Question

The partial pressure of carbon dioxide in the reaction

\[\ce{CaCO3(s) <=> CaO(s) + CO2(g)}\] is 1.017 × 10^-3 atm at 500°C. Calculate K_p at 600°C for the reaction. H for the reaction is 181 KJ mol^-1 and does not change in the given range of temperature.

Answer 1

P_CO2 = 1.017 × 10^-3 atm

T = 500°C;

K_p = P_CO2

∴ K_p1 = 1.017 × 10^-3;

T = 500 + 273 = 773 K

K_p2 = ?

T = 600 + 273 = 873 K

∆H° = 181 KJ mol^-1 

`log  (("K"_("P"_2))/("K"_("P"_1))) = (Delta "H"^0)/(2.303 "R") [("T"_2 - "T"_1)/("T"_2"T"_1)]`

`log  (("K"_("P"_2))/(1.017 xx 10^-3)) = (181 xx 10^3)/(2.303 xx 8.314)`

`= ((873 - 773)/(873 xx 773))`

`log  (("K"_("P"_2))/(1.017 xx 10^-3)) = (181 xx 10^3 xx 100)/(2.303 xx 8.314 xx 873 xx 773)`

`(("K"_("P"_2))/(1.017 xx 10^-3))` = anti log of (1. 40)

`("K"_("P"_2))/(1.017 xx 10^-3)` = 25.12

`=> "K"_("P"_2) = 25.12 xx 1.017 xx 10^-3`

`"K"_("P"_2) = 25.54 xx 10^-3`