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

Answer

C

Explanation

Given
`CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)H^(@)=890.3`
C (graphite) `+O_(2)(g)toCO_(2)(g),`
`Delta_(r)H^(@)=-393.5kJ"mol"^(-1)`…….i
`H_(2)(g)+1/2O_(2)(g)toH_(2)O(1),`
`Delta_(r)H^(@)=-285.8kJ"mol"^(-1)` ……ii
`H_(2)(g)+1/2O_(2)(g)toH_(2)O(1),`
`Delta_(r)H^(@)=-285.8kJ"mol"^(-1)` .......iii
`Delta_(r)H^(@)=sum(Delta_(r)H^(@))_("products")-sum(Delta_(f)H^(@))_("Reactants")`
`=(Delta_(CH_(4))^(@)+2xxDeltaH_(O_(2))^(@))-(DeltaH_(CO_(2))^(@)+2xxDeltaH_(H_(2)O)^(@))`
`890.3=[1xx(Delta_(f)H^(@))_(CH_(4))+2xx0]-[(1xx(-393.5)+2(-285.8)]`
`(Delta_(f)H^(@))_(CH_(4))=890.3-965.1=-74.8`kJ/mol