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EquilibriumNational 2025

At high temperatures, solid calcium carbonate (M = 100.09) is in equilibrium with solid calcium oxide and carbon dioxide gas. $\ce{CaCO3(s) <=> CaO(s) + CO2(g)}$ Into an evacuated 30.0 L chamber maintained at 1200 K is introduced 3.0 g solid $\ce{CaCO3}$ . After the system is allowed sufficient time to equilibrate, the pressure is 0.10 bar. What will the total pressure be at equilibrium if an additional 3.0 g $\ce{CaCO3}$ is added to the chamber?

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Explanation

draft

Check whether the first sample actually reached true (heterogeneous) equilibrium.

At equilibrium, $K=P_{CO_2}$ (both solids have activity 1), so as long as some solid $\ce{CaCO3}$ remains, the equilibrium pressure is fixed by $K$ regardless of how much $\ce{CaCO3}$ is present.

Compute the moles of $CO_2$ actually present at $P=0.10\text{ bar}$ :
$n_{CO_2}=\frac{PV}{RT}=\frac{(0.10\text{ bar})(30.0\text{ L})}{(0.08314\text{ L·bar/mol·K})(1200\text{ K})}=0.0301\text{ mol}$

Compare with the moles of $\ce{CaCO3}$ originally added:
$n_{CaCO_3}=\frac{3.0\text{ g}}{100.09\text{ g/mol}}=0.0300\text{ mol}$

These are essentially identical — all of the original $\ce{CaCO3}$ decomposed to reach 0.10 bar, meaning no solid $\ce{CaCO3}$ remained. The system was limited by the amount of reactant, not by the true equilibrium constant $K$ . Therefore the measured 0.10 bar is not the equilibrium partial pressure of $CO_2$ ; it's just an upper bound (the true $K$ must be $\ge 0.10$ bar).

When more $\ce{CaCO3}$ is added, there may (or may not) be enough solid remaining at equilibrium to fix $P_{CO_2}=K$ . Since we don't know the actual value of $K$ (only that $K\ge0.10$ bar), we cannot calculate the new total pressure from the given data.

Answer: (D) — the information given is insufficient to determine the new pressure.

Assumes the first trial reached genuine equilibrium with leftover solid $\ce{CaCO3}$ , so $K=0.10$ bar is fixed and adding more solid can't change $P$ . This ignores the check showing all the original solid was consumed, so 0.10 bar was not actually the equilibrium value.

Comes from an incorrect 'averaging' or square-root manipulation of the pressure/mole relationship (e.g., treating $P^2\propto$ total moles of $CaCO_3$ decomposed), a made-up scaling law with no thermodynamic basis.

Assumes $CO_2$ production scales linearly with the mass of $\ce{CaCO3}$ added (like a stoichiometric, non-equilibrium reaction going to completion), doubling the pressure to 0.20 bar — but equilibrium pressure of a gas in a solid-solid-gas equilibrium doesn't depend on the amount of solid present once true equilibrium is reached.

Correct: checking the numbers shows the first sample fully decomposed (no leftover solid $\ce{CaCO3}$ ), so 0.10 bar is not the true equilibrium pressure; the actual $K$ is unknown, so the new pressure cannot be calculated.

Concept · Heterogeneous equilibrium — solid completely consumed vs. true equilibrium