Problem 106 - Chemistry An Atoms First Approach

Question

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$.A 1.00 -L flask containing \(10.0 \mathrm{g}\) NaHCO \(_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C}\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

Step-by-Step Solution

Answer

a. The partial pressures of CO2 and H2O at equilibrium are 0.5 atm. b. The masses of NaHCO3 and Na2CO3 at equilibrium are 7.50 g and 1.61 g, respectively. c. The minimum container volume necessary for complete decomposition of NaHCO3 is 4.00 L.

Step 1: Find the moles of NaHCO3 initially in the flask

Given the mass of NaHCO3, we need to find its moles by dividing the mass by its molar mass. Molar mass of NaHCO3 = 23 (Na) + 1 (H) + 12 (C) + 16 * 3 (O) = 84 g/mol Initial mass of NaHCO3 = 10 g So, the initial moles of NaHCO3 = 10 g ÷ 84 g/mol = \( \frac{5}{42}\) mol

Step 2: Set up the equilibrium expression

We know the Kp for the given reaction at 125°C. We can create the equilibrium expression according to the balanced equation as follows: \[ K_{p} = \dfrac{P_{CO_{2}} \times P_{H_{2}O}}{(P_{NaHCO_{3}})^{2}}\] Let x represent the partial pressures of CO2 and H2O (since their stoichiometry is 1:1). Since the NaHCO3 is a solid, its activity does not depend on its partial pressure, and we will assume P_NaHCO3 = 1. At equilibrium: Kp = 0.25 The equilibrium expression becomes: \( 0.25 = \dfrac{x^2 \times 1}{1^2}\)

Step 3: Calculate the partial pressures of CO2 and H2O at equilibrium

We can use the equation derived in step 2 to find the partial pressures: \[ x^2 = 0.25\] Solving for x we get, \[ x = \sqrt{0.25} = 0.5 \] So, the partial pressures of CO2 and H2O at equilibrium are 0.5 atm.

Step 4: Calculate the moles of products at equilibrium

We can use the ideal gas law to find the moles of CO2 and H2O at equilibrium: \[ PV = nRT\] Given the volume of the flask (1.00 L) and the temperature (125°C = 398.15K), we can calculate the moles: \[ n_{CO_{2}} = n_{H_{2}O} = \dfrac{P_{CO_{2}} V_{flask}}{RT} =\dfrac{0.5\, atm \times 1.00\, L}{0.0821\, L\, atm\, mol^{-1} K^{-1} \times 398.15\, K}\] n_CO2 = n_H2O = 0.01518 mol

Step 5: Calculate the remaining moles of NaHCO3 and moles of Na2CO3 at equilibrium

Since for every mole of CO2 and H2O produced, 2 moles of NaHCO3 decompose, we can calculate the remaining moles of NaHCO3 and moles of Na2CO3: Moles of NaHCO3 decomposed = 2 * (moles of CO2) = 0.03036 mol Remaining moles of NaHCO3 = initial moles of NaHCO3 - moles of NaHCO3 decomposed = \( \frac{5}{42}\) - 0.03036 = 0.08929 mol Moles of Na2CO3 = moles of CO2 = 0.01518 mol

Step 6: Calculate the masses of NaHCO3 and Na2CO3 at equilibrium

Now that we have the moles of NaHCO3 and Na2CO3 at equilibrium, we can find their masses: Mass of NaHCO3 = moles of NaHCO3 * molar mass of NaHCO3 = 0.08929 mol * 84 g/mol = 7.50 g Molar mass of Na2CO3 = 2 * 23 (Na) + 12 (C) + 16 * 3 (O) = 106 g/mol Mass of Na2CO3 = moles of Na2CO3 * molar mass of Na2CO3 = 0.01518 mol * 106 g/mol = 1.61 g

Step 7: Calculate the minimum container volume necessary for complete decomposition

Finally, we will calculate the minimum volume required for complete decomposition of NaHCO3. For this, we should find the volume needed to convert all of the initial NaHCO3 into Na2CO3: Initial moles of NaHCO3 = \( \frac{5}{42} \) mol 2 moles of NaHCO3 produce 1 mole of CO2, so the moles of CO2 produced = \( \frac{5}{84} \) mol Now we can calculate the minimum volume required for complete decomposition using the ideal gas law at 125°C: \[ V = \dfrac{nRT}{P} = \dfrac{(\frac{5}{84}\,mol) \times (0.0821\, L\,atm\,mol^{-1}\,K^{-1}) \times (398.15\,K)}{0.5\, atm}\] Minimum volume required = 4.00 L In summary, the results for each part are: a. Partial pressures of CO2 and H2O at equilibrium are 0.5 atm. b. Masses of NaHCO3 and Na2CO3 at equilibrium are 7.50 g and 1.61 g, respectively. c. The minimum container volume necessary for complete decomposition of NaHCO3 is 4.00 L.

Key Concepts

Partial Pressure

In a chemical reaction where gases are involved, it's essential to understand the concept of partial pressure. Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. When dealing with chemical equilibria that include gaseous components, such as in the decomposition of sodium bicarbonate (NaHCO₃), partial pressures play a significant role.

Each gas present in the system contributes to the total pressure independently of other gases. For example, in the decomposition reaction given, the partial pressures of carbon dioxide (CO₂) and water vapor (H₂O) were calculated. Each of these gases exerts its pressure independently, leading to an identifiable partial pressure in the reaction flask.

• The sum of all partial pressures equals the total pressure in the container.
• In the given equilibrium, both CO₂ and H₂O have equal partial pressures of 0.5 atm due to the stoichiometry of the balanced reaction being 1:1.

This understanding helps predict how changes in conditions can affect the equilibrium position.

Molar Mass Calculation

To perform calculations within a chemical reaction, we must first be familiar with calculating molar masses. Molar mass is the mass of one mole of a substance and is given in grams per mole (g/mol). Understanding and correctly calculating molar masses are fundamental in converting between masses and moles in chemistry.

For sodium bicarbonate (\(\text{NaHCO}_3\)), we calculated the molar mass by adding together the atomic masses of each element present in the compound: \(23 \, \text{g/mol (Na)} + 1 \, \text{g/mol (H)} + 12 \, \text{g/mol (C)} + 3 \times 16 \, \text{g/mol (O)} = 84 \, \text{g/mol}\).

• It allows us to convert the known mass of a substance into moles, which is crucial for further calculations, such as determining the amount of reactants or products involved.
• Accurate molar mass values enable us to derive the remaining mass or moles of substances at equilibrium accurately.

Calculating molar mass accurately is the foundation for making informed and precise chemical calculations.

Ideal Gas Law

The ideal gas law is an essential principle in chemistry that relates the macroscopic properties of gases: pressure (P), volume (V), temperature (T), and the number of moles (n). It is given by the equation \(PV = nRT\), where \(R\) is the ideal gas constant (0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1}).

This law assumes that gases consist of many small particles with negligible volume, no intermolecular forces, and perfectly elastic collisions. Although real gases may deviate under certain conditions, the ideal gas law is an excellent approximation for many everyday situations.

In the exercise, we used the ideal gas law to calculate the number of moles of gaseous products at equilibrium:

• Knowing the partial pressures and the volume of the system allowed us to determine the moles of CO₂ and H₂O.
• The formula helped us connect these quantities with the known temperature, enabling precise calculations of changes throughout the chemical reaction.

This ability to relate changes in pressure, volume, and temperature through the ideal gas law is a powerful tool when examining chemical reactions involving gases.

Chemical Decomposition

Chemical decomposition occurs when a single compound breaks down into two or more simpler substances. This type of reaction is common in many industrial and natural processes. Decomposition reactions are essential for understanding how reactions progress and how to manipulate conditions to achieve desired outcomes.

In this exercise, we examined the decomposition of sodium bicarbonate into sodium carbonate (Na₂CO₃), carbon dioxide (CO₂), and water (H₂O). This process was influenced by several variables, leading to the establishment of equilibrium.

• The initial compound, sodium bicarbonate, breaks down through heat-induced decomposition.
• The rate of formation and breakdown of products and reactants eventually balances, achieving chemical equilibrium.

In understanding chemical decomposition, it's crucial to consider how different factors affect reaction rates and equilibria. These insights can then be utilized to forecast changes and adjust conditions to reach specific goals in chemical processing.