Question

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?

Short answer

The percentage of calcium carbonate reacting to reach equilibrium by mass is approximately 25.26% at 800°C with Kp = 1.16.

Step-by-step solution

Calculate the initial amount of calcium carbonate in moles

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

Write the expression for Kp definition and ICE table

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.

Calculate the equilibrium partial pressure of CO2

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: \(P_{CO2} = (1.16)(x)(0.0821)(1073.15)\)

Calculate the amount of reacted CaCO3 to reach equilibrium

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

Calculate the percentage of reacted CaCO3 and express it by mass

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%.

Key concepts

Kp (equilibrium constant)

The equilibrium constant, specifically for gas-phase reactions, is denoted as \( K_p \). It represents the ratio of the products' partial pressures to the reactants' partial pressures, each raised to the power of their stoichiometric coefficients at equilibrium. In our given reaction, \[ \text{CaCO}_{3}(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_{2}(g) \] we know that both \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids. The concentration of solids does not appear in the expression for the equilibrium constant because their activity is considered to be 1. Thus, for this reaction, the equilibrium constant \( K_p \) is simply equal to the partial pressure of CO\(_2\). This means \( K_p = P_{\text{CO}_2} \). For this problem, \( K_p \) is given as 1.16, reflecting the state of equilibrium at 800°C, meaning that at 800°C, the partial pressure of CO\(_2\) at equilibrium will equal this value. This makes \( K_p \) a handy tool for predicting how a gaseous reaction will shift under various conditions. Understanding \( K_p \) helps us to determine the extent of a reaction and whether it will favor products or reactants under a given set of conditions.

ICE table

The ICE table is a systematic way to track changes in concentrations or pressures of reactants and products throughout a chemical reaction, moving from the Initial state, through the Change, to the Equilibrium state. It is incredibly useful when dealing with equilibrium problems. Let's take a closer look at how it applies to our given reaction.Initially, we have 0.2 mol of \( \text{CaCO}_3 \), and both \( \text{CaO} \) and \( \text{CO}_2 \) start at 0. As the reaction progresses toward equilibrium, \( x \) amount of \( \text{CaCO}_3 \) will decompose. Thus, at equilibrium, the amount of remaining \( \text{CaCO}_3 \) is \( 0.2 - x \), whereas \( \text{CaO} \) and \( \text{CO}_2 \) each have \( x \) moles.Applied to our reaction:

• Initial: 0.2 moles \( \text{CaCO}_3 \); 0 for \( \text{CaO} \) and \( \text{CO}_2 \).
• Change: -\( x \) moles for \( \text{CaCO}_3 \); +\( x \) for both \( \text{CaO} \) and \( \text{CO}_2 \).
• Equilibrium: \( 0.2 - x \) for \( \text{CaCO}_3 \); \( x \) for \( \text{CaO} \) and \( \text{CO}_2 \).

This method clarifies how much of each species is present at equilibrium, paving the way for further calculations like determining the equilibrium constant or the exact change in concentration/pressure.

Partial pressure of gases

The term 'partial pressure' refers to the pressure exerted by an individual gas within a mixture of gases. For chemical reactions involving gases, understanding partial pressures is crucial. The sum of the partial pressures of all gases in a system gives the total pressure.In the context of our equilibrium reaction, only \( CO_2 \) is in the gaseous state. Its partial pressure is crucial for calculating \( K_p \). From the ICE table analysis, we find that at equilibrium, the partial pressure of \( CO_2 \) is represented by \( x \), with \( K_p = x = 1.16 \). This relates to the number of moles and volume using the ideal gas law: \[ P_{\text{CO}_2} = \frac{n_{\text{CO}_2}RT}{V} \]Here, \( n_{\text{CO}_2} \) corresponds to the moles of \( CO_2 \), \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This equation connects the tangible conditions (temperature, volume, and moles) to the more abstract, but critically important, concept of partial pressure.

Reaction stoichiometry

Stoichiometry is the calculation relationship between reactants and products in chemical reactions, often coming in handy to make predictions. It ensures the reaction's mass conservation. In the exercise where we decompose \( \text{CaCO}_3 \), the stoichiometry is relatively straightforward due to the 1:1 ratio between \( \text{CaCO}_3 \), \( \text{CaO} \), and \( \text{CO}_2 \).This stoichiometric simplicity translates to each mole of \( \text{CaCO}_3 \) producing one mole of \( \text{CaO} \) and one mole of \( \text{CO}_2 \). The stoichiometry helps us interpret the results from the ICE table, simplifying calculations of how much \( \text{CaCO}_3 \) decomposes to achieve equilibrium, based on the change, \( x \).Understanding this 1:1:1 stoichiometric relationship ensures that the reaction maintains balanced proportions, facilitating calculations of moles transitioning to equilibrium. This common tool allows predictions about product formation from given reactant amounts, which is essential when handling reactions like this that depend on precise quantities for accuracy.