Problem 116 - Chemistry

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 \(1.00-\mathrm{L}\) flask containing 10.0 \(\mathrm{g} \mathrm{NaHCO}_{3}\) 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

In summary, after the equilibrium is established at \(125^{\circ}\mathrm{C}\), the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) are 0.0297 atm. At equilibrium, there are 5.01 g of \(\mathrm{NaHCO}_{3}\) and 3.99 g of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) present. The minimum container volume required for all of the \(\mathrm{NaHCO}_{3}\) to decompose is 0.0141 L.

Step 1: Calculate the moles of NaHCO3

First, we need to find the moles of NaHCO3. We are given that the sample contains 10.0 g of NaHCO3. Using the molar mass of NaHCO3, we can find the moles: Moles of NaHCO3 = Mass of NaHCO3 / Molar mass of NaHCO3 The molar mass of NaHCO3 = 23 (Na) + 1 (H) + 12 (C) + 16 × 3 (O_3) = 84 g/mol Moles of NaHCO3 = 10.0 g / 84 g/mol = 0.119 moles

Step 2: Set up the ICE table

Now that we have the moles of NaHCO3, we can set up an ICE (Initial, Change, Equilibrium) table to relate the amounts of all species involved. $$ \begin{array}{c|c|c|c|c} & \mathrm{Initial} & \mathrm{Change} & \mathrm{Equilibrium} \\ \hline \mathrm{NaHCO}_{3}(s) & 0.119 & -2x & 0.119-2x \\ \mathrm{Na}_{2}\mathrm{CO}_{3}(s) & 0 & x & x \\ \mathrm{CO}_{2}(g) & 0 & x & x \\ \mathrm{H}_{2}\mathrm{O}(g) & 0 & x & x \\ \end{array} $$

Step 3: Use Kp to find the partial pressures

Now we can use the given Kp value (0.25) and the equilibrium amounts to find the partial pressures of CO2 and H2O. \(\mathrm{Kp}=\dfrac{[\mathrm{CO}_{2}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{NaHCO}_{3}]^2}=0.25\) Since the reaction is in a 1.00-L flask, the partial pressures will be equal to the molar concentrations. Thus, we can rewrite the equation as follows: \(0.25=\dfrac{x^2}{(0.119-2x)^2}\) To simplify the problem, we can assume that 2x is very small compared to 0.119. Thus, we can write the equation as: \(0.25=\dfrac{x^2}{(0.119)^2}\) Now, we can solve for x: \(x=\sqrt{0.25\times(0.119)^2}\) \(x=0.0297\) Now, we can find the partial pressures of CO2 and H2O: P(CO2) = P(H2O) = x = 0.0297 atm

Step 4: Calculate the masses of NaHCO3 and Na2CO3 at equilibrium

Now that we have the amount of CO2 and H2O in moles at equilibrium, we can find the moles of NaHCO3 and Na2CO3 at equilibrium: Moles of NaHCO3 at equilibrium = 0.119 - 2x Moles of Na2CO3 at equilibrium = x Moles of NaHCO3 at equilibrium = 0.119 - 2(0.0297) = 0.0596 moles Moles of Na2CO3 at equilibrium = 0.0297 moles Now, we can find the masses of NaHCO3 and Na2CO3 at equilibrium: Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3 Mass of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3 Mass of NaHCO3 = 0.0596 moles × 84 g/mol = 5.01 g Mass of Na2CO3 = 0.0297 moles × (46 + 12 + 48) g/mol = 3.99 g

Step 5: Calculate the minimum container volume

To find the minimum container volume for all of the NaHCO3 to decompose, we can use the reaction stoichiometry and the Kp value. All of the NaHCO3 must decompose, so we will need to find the minimum volume that would allow the Kp value to be equal to 0.25. Let V be the required minimum volume. So, after the complete decomposition of 0.119 moles of NaHCO3: Moles of CO2 = Moles of H2O = Moles of Na2CO3 = 0.0595 moles Now, we can write the equation using Kp and V: \(0.25=\dfrac{[0.0595/V]^2}{[(\text{0.0595}/V)]^2}\) From here, we can solve for V: \(V=\dfrac{0.0595^2}{0.25}=0.0141\mathrm{L}\) So, the minimum container volume necessary for all of the NaHCO3 to decompose is 0.0141 L.

Key Concepts

Partial Pressure

Partial pressure is a concept in chemistry that describes the pressure exerted by a single gas in a mixture of gases. It's a crucial idea when discussing chemical equilibrium, especially when reactions involve gases. For a given gas in a mixture, the partial pressure is the pressure it would exert if it occupied the entire volume on its own at the same temperature. In the context of the given reaction, we calculate the partial pressures of carbon dioxide (CO₂) and water vapor (H₂O) when the system reaches equilibrium.
To determine the partial pressure of these gases in the flask, we use the equilibrium constant for pressure, known as Kp. Since the flask is 1.00 L, the partial pressures are equal to the molar concentrations of the gases at equilibrium: **P(CO₂) = P(H₂O) = x atm**, where **x** is derived from solving the equation using Kp.

ICE Table

An ICE table is a fundamental tool used to track changes in concentrations or pressures of reactants and products as a reaction progresses towards equilibrium. ICE stands for Initial, Change, and Equilibrium. It's structured to depict how each component in a chemical reaction starts, changes, and finally what amount remains at equilibrium.
In this reaction involving sodium bicarbonate (\(\text{NaHCO}_3\)) and its decomposition products (\(\text{Na}_2\text{CO}_3, \text{CO}_2, \text{and } \text{H}_2\text{O}\)), the ICE table demonstrates these processes:

• **Initial**: The starting amount of substances, i.e., 0.119 moles for \(\text{NaHCO}_3\) and zero for the products.
• **Change**: The change, denoted by x, represents the shift in quantities as the reaction proceeds to equilibrium.
• **Equilibrium**: The final quantities adjusted for the reaction, such as \(0.119 - 2x\) for \(\text{NaHCO}_3\).

The table becomes particularly useful when calculating Kp and solving for equilibrium pressures or concentrations.

Kp Calculation

The equilibrium constant for pressure, denoted as Kp, measures a reaction's tendency to reach equilibrium. It relates to the pressures (for gases) of the reactants and products. For our specific reaction, Kp is given as 0.25 at \(125^\circ \text{C}\).
Using Kp, we can calculate the partial pressures of the products at equilibrium. The equation for Kp in our decomposition reaction is:\[ Kp = \frac{[\text{CO}_2][\text{H}_2\text{O}]}{[\text{NaHCO}_3]^2} = 0.25 \]Since both \(\text{CO}_2\) and \(\text{H}_2\text{O}\) have equilibrium pressures of x, and the reaction occurs in a 1.00-L flask, the pressures equal their concentrations. By assuming 2x is negligible compared to 0.119 (the original moles of \(\text{NaHCO}_3\)), we simplify the math to solve for x. This simplification leads to an easier calculation to determine partial pressures, which provide valuable insight into the state of the system at equilibrium.

Decomposition Reaction

A decomposition reaction involves the breakdown of a compound into two or more simpler substances. The exercise involves the decomposition of sodium bicarbonate (\(\text{NaHCO}_3\)) into sodium carbonate (\(\text{Na}_2\text{CO}_3\)), carbon dioxide (\(\text{CO}_2\)), and water (\(\text{H}_2\text{O}\)).
In decomposition reactions, understanding the stoichiometry is critical. This means knowing how many moles of each product are generated from the reactants. The balanced equation shows us that two moles of \(\text{NaHCO}_3\) decompose to form one mole each of \(\text{Na}_2\text{CO}_3\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\).
The process reveals the interplay between the chemical equilibrium and decomposition. At equilibrium, not all given \(\text{NaHCO}_3\) will convert completely; some remains. By understanding these concepts and using calculated Kp and ICE tables, you can predict how much of each product is formed.