Question

When $20\,gm$ of $CaC{O_3}$ were put into $10$ litre flask and heated to ${794^O}C$, $40\%$ of $CaC{O_3}$ decomposition at equilibrium. ${K_P}$ for the decomposition of $CaC{O_3}$ is:

Answer 1

Calcium carbonate is a white insoluble substance with chemical formula $CaC{O_3}$. Calcium carbonate is obtained by passing carbon dioxide gas through slaked lime. It is found in earth’s crust in many forms like marble, limestone, etc. It is known to us that ${K_P}$ is also a form of equilibrium constant which is used with partial fraction is real gas constant.

Complete step by step solution
When $20\,gm$ of $CaC{O_3}$ is put into $10$ litre flask and heated to ${794^O}C$ the decomposition reaction takes place. Calcium carbonate on heating decomposes into carbon dioxide and calcium oxide. This reaction can be represented with the help of following chemical reaction;
Given that,
Mass of calcium carbonate is $20\,gm$.
Volume capacity of flask is $10$ litre.
Percentage decomposition of calcium carbonate is $40\%$.
Temperature is ${794^{\circ}}C$.
Let alpha be the degree of decomposition
To find the ${K_P}$ for decomposition, first of all, we have to find the number of moles of calcium carbonate; which is calculated by $No. \, of \, moles\, = \dfrac{{weight}}{{molar\,mass}}$.
Molar mass of calcium carbonate is $100\,gm/mol$.
So, the number of moles of calcium carbonate is $= \dfrac{{20}}{{100}} = 0.2\,mole$.
Since only $40\%$ of calcium carbonate is at equilibrium so $40\%$ of calcium carbonate will be unreacted.
So, total number of moles unreacted is $= \dfrac{{40}}{{100}} \times 0.2 = 0.08\,mole$
The degree of dissociation will be equal to $\alpha = 0.2 - 0.08 = 0.12$.
So, the molecules dissociated at equilibrium is equal to:
The equilibrium constant is ${K_c} = \left[ {C{O_2}} \right] = 0.16$.
Since, For the calculation we take only the concentration of gaseous atoms not of solids.
We know that,
${K_p} = {K_C}{\left( {RT} \right)^{\Delta n}}$
Where, ${K_P}$ is also a form of equilibrium constant which is used with partial fraction ,R is the real gas constant and $\Delta n$ is the difference between gaseous products and reactants(here only one product is in solid so it is equal to one).
Putting all the values we get,
${K_p} = 0.16 \times 0.082 \times 1067 = 13.99\,atm$

**Hence, the value of ${K_P}$ is equal to $13.99\,atm$.

Note:**
${K_p}$ and ${K_c}$ are dimensionless in nature. The value of equilibrium constant indicates the direction of reaction. If its value is greater than one then there will be formation of product while if the value of equilibrium constant is less than one then reactant will be favoured. And if its value is equal to one then reaction is in equilibrium.