Heat of combustion `DeltaH^(@)` for `C(s),H_(2)(g)` and `CH_(4)(g)` are `94, -68` and `-213Kcal//mol` . Then `DeltaH^(@)` for `C(s)+2H_(2)(g)rarrDeltaCH_(4)(g)` is

Options

A. `-17 kcal`
B. `-111 kcal`
C. `-170 kcal`
D. `85 kcal`

Answer

A

Explanation

`(i) C(s)+2H_(2)(g)toCH_(4)(g)`
`(ii) C(s)+O_(2)(g)toCO_(2)(g)` , `DeltaH=-94 kcal mol^(-1)`
`(iii) H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)`, `DeltaH=-68kcal mol^(-1)`
`(iv) CH_(4)+3//2O_(2)toCO_(2)+2H_(2)O`, `DeltaH=-213 kcal mol^(-1)`
To obtain eqn. `(i)` operate
`(ii)+2xx(iii)-(iv)`
`:. DeltaH=-94+2(-68)-(-213)`
`=-94-136+213`
`=-230+213=-17 kcal`