Problem 81 - Chemistry The Central Science In Si Units

Question

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C}\). The following reaction takes place: $$ \mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) taken together) is \(62.21 \mathrm{kPa}\). What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The value of \( K_p \) is \( 967.51 \text{ kPa}^2 \).

Step 1: Understand Kp Expression

For the given equilibrium reaction \( \mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{H}_{2} \mathrm{~S}(g) \), the equilibrium constant in terms of pressure, \( K_p \), is expressed as \( K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2S} \). Solid substances do not appear in the equilibrium expression, so only the gases \( \mathrm{NH}_3 \) and \( \mathrm{H}_2 \mathrm{~S} \) are included.

Step 2: Calculate Partial Pressures

At equilibrium, the total pressure of the gases \( \mathrm{NH}_3 \) and \( \mathrm{H}_2 \mathrm{~S} \) is given as \( 62.21 \) kPa. Assuming the reaction produces these gases in a 1:1 molar ratio, let \( x \) represent the partial pressure of each gas. Thus, \( 2x = 62.21 \text{ kPa} \), leading to \( x = 31.105 \text{ kPa} \) for each gas.

Step 3: Substitute to Find Kp

Now substitute the calculated partial pressures into the expression for \( K_p \): \[ K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2S} = 31.105 \text{ kPa} \times 31.105 \text{ kPa} = 967.51 \text{ kPa}^2. \]

Step 4: Conclusion and Verification

The \( K_p \) value at \( 24^{\circ} \text{C} \) is \( 967.51 \text{ kPa}^2 \). Verify if the math aligns logically based on initial assumptions about the reaction and gas behavior, validating the calculation.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that predicts how a change in conditions can affect chemical equilibrium. When a system at equilibrium experiences a change in concentration, temperature, or pressure, the system adjusts in such a way as to counteract the change and restore a new equilibrium. This principle is often used to understand and predict the direction of reaction shifts.

• If the pressure increases due to a shift in volume, the system will favor the side of the reaction with fewer moles of gas.
• For instance, in our equilibrium, the addition of gas pressure in the system might cause the reaction to shift to the left where there's one solid rather than two gas molecules.
• If you increase the concentration of reactants or products, the equilibrium will shift to use up the added substances.

This principle, while not directly affecting the calculation of equilibrium constants, provides important insights into how and why a system might move away from or maintain its equilibrium position under various conditions.

Partial Pressure

Partial pressure is a vital concept for understanding gas mixtures in equilibrium. The partial pressure of a gas in a mixture is the pressure it would exert if it were alone in the container at the same temperature. In the equilibrium reaction involving \( \mathrm{NH}_4 \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2 \mathrm{~S}(g) \), each gas contributes to the total pressure proportionally.

• For a reaction at equilibrium, the sum of the partial pressures of the gases will equal the total pressure of the system.
• In the given example, with a total pressure of 62.21 kPa from both gases, assuming a 1:1 molar production, each gas has a partial pressure of 31.105 kPa.
• These partial pressures are crucial for calculating the equilibrium constant \( K_p \).

By understanding partial pressures, one can decipher the dynamics of gaseous systems and predict how gases might behave under altered conditions.

Kp Calculation

Calculating \( K_p \), the equilibrium constant in terms of partial pressures, requires a clear understanding of how the pressures of each gas relate to the overall reaction. For the given equilibrium reaction \( \mathrm{NH}_4 \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_3(g) + \mathrm{H}_2 \mathrm{~S}(g) \), recall:

• The equilibrium constant \( K_p \) is the product of the partial pressures of the gases raised to the power of their stoichiometric coefficients in the balanced equation.
• Here, \( K_p = P_{\mathrm{NH}_3} \times P_{\mathrm{H}_2 S} \), since both gases are balanced as \(1:1\) in the equation.
• Given each gas has a partial pressure of 31.105 kPa, substituting these into the formula gives \( K_p = 31.105 \text{ kPa} \times 31.105 \text{ kPa} = 967.51 \text{ kPa}^2 \).

Understanding this step ensures clarity in processing these calculations and is crucial for anyone dealing with reactions involving gaseous equilibria.