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Using the data given below (all values a...

Using the data given below (all values are kilories per mol at `25^(@)C)`, calculate the bond energy of `C -C` and `C -H` bonds.
`C(s) rarr C(g), DeltaH = 172`
`H_(2) rarr 2H, DeltaH = 104`
`H_(2)+(1)/(2)O_(2) rarr H_(2)O(l), DeltaH =- 68.0`
`C(s) +O_(2) rarr CO_(2), DeltaH =- 94.0`
Heta of combustion of `C_(2)H_(6) =- 372.0`
Heat of combustion of `C_(3)H_(8) =- 530.0`

Text Solution

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For the enthalpy of combustion of ethane and propane, we write
`C_2H_6(g) + 7/2O_2 (g) to 2CO_(2) + 3H_(2)O(l)`
`DeltaH_("comb") = 3Delta H_(f(H_2O,l)) + 2DeltaH_(f(CO_2,g)) - DeltaH_(f(C_2H_6,g))`
`DeltaH_(f(C_2H_6,g)) = -DeltaH_("comb")+ 3DeltaH_(f(H_2O,l)) + 2DeltaH_(f(CO_2,g))`
`=372 - 3 xx 68 - 2 xx 94) kcal mol^(-1) = -20 kcal mol^(-1)`
`C_(3)H_(8)(g) + 5O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(l)`
`Delta H_("Comb") = 3DeltaH_(f(CO_2,g)) + 4DeltaH_(f(H_2O,l)) - DeltaH_(f(C_3H_8,g))`
Thus , `Delta H_(f(C_3H_8,g)) = -Delta H_("comb") + 3Delta H_(f(CO_2,g)) + 4Delta H_(f(H_2O,l))`
`=(530 - 3 xx 94 - 4 xx 60) kcal mol^(-1) = -24 kcal mol^(-1)`,
To calculate `epsilon_(C-H) and epsilon_(C - C)`, we carry out the following operations.
(i) `2C_("graphite") + 3H_2(g) to C_2H_6(g)," "DeltaH = -20 kcal mol^(-1)`
`2C(g) + 6H(g) to C_(2)H_(6)(g)," "DeltaH = -2 xx 172 kcal mol^(-1)`
On Adding, `2C(g) + 6H(g) to C_(2)H_(6)(g), " "DeltaH = -3 xx 104 kcal mol^(-1)`
We get `2C(g) + 6H(g) to C_(2)H_(6)(g)`
`DeltaH_(i) = (-20 - 2 xx 172 - 3 xx 104) kcal mol^(-1) = -676 kcal mol^(-1)`
(ii) `3C_("graphite") + 4H_2(g) to C_3H_8(g)," "DeltaH = -24 kcal mol^(-1)`
`3C(g) to 3C("graphite")," "DeltaH = -3 xx 172 kcal mol^(-1)`
On Adding, `8H(g) to 4H_(2)(g), " "DeltaH = -4 xx 104 kcal mol^(-1)`
We get `3C (g) + 8H(g) to C_(3)H_(8)(g)`
`DeltaH_(ii) = (-24 - 3 xx 172 - 4 xx 104) kCalmol^(-1) = -956 kcal mol^(-1)`
Now, `DeltaH_(i) = -epsilon_(C-C) - 6epsilon_(C-H) = -676 kcal mol^(-1)`
`Delta H_(ii) = -2epsilon_(C- C) - 8epsilon_(C-H) = -956 kcal mol^(-1)`
Solving for `epsilon_(C-C) and epsilon_(C-H) `, we get
`epsilon_(C-H) = 99 kcal mol^(-1)`
`epsilon_(C-H) = 82 K cal mol^(-1)`.
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