Question

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.

Answer 1

`CaCO_(3)(s) hArr CaO(s)+CO_(2)(g)`
`K_(p)=p_(CO_(2))`
Here `K_(p)` includes the constant active masses of `CaCO_(3)` and `CaO` which are solids.
Thus, `K_(p1)=0.773 mm` at `500^(@)C`
`K_(p2)=p mm (say) at 600^(@)C`
`DeltaH=43.2 kcal "mol"^(-1)=43200 cal`
Now we have,
`"log" K_(p2)/K_(p1)=(DeltaH)/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`"log"p/0.773=43200/(2.303xx1.98)((873-773)/(873xx773)), (R=1.98 cal deg^(-1) "mol"^(-1))`
`p=19.6 mm`
`:. K_(p)` at `600^(@)C` is `19.6 mm`.