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. `-144.8kJ:" " mol^(-1)`

Answer

C

Explanation

From the given equations:
`C_(("graphite"))+O_(2)(g)toCO_(2)(g),Delta_(r)H^(@)=-393.5Kj" "Mol^(-1)`
`2H_(2)(g)+O_(2)(g)to2H_(2)O(l),Delta_(r)H^(@)=-285.8xx2kJ" "mol^(-1)`
`CO_(2)(g)+2H_(2)O(l)toCH_(4)(g)+2O_(2)(g),Delta_(r)H^(@)=+890.3kJ" "mol^(-1)`
Adding above three equations we get
`C_(("Graphite"))+2H_(2)(g)toCH_(4)(g)`
`Delta_(r)H^(@)=-393.5-285.8xx2+890.3`
`=74.8kJ" "mol^(-1)`.