Calculate the value of `log K_(p)` for the reaction, `N_(2(g))+3H_(2(g))hArr2NH_(3(g))` at `25^(@)C`. The standard enthalpy of formation of `NH_(3(g))` is `-46kJ` and standard entropies of `N_(2(g))`, `H_(2(g))` and `NH_(3(g))` are `191`, `130`, `192` `JK^(-1) mol^(-1)`. respectively. (`R=8.3JK^(-1)mol^(-1)`)

Explanation

`N_(2(g))+3H_(2(g))hArr2NH_(3(g))`
At equilibrium, `-DeltaG^(@)=2.303RT log_(10)K_(p)` ….(`1`)
Also `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)`
(Given : `DeltaH^(@)` for `NH_(3)=-46 kJ`)
and `DeltaS_(reaction)^(@)=2xxS_(NH_(3))^(@)-S_(N_(2))^(@)-3xxS_(H_(2))^(@)`
`=2xx192-191-3xx130=-197J`
Also `T=273+25=298 K`
Thus, `DeltaG^(@)= -92xx10^(3)-298xx(-197)`
(`DeltaH^(@)` for reaction`= -46xx2 kJ`)
`=-92000+58706= -33294 J`
Thus, from eq. (`1`),
`+33294=2.303xx298xx8.3 log_(10) K_(p)`
`:. log_(10) K_(p)=5.845`