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From the following data, calculate the standard enthalpy of formation of propane `Delta_(f)H^(Theta) CH_(4) = - 17 kcal mol^(-1)` `Delta_(f)H^(Theta)C_(2)H_(6) =- 24 kcal mol^(-1), BE (C-H) = 99 kcal mol^(-1)` `(C- C) = 84 kcal mol^(-1)`. |
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Answer» `C(s) +2H_(2)(g) rarr CH_(4)(g), DeltaH_(1) = - 17 kcal …(i)` `2C(s)+3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH_(2) =- 24 kcal …(ii)` ltbr. `3C(s) +4H_(2)(g) rarr C_(3)H_(8)(g), Delta_(f)H = ?....(iii)` From equations (i) and (ii), we get Let `x kcal mol^(-1)` is the energy of `C(s) rarr C(g)` Let `y kcal mol^(-1)` is the `BE` energy of `(H -H)` From equation (i), we get `-17 = x +2y - 4 xx 99` From equation (ii), we get `-24 = 2x +3y -(84 +6 xx 99)` SOlve for `x` and `y`, `x = 171 kcal, y = 104 kcal` From equation (iii), we get `Delta_(f)H = 3x +4y - [2 xx 84 +8 xx 99]` `=- 31 kcal mol^(-1)` |
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