Question 17.69: The equilibrium constant for the reaction is Kp = PCO2 = 0.2......
The equilibrium constant for the reaction is K_\text{p} = \text{P}_{\text{CO}_2} = 0.220\text{ atm}
The amount of calcium carbonate solid in the container at the first equilibrium equals the original amount, 0.100 mol, minus the amount reacted to form 0.220 atm of carbon dioxide. The moles of \text{CaCO}_3 reacted is equal to the number of moles of carbon dioxide produced. Use the pressure of \text{CO}_2 and the Ideal Gas Equation to calculate the moles of \text{CaCO}_3 reacted:
0. 100\text{ mol CaCO}_3 – 0.06960\text{ mol CaCO}_3 = 0.0304\text{ mol CaCO}_3 at first equilibrium
As more carbon dioxide gas is added, the system returns to equilibrium by shifting to the left to convert the added carbon dioxide to calcium carbonate to maintain the partial pressure of carbon dioxide at 0.220 atm (K_\text{p}). Convert the added 0.300 atm of \text{CO}_2 to moles using the Ideal Gas Equation. The moles of \text{CO}_2 reacted equals the moles of \text{CaCO}_3 formed.
\text{Moles CaCO}_3 = \text{moles CO}_2 = \text{n = PV / RT} \\ \text{n}=\frac{(0.300\text{ atm})(10.0\text{ L})}{\left\lgroup 0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}} \right\rgroup (385\text{ K})} =0.09491\text{ mol CaCO}_3\text{ formed (unrounded)}
Add the moles of \text{CaCO}_3 formed in the second equilibrium to the moles of \text{CaCO}_3 at the first equilibrium position and convert to grams.
\text{Mass CaCO}_3=\left\lgroup\frac{0.0304\text{ mol}+0.09491\text{ mol CaCO}_3}{} \right\rgroup \left\lgroup\frac{100.09\text{ g CaCO}_3}{1\text{ mol CaCO}_3} \right\rgroup =12.542=12.5\text{ g CaCO}_3