Question

When 20g of $CaC{{O}_{3}}$ were put in a 10 litre flask and heated to ${{800}^{\circ }}C$, 30% of $CaC{{O}_{3}}$ remained unreacted equilibrium. ${{K}_{p}}$ for decomposition of $CaC{{O}_{3}}$ will be:

Answer 1

HINT: To solve this firstly write down the decomposition reaction of calcium carbonate. Then find the concentration of the species that will affect the equilibrium constant in terms of concentration. Use the relation, ${{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}$ to find the required answer.

COMPLETE STEP BY STEP SOLUTION: We know that calcium carbonate decomposes as-
\[CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g)\]
We can write that number of moles of calcium carbonate = 0.2 moles (as no. of moles = $\dfrac{mass}{molar\text{ mass}}$ ).
Now, it is given that 30% of the calcium carbonate remained unreacted.
Therefore, the number of moles of calcium carbonate that remained unreacted will be 30% of 0.2 i.e. 0.06 moles.
And, the number of moles of products thus formed will be 0.2 – 0.06 = 0.14 (as only 70% reacted and moles of reactants will be equal to number of moles of products)
Therefore, we can write the equilibrium equation and the number of moles before and after dissociation as-
\[CaC{{O}_{3}}(s)\rightleftharpoons CaO(s)+C{{O}_{2}}(g)\]

	\[CaC{{O}_{3}}\]	$CaO$	$C{{O}_{2}}$
Moles before dissociation:	0.2	0	0
Moles after dissociation:	0.06	0.14	0.14

Now, here for the equilibrium constant we will only consider carbon dioxide gas. Therefore, the equation of ${{K}_{C}}$ will be:
\[\begin{align}
  & {{K}_{C}}=\left[ C{{O}_{2}} \right] \\
 & or,{{K}_{c}}=0.14 \\
\end{align}\]
i.e. for 10 litre so for 1 litre, it will be 0.014.
Also, we know that ${{K}_{p}}={{K}_{c}}{{\left( RT \right)}^{\Delta n}}$
Temperature of given to us as ${{800}^{\circ }}C$ i.e. 800 + 273 = 1073 K.
And R is the universal gas constant whose value is 0.082 L atm/ K and n is 1 as we have 1 mole of calcium carbonate.
Therefore, putting these values we will get ${{K}_{p}}=0.14mol\text{ }{{\text{L}}^{-1}}\times 0.082L\text{ atm mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}\times 1073K=12.318\text{ }atm$
Therefore, the correct answer is option [B] 1.231 atm.
NOTE: We can see that the reaction here is in equilibrium. The starting compound is a pure solid and so is the calcium oxide that is produced. We know that the amount of pure solid or liquid is excluded from the equilibrium constant as their concentration stays the same throughout the reaction. Also, if we write the equilibrium constant in terms of partial pressure, since calcium carbonate and calcium oxide are pure solids so they are excluded from the equation. So K is dependent only on the partial pressure of carbon dioxide. \[K={{P}_{C{{O}_{2}}}}\]