Question

For the reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$, complete conversion of $CaCO_3$ can be achieved by:

• A. Loading more amount of CaCO\(_3\)
• B. Removing CO\(_2\) continuously
• C. Increasing pressure
• D. Increasing temperature

Hint:

Continuous removal of product drives equilibrium completely to the product side.

Answer 1

The Correct Option is B

Step 1: Write the equilibrium constant expression for this heterogeneous system.
\[ CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \]
Pure solids have a fixed activity of $1$ and drop out of the equilibrium expression, so:
\[ K_p = P_{CO_2} \]
Equilibrium is reached the moment the $CO_2$ pressure in the container climbs up to this fixed value $K_p$, no matter how much solid is present.

Step 2: Ask what "complete conversion" really requires.
Complete conversion means the reaction quotient $Q_p = P_{CO_2}$ must stay below $K_p$ for the whole process, because the moment $Q_p$ reaches $K_p$, the system sits at equilibrium and net decomposition stops.

Step 3: Test loading more $CaCO_3$ against this condition.
Since solids do not appear in $K_p$ at all, adding more $CaCO_3$ cannot change $K_p$ or the equilibrium $CO_2$ pressure. The reaction still stops at the same $P_{CO_2}$, so this does not help.

Step 4: Test increasing pressure.
Raising the external pressure raises $P_{CO_2}$ towards $K_p$ faster, which only brings the system to equilibrium sooner and blocks further decomposition. This works against complete conversion, not for it.

Step 5: Test increasing temperature.
A higher temperature raises $K_p$ itself, since the decomposition is endothermic, which does push the equilibrium further, but $Q_p$ still eventually catches up to the new, higher $K_p$ and the reaction stops there. Temperature alone does not keep $Q_p$ below $K_p$ forever.

Step 6: Test removing $CO_2$ continuously.
If $CO_2$ is pumped out as fast as it forms, $Q_p = P_{CO_2}$ never gets the chance to rise up to $K_p$. The system keeps chasing an equilibrium it never reaches, so $CaCO_3$ keeps decomposing until none is left.

Final Answer:
\[ \boxed{\text{Removing } CO_2 \text{ continuously}} \]