Question

\(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) decomposes at elevated temperatures according to the stoichiometric equation $$\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)$$ a. If pure \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) is put into a sealed vessel, the air is pumped out, and the vessel and its contents are heated, the total pressure is 0.290 bar. Determine \(K_{P}\) under these conditions. b. If the vessel also contains 0.120 bar \(\mathrm{H}_{2} \mathrm{O}(g)\) at the final temperature, what is the partial pressure of \(\mathrm{CO}_{2}(g)\) at equilibrium?

Short answer

The equilibrium constant \(K_P\) under the given conditions is 0.021. When the vessel also contains additional \(\mathrm{H}_{2} \mathrm{O}(g)\) with a pressure of 0.120 bar, the partial pressure of \(\mathrm{CO}_{2}(g)\) at equilibrium is 0.079 bar.

Step-by-step solution

Set up the initial conditions

At the beginning, we have the following: \[\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\] Initial state: \[P_{\mathrm{H}_{2} \mathrm{O}} = 0 \quad , \quad P_{\mathrm{CO}_{2}} = 0\]

Set up expressions

Since we have a reversible reaction, we can use Dalton's law of partial pressures to find the pressures of the individual gases at equilibrium: \[P_{\mathrm{H}_{2} \mathrm{O}} = x \quad , \quad P_{\mathrm{CO}_{2}} = x\] The total pressure at equilibrium is the sum of partial pressures: \[P_{total} = P_{\mathrm{H}_{2} \mathrm{O}} + P_{\mathrm{CO}_{2}} \quad , \quad P_{total} = 0.290\,\text{bar}\]

Calculate partial pressures

Now we can solve the equation for the partial pressures: \[0.290\,\text{bar} = x + x\] \[x = \frac{0.290}{2}\,\text{bar}\] \[P_{\mathrm{H}_{2} \mathrm{O}} = P_{\mathrm{CO}_{2}} = 0.145\,\text{bar}\]

Calculate \(K_{P}\)

Using the partial pressures, we can calculate the equilibrium constant \(K_P\) under these conditions: \[K_P = \frac{P_{\mathrm{H}_{2} \mathrm{O}} \cdot P_{\mathrm{CO}_{2}}}{P_{\mathrm{Ca}( \mathrm{HCO}_{3})_{2}} }\] Since pure \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) is a solid, its pressure is omitted from the equilibrium expression: \[K_P = P_{\mathrm{H}_{2} \mathrm{O}} \cdot P_{\mathrm{CO}_{2}}\] \[K_P = (0.145)^2\] \[K_P = 0.021\] So, under this condition, \(K_P = 0.021\).

Calculate the partial pressure of \(\mathrm{CO}_{2}(g)\) at equilibrium when the vessel also contains additional pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\)

Given that the partial pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\) is initially 0.120 bar, we can add it to the initial equilibrium: \[P'_{\mathrm{H}_{2} \mathrm{O}} = 0.145 + 0.120 = 0.265 \, \text{bar}\] Now, we can set up this reaction expression: \[K_P = \frac{P'_{\mathrm{H}_{2} \mathrm{O}} \cdot P_{\mathrm{CO}_{2}}}{P_{\mathrm{Ca}( \mathrm{HCO}_{3})_{2}} }\] Since \(K_P = 0.021\), the expression becomes: \[0.021 = (0.265) \cdot P_{\mathrm{CO}_{2}}\] Now we can solve the expression for \(P_{\mathrm{CO}_{2}}\): \[P_{\mathrm{CO}_{2}} = \frac{0.021}{0.265}\] \[P_{\mathrm{CO}_{2}} = 0.079\,\text{bar}\] So, the partial pressure of \(\mathrm{CO}_{2}(g)\) at equilibrium when the vessel also contains an additional pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\) is 0.079 bar.

Key concepts

Partial Pressure

Partial pressure is a measure of the contribution each gas in a mixture of gases makes to the total pressure. It's based on the idea that gases in a mixture each act independently of each other. In the context of chemical reactions involving gases, understanding partial pressure is crucial for predicting how reactions proceed under different conditions.

For a single gas, its partial pressure can be calculated by multiplying the mole fraction of the gas by the total pressure of the gas mixture. This is particularly useful when dealing with gases produced or consumed in a chemical reaction, as seen in the given exercise with \(\mathrm{H}_{2}\mathrm{O}(g)\) and \(\mathrm{CO}_{2}(g)\).

Dalton's Law of Partial Pressures

Dalton's law of partial pressures states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas. This law allows us to calculate the total pressure simply by adding up the pressure contributed by each gas in a mixture, assuming ideal behavior.

In the exercise's step-by-step solution, Dalton's law is employed to determine the total pressure at equilibrium, which is the sum of the partial pressures of water vapor and carbon dioxide. Remember, though, Dalton's law assumes gases don't interact, which is a good approximation for many situations but not all.

Decomposition Reaction

A decomposition reaction is a type of chemical reaction where one compound breaks down into two or more simpler substances. In the provided exercise, \(\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}(s)\) decomposes to form \(\mathrm{CaCO}_{3}(s)\), \(\mathrm{H}_{2}\mathrm{O}(g)\), and \(\mathrm{CO}_{2}(g)\).

Understanding the nature of decomposition is essential when predicting the products and changes in a reaction system. Decomposition reactions often require an input of energy, such as heat, to proceed, as is the case with the heating of the vessel in the exercise.

Chemical Equilibrium

Chemical equilibrium is the state in which the rates of the forward and reverse reactions are equal, resulting in no net change in the amounts of products and reactants. It's essential to note that equilibrium does not mean the reactants and products are present in equal amounts but that their concentrations remain constant over time.

For gases, the position of equilibrium can be expressed using an equilibrium constant (\(K_P\)) based on partial pressures. This constant gives us valuable information about the reaction conditions at equilibrium, like in our exercise where \(K_P\) helps determine the partial pressures of the gases involved.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the calculation of reactants and products in chemical reactions. It is rooted in the conservation of mass where the quantity of each element must be the same before and after a chemical reaction.

In practice, stoichiometry provides the tools to calculate how much product can be formed from a given amount of reactants, or how much of one reactant is needed to completely react with another. The exercise shows a stoichiometric calculation, where the mole ratio of reactants and products is used to determine the amount of gases produced at equilibrium.