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Calculate DeltaG^@ and the equilibrium c...

Calculate `DeltaG^@` and the equilibrium constant for the formation of `NO_2` from `NO `and `O_2` at `298 K NO_((g)) + 1/2 O_(2(g)) Where `Delta_f G^@ (NO_2) = 52.0 kJ mol^(-1)`
`Delta_f G^@ (NO) = 87.0 kJ mol^(-1)`
`Delta_f G^@ (O_2) = 0 kJ mol^(-1)`

Text Solution

Verified by Experts

`NO_((g)) + 1/2 O_(2(g)) `DeltaG^@ = Delta_f G^@ (NO_2) - [Delta_f G^@(NO) + 1/2 Delta_f G^@ (O_2)]`
`= 52.0 - 87.0 - 1/2 xx 0 = - 35 kJ mol^(-1)`
Now , `log K = - (DeltaG^@)/(2.303 RT)`
`= - (-35 xx 10^3 J mol^(-1))/(2.303 xx 8.314 J K^(-1) mol^(-1) xx 298 K) = 6.134`
`implies K = 1.362 xx 10^6`
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