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) to CO_2(g)`
`DeltaH_r = -393.5 kJ//mol = DeltaH_f CO_2(g)`
`H_2(g) + 1/2 O_2(g) to H_2O (l)`
`DeltaH_r = -285.8 kJ//mol = DeltaH_f H_2O(l)` `DeltaH_r = DeltaH_f (CH_4) - DeltaH_f CO_2(g) -2DeltaH_f H_2O(l) = 890.3`
`rArr DeltaH_f CH_4 + 393.5 + 2 xx 285.8 = 890.3`
`rArr DeltaH_f CH_4 (g) = -74.8kJ//mol.`