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Calculate the bond energy of C-=C in C_(2)H_(2) from the following data : (i)C_(2)H_(2)(g)+2(1)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O,DeltaH=-310"kcal" (ii)C(s)+O_(2)(g) to CO_(2)(g),DeltaH=-94"kcal" (iii) H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-68 "kcal" Bond energy of C-H bonds =99 kcal Heat of atomisation of C=171 kcal Heat of atomisation of H=52 kcal |
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Answer» SOLUTION :Let us first calculate heat of formation of `C_(2)H_(6)` from which we can calculate bond energy of `C-=C` bond. `2C(s)+H_(2) to C_(2)H_(2)(g)` , `DeltaH=?` Applying the inspection method , i.e., `[2xx` Eqn. (ii) `+` Eqn. (iii) - Eqn. (i)] we get `2C(s)+2O_(2)(g)+H_(2)(g)+(1)/(2)O_(2)(g)-C_(2)H_(2)(g)-2(1)/(2)O_(2) to 2CO_(2)(g)+H_(2)O(g)-2CO_(2)(g)-H_(2)O(g), DeltaH=2xx(-94)+(-68)-(-310)kcal` or `2C(s)+H_(2)(g) to `C_(2)H_(2)(g) `, `DeltaH=54` kcal Now heat CHANGES for reactants Heat of atomisation of 2 moles of `C=2xx171` kcal Heat of atomisation of 2 moles of `H=2xx52` kcal And heat change for the product `(H-C-=C-H)` Heat of formation of 2 moles of C-H bonds `=-2xx99` kcal Heat of formation of 1 MOLE of `C-=C` bonds `=x` (say) Summing up, we get heat of formation of `C_(2)H_(2)`, `2xx171+2xx52-2xx99+x=54` `x=-194` kcal Hence bond energy of `C-=C` bond in `C_(2)H_(2)` is `+194` kcal |
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