Given `C_(("graphite"))+O_(2)(g)toCO_(2)(g),`
`Delta_(r)H^(0)=-393.5kJ" "mol^(-1)`
`H_(2)(g)=+(1)/(2)O_(2)(g)toH_(2)O(1),`
`Delta_(r)H^(0)=-285.8" kJ "mol^(-1)`
`CO_(2)(g)+2H_(2)O(1)toCH_(4)(g)+2O_(2)(g)`,
`Delta_(r)H^(0)=+890.3kJ" "mol^(-1)`
Based on the above thermochemical equations, the value of `Delta_(r)H^(0)` at at 298 K for the reaction
`C_(("graphite"))+2H_(2)(g)toCH_(4)(g)` will be:

Options

A. `- 74.8 kJ mol^(-1)`
B. ` -144.0 kJ mol^(-1)`
C. ` + 74.8 kJ mol^(-1)`
D. `+ 144.0 kJ mol^(-1)`

Answer

A

Explanation

Eqn. (i) `+2 xx ` Eqn. (ii)`+` Eqn. (iii) gives the required equation with
`DeltaH = ( - 393.5)+ 2 ( - 285.8 )+(+ 890.3)kJ mol^(-1)`
`= - 74.8 kJ mol^(-1)`