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Calculate `(a) DeltaG^(Theta)` and (b) the equilibrium constant for the formation of `NO_(2)` from NO and `O_(2)` at 298 K `NO(g) +1//2 O_(2) (g) hArr NO_(2)(g)` where `Delta_(f) G^(Theta) (NO_(2)) =52 .0 kJ//mol, Delta_(f) G^(Theta) (NO) =87.0 kJ//mol, Delta_(f) G^(Theta) (O_(2)) =0kJ//mol.` |
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Answer» `Delta_(r)G^(Θ)=SigmaDelta_(f)G^(Θ)("Products")-SigmaDelta_(f)G^(Θ)("Reactants")` `Delta_(f)G^(Θ)(NO_(2))-[SigmaDelta_(f)G^(Θ)(NO)+1/2Delta_(f)G^(@)(O_(2))]` `=52.0-(87.0+1/2xx0)=-35.0 kJ mol^(-1)` b. `-DeltaG^(Θ)=2.303RT log K. "Hence", -(-35000)` `=2.303xx8.314xx298xxlog K` or `log K=6.1341 or K=1.361xx10^(6)` |
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