The partial pressure of `CO_(2)` in the reaction
`CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`
Is `0.773` mm at `500^(@)C`. Calculate `K_(p)` at `600^(@)C` for the above reaction, `DeltaH` of the reaction is `43.2` kcal per mole and does not change in the given range of temperature.

Explanation

`CaCO_3 (s) Leftrightarrow CaO(s)+CO_(2) (g)`
`K_(p)=p_(CO_(2))`
Hence, `K_(p)` includes the constant active masses of `CaCO_3 and CaO` which are solids.
Thus, `K_(p_(1))="0.773 mm at 500"^@C`
`K_(p_(2))="p mm (say) at 600"^@C`
`triangleH^(0)="43.2 kcal per mole"`
=43200 cal
Now we have
`log (K_(p_(2)))/(K_(p_(1)))=(triangleH^(0))/(2.303 R) (T_(2)-T_(1))/(T_(1)T_(2))`.........(Eqn. 11)
`log (p)/(0.773) =(43200)/(2.303 xx 1.98) ((873-773)/(873 xx 773)) (R=1.98" cal deg"^(-1) "mol"^(-1))`
p=19.6 mm
`K_(p)" at "600^@C` is 19.6mm