Using the data provided, calculate the multiple bond energy `(kJ mol^(-1))` of a `C-=C` bond in `C_(2)H_(2)`. That energy is `(` take the bond energy of a `C-H` bond as `350kJ mol^(-1))`.
`2C_((s))+H_(2(g))rarr C_(2)H_(2(g)), Delta=225kJ mol^(-1)`
`2C_((s))rarr 2C_(g)), DeltaH=1410kJ mol^(-1)`
`H_(2(g)) rarr 2H_((g)),DeltaH=330kJ mol^(-1)`

Options

A. 1165
B. 837
C. 865
D. 815

Answer

D

Explanation

For calculation of `C -= C` bond energy, we must first calculate dissociation energy of `C_(2)H_(2)` as
`C_(2)H_(2 )" " … (i)`
Using the given bond energies and enthalpies:
`C_2 H_2 (g) rarr 2C(s) + H_(2)(g) , Delta H = -225 kJ" "……(ii)`
`2C (s) rarr 2C(g) , " "Delta H = 1410 kJ " "....(iii)`
`H_(2) (g ) rarr 2H(g) , " "Delta H = 330 kJ" "....(iv)`
Adding Eqs. (ii),(iii) and (iv) gives Eq. (i)
`implies C_2H_2(g) rarr 2C(g) + 2H(g) , " "Delta = 1515 kJ`
`implies 1515 kJ = 2 xx (C - H) BE + (C -= C) BE = 2 xx 350 + (C -= C) BE`
`implies (C -= C) BE = 1515 - 700 = 815 kJ//mol".