Enthalpy of atomization of `C_(2)H_(6) (g)` and `C_(3)H_(8) (g)` are 620 and `880 kJ "mol"^(-1)` respectively. The C-C and C-H. bond energies are respectively.

Options

A. 80 and 60 `"kJ mol"^(-1)`
B. 80 and 90 `"kJ mol"^(-1)`
C. 70 and 90 `"kJ mol"^(-1)`
D. 200 and 80 `"kJ mol"^(-1)`

Answer

B

Explanation

Since, `C_(2)H_(6)` contains one C-C and 6 C-H bonds, thus 620 is the sum of Bond Energy of one C-C and 6 C-H bonds. Thus,
`C_(2)H_(6) to C - C + 6C - H DeltaH = 620` kJ/mol………..(i)
Similarly,
`C_(3)H_(8) to 2C -C + 8C - H DeltaH = 880` kJ/mol............(ii)
Multiply Eq. (i) by 2 then we get
`2C_(2)H_(6) to 2C - C + 12 C - H DeltaH = 1240` kJ/mol......(iii)
Equation (iii)- (ii)
2C – C + 8C – H = 880 kJ/mol
2C – C + 12C – H = 1240 kJ/mol.
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`4C - H = 360` kJ/mol
`therefore C - H = 90.0` kJ/mol
On putting the value of C-H, in equation (i) we get
`6 xx 90.0 + C - C = 620`
`C - C = 620 - 540 = 80` kJ/mol