Question

In the preparation of quick lime from limestone, the reaction is

CaCO₃ (s) ⇋ CaO(s) + CO₂ (g) 

Experiments carried out between 850ºC and 950ºC led to a set of K_p values fitting an empirical equation

log k_p = 7.282 - 8500/T

where T is the absolute temperature. If the reaction is carried out in quiet air, what temperature would be predicted from this equation for complete decomposition of the limestone?

Answer 1

For the equilibrium CaCO₃ (s) ⇋ CaO(s) + CO₂ (g)

Kp = CO₂

the decomposition of CaCO₃ in quiet air, would continue till the pressure developed due to CO₂ equals 1 atm (atmospheric pressure).

the decomposition of CaCO₃ in quiet air, would continue till the pressure developed due to CO₂ equals 1 atm (atmospheric pressure).

∴ when the decomposition is complete K_p = 1atm Substituting K_p in the given empirical equation,

log 1 = 7.282 - 8500/T = 0.

T = 1167 K = 894º C