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The equilibrium constant K(p), for the r...

The equilibrium constant `K_(p)`, for the reaction `N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)` is `1.6xx10^(-4)` at `400^(@)C`. What will be the equilibrium constant at `500^(@)C` if the heat of reaction in this temperature range is `-25.14` kcal?

Text Solution

Verified by Experts

Using the reaction,
`"log" K_(p_(2))/K_(p_(1))=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(2)T_(1))]`
Given
`K_(P1)=1.6xx10^(-4),DeltaH=-25.14 kcal`,
`R=2xx10^(-3) kcal deg^(-1) "mol"^(-1)`
`T_(1)=400+273=673 K, T_(2)=500+273=773 K`
`"log" K_(p2)/((1.6xx10^(-4)))=(-25.14)/(2.303xx2xx10^(-3))[(773-673)/(773xx673)]`
`"log" K_(p2)=log (1.6xx10^(-4))-(25.14xx10^(3)xx100)/(2.303xx2xx773xx673)`
`=-3.7960-1.049=-4.8450`
`K_(p2)=1.429xx10^(-5) "atm"^(-2)`
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