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

If all thermochemical data to used in predicting the reaction enthalpy for a desired equation are heats of combustion, Hess's law can be expressed as
`Delta_(r) H^(@) = sum_(i) a_(i) Delta_(c ) H^(@)` (products) `+ sum_(i) b_(i) Delta_(c ) H^(@)` (reactants)
`= - (-213) + [(-94) + (2) (-68)]`
`= - 17 cal`
Alternatively,
`{:((a),C(s)+O_(2)(g)rarr(g),,,Delta_(C )H^(@)=-94 kcal),((b),H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l),,,Delta_(C )H^(@)=-68 kcal),((c ),CH_(4)(g)+2O_(2)(g)rarrCO_(2)(g)+2H_(2)O(l),,,Delta_(C )H^(@)=-213 kcal):}`
To obtain
`{:(C(s)+2H_(2)(g)rarrCH_(4)(g),,,Delta_(r )H^(@)=?):}`
We perform
`(a) + 2(b) - (C )`
`:. Delta_(r) H^(@) = (-94) + 2 (-68) - (-213) = - 17 kcal`