For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .^{\circ} \mathrm{C}\) $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ If a 20.0 -g sample of \(\mathrm{CaCO}_{3}\) is put into a 10.0 -L container and heated to \(800 .^{\circ} \mathrm{C},\) what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

We are given a 20.0 g sample of CaCO3, using its molar mass, we can convert this mass to moles: Molar mass of CaCO3 = 40.08 (for Ca) + 12.01 (for C) + 3 * 16.00 (for O) = 100.09 g/mol Now, calculate the moles of CaCO3: moles of CaCO3 = (20.0 g) / (100.09 g/mol) = 0.2 mol

The reaction at equilibrium can be described by the equilibrium constant Kp, which is given as 1.16. Write the expression for Kp: \(K_p = \frac{[CO_2]^{\delta n_{CO_2}}}{[CaCO_3]^{\delta n_{CaCO_3}}[CaO]^{\delta n_{CaO}}}\) In this case, since CaCO3 and CaO are in solid state and their concentrations remain constant, their activity is considered as 1. We can simplify the expression: \(K_p = [CO_2]\) Now, let's create an ICE (Initial - Change - Equilibrium) table for the reaction: | | CaCO3 | CaO | CO2 | |:----------:|:-----:|:------:|:-----:| | Initial | 0.2 | 0 | 0 | | Change | -x | x | x | | Equilibrium| 0.2-x | x | x | Where x is the amount of CaCO3 that reacts to reach equilibrium.

We can now calculate the equilibrium partial pressure of CO2. We know that Kp = [CO2] and that the volume of the container is 10.0 L. The relationship between partial pressure and concentration is given by: \(P_{CO2} = \frac{n_{CO2}RT}{V}\) Where R is the ideal gas constant = 0.0821 L atm/mol K and T is the temperature in Kelvin (in this case, 800°C = 1073.15 K). We can express the equilibrium concentration of CO2 in terms of the partial pressure as: \([CO_2] = \frac{P_{CO2}V}{n_{CO2}R}\) Substitute this into the Kp expression: \(K_p = \frac{P_{CO2}V}{n_{CO2}R}\) Solve for P_CO2: \(P_{CO2} = K_p n_{CO2}R\) Now, substitute the values of Kp, n_CO2 (which is x), R, and V: \(P_{CO2} = (1.16)(x)(0.0821)(1073.15)\)

We can use the relationship between partial pressure and concentration to find the value of x, which represents the amount of CaCO3 that reacted to reach equilibrium: \(P_{CO2} = x*RT/V\) \(x = \frac{P_{CO2}V}{RT}\) Plug in the values for P_CO2, R, T, and V: \(x = \frac{(1.16)(0.0821)(1073.15)(10.0)}{0.2}\) x = 0.05051 mol

Now that we have the moles of reacted CaCO3, we can calculate the percentage of the initial 20.0 g sample that reacted: Percentage reacted = \(\frac{reacted \:moles}{initial\: moles} \times 100\) Percentage reacted = \(\frac{0.05051}{0.2} \times 100 \approx 25.26\% \) The percentage of calcium carbonate reacting to reach equilibrium by mass is 25.26%.