Calculate (a) DeltaG^0and (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)

Calculate (a) DeltaG^0and (b) the equilibrium constant for the formati

1.	Calculate (a) DeltaG^0and (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)
Answer» Solution :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`