Problem 2 - Chemistry in 30 days

Question

Baking soda, \(\mathrm{NaHCO}_{3}\), decomposes as \(2 \mathrm{NaHCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). The equilibrium pressure is \(1.04 \mathrm{~atm}\). The \(K_{p}\) for the reaction is (a) \(0.2704\) (b) \(2.704\) (c) \(27.04\) (d) \(270.4\)

Step-by-Step Solution

Answer

The correct answer is (a) 0.2704.

Step 1: Understand the Reaction

We are given the decomposition reaction of baking soda, \(\text{NaHCO}_3\), which is an equilibrium between solid \(\text{NaHCO}_3\) and \(\text{Na}_2\text{CO}_3(s)\), \(\text{CO}_2(g)\), and \(\text{H}_2\text{O}(g)\). Solids do not appear in the expression for the equilibrium constant \(K_p\), so we only consider the gaseous products, \(\text{CO}_2\) and \(\text{H}_2\text{O}\).

Step 2: Write the Expression for Kp

The expression for the equilibrium constant \(K_p\) in terms of the partial pressures of the gases is: \[ K_p = P_{\text{CO}_2} \times P_{\text{H}_2\text{O}} \]Since the stoichiometry of the reaction shows that one mole of \(\text{CO}_2\) and one mole of \(\text{H}_2\text{O}\) are produced, their partial pressures will be equal and can be denoted as \(P_{\text{CO}_2} = P_{\text{H}_2\text{O}} = x\). Thus, the expression becomes: \[ K_p = x \times x = x^2 \]

Step 3: Relate Partial Pressure to Total Pressure

Given the total equilibrium pressure \(1.04 \text{ atm}\), which is the sum of the partial pressures of \(\text{CO}_2\) and \(\text{H}_2\text{O}\):\[ P_{\text{CO}_2} + P_{\text{H}_2\text{O}} = 1.04 \text{ atm} \]Since \(P_{\text{CO}_2} = P_{\text{H}_2\text{O}} = x\), we have:\[ 2x = 1.04 \text{ atm} \]

Step 4: Solve for x

Solve the equation from step 3 to find the value of \(x\):\[ x = \frac{1.04}{2} = 0.52 \text{ atm} \]

Step 5: Calculate Kp

Now use the value of \(x\) to find \(K_p\):\[ K_p = x^2 = (0.52)^2 = 0.2704 \]

Step 6: Determine the Correct Answer

The calculated \(K_p = 0.2704\) corresponds to option (a).

Key Concepts

Le Chatelier's Principle

In a chemical reaction, understanding how changes in conditions affect the system is crucial. Le Chatelier's Principle helps us predict these effects. This principle states that if a system at equilibrium is disturbed, it will adjust itself to counteract the change and re-establish equilibrium.

For example, in the reaction where baking soda decomposes, if you increase the pressure, the equilibrium will shift to mitigate this change. Since the decomposition reaction of baking soda produces gases, increasing pressure will favor the side of the reaction with fewer gas molecules. When considering Le Chatelier's Principle:

• If temperature increases, the system will favor the endothermic direction (absorbing heat).
• If pressure increases by adding an inert gas, there will be no shift if the number of moles of gas on both sides is the same.
• Changes to concentration of reactants or products will cause the system to shift in a way that counteracts the change.

Understanding these effects is crucial for predicting and controlling chemical reactions in laboratory and industrial settings.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time.
In the context of the decomposition of baking soda, the balanced chemical equation helps determine when equilibrium has been reached. For solids and pure liquids, their concentrations remain constant and do not appear in the equilibrium constant expressions.Equilibrium can be expressed using the equilibrium constant, such as the equilibrium constant for partial pressures, denoted as \(K_p\).
Factors that can influence equilibrium include:

• Temperature: Changing the temperature can favor the endothermic or exothermic direction.
• Pressure: Changes can shift the equilibrium position in gas-phase reactions.
• Concentration: Altering the amount of reactants or products affects the equilibrium.

Each of these factors can alter the state of equilibrium, shifting it toward more reactants or products depending on the external changes applied.

Partial Pressure

In gaseous reactions, partial pressure plays a key role in understanding and calculating equilibrium constants.Each gas in a mixture exerts its own pressure, known as its partial pressure, and the sum of these pressures gives the total pressure of the system.
For the baking soda decomposition reaction, both carbon dioxide and water vapor, which are products, contribute to the total pressure in the system.We denote their partial pressures as \(P_{\text{CO}_2}\) and \(P_{\text{H}_2O}\), respectively.Some key points about partial pressure:

• It is proportional to the amount of gas present at constant volume and temperature.
• The total pressure is the sum of all individual partial pressures in a mixture.
• Understanding partial pressures aids in expressing equilibrium constant \(K_p\).

By knowing the total pressure and using stoichiometry, as in the baking soda problem, you can calculate individual partial pressures.

Stoichiometry

Stoichiometry is essential in solving equilibrium problems because it involves using the balanced chemical equation to relate the quantities of reactants and products.In the decomposition of baking soda, stoichiometry helps derive the relationship between the quantities of carbon dioxide and water vapor.
For the balanced reaction, \(2\text{NaHCO}_3\) decomposes to form \(\text{Na}_2\text{CO}_3\), \(\text{CO}_2\), and \(\text{H}_2\text{O}\), with the stoichiometric coefficients indicating the mole ratios.Stoichiometry allows us to calculate partial pressures:

• Identify the stoichiometric coefficients from the balanced equation.
• Use these coefficients to set up equations for the partial pressures of products in terms of a single variable \(x\).
• Solve for this variable to find the partial pressures and ultimately the equilibrium constant.

By applying stoichiometry, you can simplify complex equilibrium scenarios, predict outcomes, and find solutions effectively.