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

D

Explanation

`C("graphite") + O_(2)(g) rarr CO_(2)(g), Delta_(r )H^(@) = -393.5 kJ mol^(-1) " " ….(i)`
`H_(2)(g) + 1/2O_(2)(g) rarr H_2O (l) , Delta_(r )H^@ = -285.8 kJ mol^(-1)" "…….(ii)`
`CO_(2)(g) + 2H_(2)O (l) rarr CH_(4)(g) + 2O_(2)(g), Delta_(r )H^(@) = +890.3 kJ mol^(-1) " ".....(iii)`
`CO_(2)(g) + 2H_(2)O(l) rarr CH_(4)(g) + 2O_(2)(g) " ".....(iv)`
`Delta_(r )H_(4) = Delta_(r )H_(1) + 2Delta_(r )H_(2) + Delta_(r )H_(3) = -393.5 + (-285. 8 xx 2) + 890.3 = -74.8 KJ//mol`.