Calculate the value of log K_(p) for the N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g) at 25^(@)C. The standard enthalpy of formation of nh_(3) is -46 kJ and standard entropies of N_(2)(g), H_(2)(g) and NH_(3)(g) are 191, 130 and 192 J K^(-1) "mol"^(-1) respectively.

Calculate the value of log K_(p) for the N_(2)(g) + 3H_(2)(g) iff 2NH_

1.	Calculate the value of log K_(p) for the N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g) at 25^(@)C. The standard enthalpy of formation of nh_(3) is -46 kJ and standard entropies of N_(2)(g), H_(2)(g) and NH_(3)(g) are 191, 130 and 192 J K^(-1) "mol"^(-1) respectively.
Answer» Solution :`N_(2)(g) + 3H_(2)(g) iff 2NH_(3)(g)` `DeltaH^(@) = 2 Delta_(f)H^(@)(NH_(3)) - [Delta_(f)H^(@)(N_(2)) + 3Delta_(f)H^(@)(H_(2))]` `= 2 xx -46 = -92 KJ` Now , `DeltaS^(@) = 2S^(@)(NH_(3))-[S^(@)(N_(2))+ 3S^(@)(H_(2))]` ` = 2 xx 192 - [191 + 3 xx 130]` ` = -197 JK^(-1) or = -0.197 kJ K^(-1)` `DeltaG^(@) = DeltaH^(@) -TDeltaS^(@)` ` = -92 -298 xx (-0.197)` ` =-33.294 kJ = - 33.294 xx 10^(3)J` Now , `DeltaG^(@)= -2.3030 RT "log K_(p)` or `log K_(p) = -(DeltaG)/(2.3030RT)` `= - (-33.294 xx 10^(3))/(2.303 xx 8.314 xx 298) = 5.835`.