Question

29.
he heat of combustion of acetylene is 312 kcal. If heat of formation of co, and
20 are 94.38 and 68.38 kcal respectively, calculate C=C bond energy. Given
e heat of atomisation of C and M are 150.0 and 51.5 kcal respectively and C-H
bondenergy is 93.54 kcal.​

Answer 1

Answer:

Given

$C_{2} H_{2} +(5/2)O_{2} ---->2CO_{2} +H_{2} O$ ΔH=312kcal---(i)

(heat of combustion are exothermic)

$C+O_{2} --->CO_{2}$;  ΔH=-94.38kcal---(ii)

$H_{2} +\frac{1}{2} O_{2} ----->H_{2} O$;=-68.38kcal----(iii)

multiply equation (ii) by 2 and add in equation (iii)

$2C+H_{2} +(5/2)O_{2} --->2CO_{2} +H_{2} O$ΔH=-257.14----(iv)

subtracting equation (i) from equation (iv)

$C_{2} H_{2} +(5/2)O_{2} --->2CO_{2} +H_{2} O$ΔH=-312kcal---(i)

-                          -               -        -                                             +

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                        $2C+H_{2} ---->C_{2} H_{2}$ΔH=+54.86kcal---(v)

And also ΔH for $C+H_{2} ----->C_{2} H_{2}$

ΔH=Bond energy data of formation of bond+Bond energy data of    

                                                                           dissociation of bond

ΔH=-[$e_{(c=c)} +2*e_{(C-H)} ]+[2C_{s-->g} +e_{(H-H)} ]$

given $C_{s-->g} =150kcal$

$\frac{1}{2} H_{2} (g)--->H(g)---51.5kcal$

$e_{H-H} =51.5*2kcal=103.0kcal$

$e_{C-H} =93.64kcal$

$54.86=-[e_{(c=c)} +2*93.64]+[2*150+1*103.0]$

$e_{c=c} =160.86kcal$