

Similarly, 

 3 × Eq. 1 + 4 × Eq. 2 - Eq. 4, gives

3 C + 4 H2  C3H8; H1 = -24 k cal   ...8

By equations (5), (6), (7) and (8),

x + 6y = 6762x + 8b = 956x = 82 k caland y = 99 k cal.

Hope this information will clear your doubts.

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