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The equilibrium constant K(p) for the re...

The equilibrium constant `K_(p)` for the reaction,
`N_(2)+3H_(2) hArr 2NH_(3)`
is `1.64xx10^(-4)` at `400^(@)C` and `0.144xx10^(-4)` at `500^(@)C`. Calculate the mean heat of formation of `1` mol of `NH_(3)` from its elements in this temperature range.

Text Solution

Verified by Experts

We have `log_(10) K_(2)/K_(1)=(DeltaH)/(2.303R)(1/T_(1)-1/T_(2))`
`"log" 0.144/1.64=(DeltaH)/(2.303xx1.987xx10^(-3))(1/673-1/773)`
`DeltaH=-25.14 kcal for 2 "mol" =-12.57 kcal "mol"^(-1)`
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