Given
`C_("graphite") + O_(2)(g) to CO_(2)(g), Delta_(r)H^(@) = -393.5 kJ "mol"^(-1)`
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l), DeltaH^(@) = -285.8 kJ "mol"^(-1)`
`CO_(2)(g) + 2H_(2)O(l) to CH_(4)(g) + 2O_(2)(g), DeltaH^(@) = +890.3 kJ "mol"^(-1)`
Based on the above thermochemical equations, the value of `Delta H^(@)` at 298 K for the reaction
`C_("graphite") + 2H_(2)(g) to CH_(4)(g)` will be :

Options

A. `+ 748 kJ`
B. `+ 144.0 kJ`
C. `-74.8 kJ`
D. `-144.0 kJ`

Answer

C

Explanation

(i) `C("graphite") + O_(2) to CO_(2)(g) DeltaH^(@) = -393.5 kJ "mol"^(-1)`
(ii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH ^(@) - 285.8 kJ "mol"^(-1)`
(iii)`CO_(2)(g) + 2H_(2)O(l) to CH_(4)(g) + 2O_(2)(g) DeltaH = 890.3 kJ "mol"^(-1)`
The required equation is
(iv)` C_("graphite") + 2H_(2)(g) to CH_(4)(g)DeltaH ^(@) = `?
By applying eq.(i) + 2 xx eq.(ii) + eq.(iii)
`DeltaH^(@) = -393.5 + 2(-285.8) + 890.3 = -74.8 kJ "mol"^(-1)`