Question

Consider the reaction:
`4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ`
Calculate `Delta_(f)G^(Theta) [NO(g)] if Delta_(f)G^(Theta) (NH_(3)) = -16.6 kJ mol^(-1)` and `Delta_(f)G^(Theta) [H_(2)O(l)] =- 237.2 kJ mol^(-1)`.

Answer 1

For the reaction
`4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ mol^(-1)`
`DeltaG^(Theta) = 4Delta_(f)G^(Theta) [NO(g)] +6Delta_(f)G^(Theta) [H_(2)O(l)]`
`-{4Delta_(f)G^(Theta) [NH_(3)(g)] +5Delta_(f)G^(Theta)[O_(2)(g)]}`
`DeltaG^(Theta) =- 1010.5 kJ mol^(-1), Delta_(f)G^(Theta) [H_(2)O(l)] =- 273.2 kJ mol^(-1)`
`Delta_(f)G^(Theta) [NH_(3)(g)] =- 16.6 kJ mol^(-1)` and `Delta_(f)G^(Theta) [O_(2)(g)] =0` (conversion)
`:. -1010.5 = 4Delta_(f)G^(Theta) [NO(g)] +6 xx (-237.2)-4 xx (-16.6) - 5 xx0`
`4Delta_(f)G^(Theta) [NO(g)] =- 1010.5 +1423.2 - 66.4`
`= 346.3 kJ mol^(-1)`
`Delta_(f)G^(Theta)[NO(g)] =86.6 kJ mol^(-1)`