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

Answer

B

Explanation

`C_(("graphite"))+O_(2)(g)rarrCO_(2)(g)`
`DeltaH_(r)=-393.5kJ//mol=DeltaH_(r)CO_(2)(g)`
`H_(2)(g)+(1)/(2)O_(2)(g)rarrH_(2)O(l)`
`DeltaH_(r)=-285.8 kJ//mol=DeltaH_(f)H_(2)O(l)`
`CO_(2)(g)+2H_(2)O(l)rarrCH_(4)(g)+2O_(2)(g)`
`DeltaH_(r)=DeltaH_(f)(CH_(4))-DeltaH_(f)CO_(2)(f)-2DeltaH_(f)H_(2)O(l)=890.3`
`implies DeltaH_(f)CH_(4)+393.5+2xx285.8=890.3`
`impliesDeltaH_(f)CH_(4)(g)=-74.8 kJ//mol`