Question

In the preparation of quicklime from limestone, the reaction is, $CaCO_{3} (s) \leftrightharpoons CaO (s) + CO_{2} (g)$
Experiments can be carried out between $850^\text{0}$C and $950^\text{0}$C led to set of $K_p$ values fitting in empirical formula $lnK_p = 7.282-\dfrac{8500}{T}$ where T is absolute temperature. If the reaction is carried out in quiet air, what minimum temperature would be predicted from this equation for almost complete decomposition of lime?

Answer 1

Hint: Calcium oxide is known as quick lime. The reaction to be in equilibrium with the value of $K_p$ must be 1, it is necessary for the complete decomposition.

Complete step by step solution:
The reaction is, $CaCO_{3} (s) \leftrightharpoons CaO (s) + CO_{2} (g)$

For the equilibrium K$_p$ = CO$_{2(g)}$, the decomposition reaction of CaCO$_{3} (s)$ occurs in air, and proceeds till the pressure is equal to 1 atm (atmospheric pressure).
Given, lnK$_p =7.282-\dfrac{8500}{T}$ , we know K$_p$ = 1
Therefore, lnK$_p =7.282-\dfrac{8500}{T}$ turns out to be 7.282 = $\dfrac{8500}{T}$
By solving 7.282 = $\dfrac{8500}{T}$, the value of T is found to be 1167 K. It is considered as the minimum temperature required for the completion of reaction.
Now, $In^\text{0}$C = 1167 – 273 = $894^\text{0}$C
We can also write it as T = 1167 K = $894^\text{0}$C
The minimum temperature required for complete decomposition of lime is $894^\text{0}$C.

Note: Don’t get confused with calcium oxide and limestone. These both are different names for the same compound. For complete decomposition of compound K$_p$ value must be equal to 1.