Question

In the preparation of quick lime from lime stone, the reaction is:-
$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$.
Experiments carried out between $850$ and $950$ led to set of $K_p$ values fitting is empirical equation by $\ln K_p=7.282-\frac{8500}{T}$, where $T$ is absolute temperature. If the reaction is carried out in quiet air, minimum temperature would be predicted from this equation for complete decomposition of lime is x K. 100x is_________.

Answer 1

$\because \ln K_p=7.282-\frac{8500}{T}$
at $T=1123K,K_p=0.751$
at $T=1223K,K_p=1.394$
$\therefore$ Average $K_p=\frac{0.751+1.394}{2}=1.0725$
Quiet air means, the state when the decomposition of $CaC{O}_3$ occurs in a way that the air surrounded to it becomes saturated with a pressure of $C{O}_2$ equal to 1 atm (to pushback the air) needed to attain equilibrium.
Thus, when $K_p\ge 1$, then reaction is spontaneous in forward direction or at $950$, the reaction occurs in forward direction only. Thus, for minimum temperature $K_p=1$. Putting in given equation.
$\therefore \log K_p=7.282-\frac{8500}{T}$
or $\log 1=7.282-\frac{8500}{T}$
$\therefore T=1167.26K$ or $894.26$