Question

When limestone is heated, quicklime is formed according to the equation,
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(s\right)$
The experiment was carried out in the temperature range ${800-900}^{o}C$. Equilibrium constant $K_p$ follows the relation,
$\log K_p=7.282-8500/T$
where, $T$ is temperature in Kelvin. At what temperature the decomposition will give ${CO}_2\left(g\right)$ at $1$ atm?

Answer 1

When limestone is heated, quicklime is formed according to the equation,
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(s\right)$
The experiment was carried out in the temperature range ${800-900}^{o}C$. Equilibrium constant $K_p$ follows the relation,
$\log K_p=7.282-8500/T$
where, $T$ is temperature in Kelvin.
The decomposition gives ${CO}_2\left(g\right)$ at $1$ atm
$K_p=P_{C{O}_2}=1$
$\log K_p=7.282-\frac{8500}{T}$
$\log 1=7.282-\frac{8500}{T}$
$0=7.282-\frac{8500}{T}$
$7.282=\frac{8500}{T}$
$T=1167.26K$
$T=1167.26-{273.15}^{o}C={894.11}^{o}C$
The temperature is ${894.11}^{o}C$