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{:(C(s)+O(2)(g)rarrCO(2)(g), ,,,,DeltaH=...

`{:(C(s)+O_(2)(g)rarrCO_(2)(g), ,,,,DeltaH= -94.3 kcal//mol),(CO(g)+(1)/(2)O_(2)(g)rarrCO_(2)(g),,,,,DeltaH= -67.4 kcal//mol),(O_(2)(g)rarr2O(g),,,,,DeltaH=117.4 kcal//mol),(CO(g)rarrC(g)+O(g),,,,,DeltaH=230.6 kcal//mol):}`
Calculate `Delta H` for `C(s)rarrC(g)` in `kcal//mol`.

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C(s)+O_(2)(g)toCO_(2)(g)," "DeltaH=-94.3 kcal//mol CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)," "DeltaH=-67.4 kcal//mol O_(2)(g)to2O(g)," "DeltaH=117.4 kcal//mol CO(g)toC(g)+O(g)," "DeltaH=230.6 kcal//mol Calculate DeltaH for C(s)toC(g) in kcal//mol

If C(s)+O_(2)(g)rarrCO_(2)(g),DeltaH=X and CO(g)+1//2O_(2)(g)rarrCO_(2)(g),DeltaH=Y , then the heat of formation of CO is

C_("graphite")+O_(2)(g)rarrCO_(2)(g) deltaH=-94.05Kcalmol^(-1) C_("diamond")+O_(2)(g),DeltaH=-94.50Kcalmol^(-1) Therefore,

C(s)+O_(2)(g)toCO_(2)(g),DeltaH=-94 kcal 2CO(g)+O_(2)to2CO_(2),DeltaH=-135.2 kcal The heat of formation of CO(g) is

C(s)+O_(2)(g)toCO_(2)(g),DeltaH=-94 kcal 2CO(g)+O_(2)to2CO_(2),DeltaH=-135.2 kcal The heat of formation of CO(g) is

Given that: i. C(s) + O_(2)(g) rarr CO_(2)(g) , DeltaH =- 94.05 kcal ii. H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH =- 68.32 kcal iii. C_(2)H_(2)(g) +(5)/(2) O_(2)(g) rarr 2CO_(2)(g) +H_(2)O(l),DeltaH =- 310.62 kcal The heat of formation fo acetylene is