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)`)

Calculate the value of `log K_(p)` for the reaction, `N_(2(g))+3H_(2(g

1.	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)`)
Answer» `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`

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