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If C(s)+O(2)(g)to CO(2)(g)+94.2 Kcal ...

If `C(s)+O_(2)(g)to CO_(2)(g)+94.2` Kcal
`H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l)+68.3` Kcal
`CH_(4)(g)+2O_(2)(g)to CO_(2)(g)+2H_(2)O+210.8` Kcal
then the heat of formation of methane will be

A

47.3 kcal

B

`-20.0`kcal

C

45.9 kcal

D

`-47. 3` kcal

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The correct Answer is:
B
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If C+O_(2) to CO_(2)+94.2kcal H_(2)+(1)/(2)O_(2) to H_(2)O+68.3"kcal" CH_(4)+2O_(2) to CO_(2)+2H_(2)O+210.8"kcal" Then, the heat of formation of methane will be

Using the following thermochemical data. C(S)+O_(2)(g)rarr CO_(2)(g), Delta H=94.0 Kcal H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l), Delta H=-68.0 Kcal CH_(3)COOH(l)+2O_(2)(g)rarr 2O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(l), Delta H=-210.0 Kcal The heat of formation of acetic acid is :-

Calculate the heat of formation of methane in kcalmol^(-1) using the following thermo chemical reactions C(s)+O_(2)toCO_(2)(g) , DeltaH=-94.2 kcal mol^(-1) H_(2)(g)+1/2O_(2)(g)toH_(2)O(l) , DeltaH=-68.3 kcal mol^(-1) CH_(4)(g)+2O_(2)(g)toCO_(2)(g)+2H_(2)O(l) , DeltaH=-210.8 kcal mol^(-1)

Given C_(s)+O_(2(g))toCO_(2(g)),DeltaH=-94.2K cal H_(2(g))+½O_(2(g))toH_(2)O_((g)),Delta=-68.3K cal CH_(4(g))+2O_(2(g))+2H=-210.8K cal what will be heat of gormation of CH_(4) in (Kcal) ?

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

C_(("graphite")) + O_(2(g)) to CO_(2(g)) - 94.0 kcal

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