InterviewSolution
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InterviewSolution
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Calculate (a) `DeltaG^(Θ)` and (b) the equilibrium constant for the formation of `NO` and `O_(2)` at `298 K` `NO(g)+1//2 O_(2)(g) hArr NO_(2)(g)` where `Delta_(f)G^(Θ)(NO_(2))=52.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(NO)=87.0 kJ mol^(-1)` `Delta_(f)G^(Θ)(O_(2))=0 kJ mol^(-1)` |
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Answer» Correct Answer - a) `-35.0 kJ`, b) `1.365 xx 10^(6)` (a) For the given reaction, `DeltaG^(@) = DeltaG^(@)("Products")-DeltaG^(@)("Reactants")` `DeltaG^(@) = 52.0 - {87.0 + 0}` `= - 35.0 kJ "mol"^(-1)` (b) We know that `DeltaG^(@) = RT "log" K_(c)` `DeltaG^(@) = 2.303 RT "log" K_(c)` `K_(c) = (-35.0 xx 10^(-3))/(-2.303 xx 8.314 xx 298)` `= 6.134` `:. K_(c) = "antilog" (6.134)` `= 1.36 xx 10^(6)` Hence, the equilibrium constant for the given reaction `K_(c)` is `1.36 xx 10^(6)` |
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