Calculate (a) `DeltaG^0` and (b) the equilibrium constant for the formation of `NO_2` from NO and `O_2` at 298 K,
`NO_((g)) + 1/2O_(2(g)) hArr NO_(2(g))`
where , `DeltaG_f^ө(NO_2)`= 52.0 kJ `"mol"^(-1)`
`DeltaG_f^ө`(NO)=87.0 kJ `mol^(-1)`
`DeltaG_f^ө (O_2)` = 0.0 kJ `"mol"^(-1)`

Explanation

Calculation of `DeltaG^0` free energy change of reaction : `DeltaG_r^0 = sumDelta_f G_"(products)"^ө -sumDelta_f G_"(reactant)"^ө`
`=Delta_f G^ө (NO_2)-(Delta_f G^ө, (NO)+1/2 Delta_f G^ө(O_2))` `=52.0-[87.0+1/2(0)]`
`=-35 "kJ mol"^(-1)`
`=-35xx10^3 "J mol"^(-1)`
Calculate of equilibrium constant `K_c` :
`DeltaG_r^0=-2.303 RT log K_c`
`therefore log_10 K_c=-(Delta_r G^0)/(2.303 RT)`
`therefore log_10 K_c=-((-35xx10^3 "J mol"^(-1)))/((2.303)(8.314 "J mol"^(-1) K^(-1))(298 K))`
=6.1341
`therefore K_c`= Antilog 6.1341
`=1.3618xx10^6 approx 1.36xx10^6`