Question

When limestone is heated, quicklime is formed according to equation,
\[CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(s)\]
The experiment was carried out in the temperature range \[800-{{900}^{\circ }}C\].
Equilibrium constant ${{K}_{b}}$ follows the relation,
$\log {{K}_{p}}=7.282-\dfrac{8500}{T}$
Where, T is temperature in Kelvin. At what temperature the decomposition will give $C{{O}_{2}}(g)$ at 1 atm?

Answer 1

Hint: The equilibrium constant is the number that expresses the relationship between the amounts of the products and the reactants present at equilibrium in a reversible chemical reaction at a given temperature.


Complete step by step solution:
The experiment was carried out in the temperature range 800−900C. Equilibrium constant \[{{K}_{p}}\]​ follows the relation,
\[log{{K}_{p}}=7.282-8500/T\]
where, T is temperature in Kelvin.

The decomposition gives the\[C{{O}_{2}}(g)\]​at 1 atm
\[{{K}_{p}}=PC{{O}_{2}}=1\]
\[log{{K}_{p}}=7.282-\dfrac{8500}{T}\]
\[\log 1=7.282-\dfrac{8500}{T}\]
\[0=7.282-\dfrac{8500}{T}\]
\[7.282=\dfrac{8500}{T}\]
\[T=1167.26K\]
\[T=1167.26-{{273.15}^{0}}C={{894.11}^{0}}C\]

In a chemical reaction, chemical equilibrium is the state where the two reactants and objects are available in fixations which have no more property to change with time so that there is none of the recognizable change in the properties of the system.

This state results when the forward response continues at a similar rate as the backward reaction. The response paces of the forward and in the backward reaction are generally not zero, however equivalent. Accordingly, there are no net changes in the centralizations of the reactants and items. Such a state is known as a unique equilibrium

In the static part, when we state that a body is in equilibrium, what we mean is that the body isn't moving at all despite the fact that there might be forces following up on it. (As a rule, equilibrium implies that there is no acceleration, i.e., the body is moving with steady velocity yet in this extraordinary case we take this consistent with being zero).

Note: The equilibrium constant cannot be 0. This is because this implies that the concentration of the products is equal to 0 at equilibrium.