Problem 33 - Chemistry The Molecular Science

Question

The value of \(K_{\mathrm{c}}\) for the reaction $$\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ is 2.00 at \(400,{ }^{\circ} \mathrm{C}\). Determine the value of \(K_{\mathrm{p}}\) for this reaction at this temperature using bars as pressure units.

Step-by-Step Solution

Answer

The value of \(K_p\) is approximately 0.000638.

Step 1: Identify the Reaction and Expression for Equilibrium Constants

The balanced chemical reaction is \[ \mathrm{N}_{2}(\mathrm{~g}) + 3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \]The equilibrium constant for concentration \(K_c\) is given, and we need to find the equilibrium constant for pressure \(K_p\).

Step 2: Determine Change in Moles (Δn)

Identify the change in moles of gas from reactants to products. For this reaction:- Moles of reactants: \(1 + 3 = 4\)- Moles of products: \(2\)Thus, \(\Delta n = 2 - 4 = -2\).

Step 3: Use the Relationship Between Kc and Kp

The relationship between \(K_c\) and \(K_p\) is given by:\[ K_p = K_c (RT)^{\Delta n} \]where \(R = 0.08314 \frac{L \, bar}{K \, mol}\) is the ideal gas constant and \(T\) is the temperature in Kelvin. First, convert the temperature 400°C to Kelvin: \(T = 400 + 273.15 = 673.15 \text{ K}\).

Step 4: Calculate Kp

Substitute the values into the equation:\[ K_p = 2.00 \times (0.08314 \times 673.15)^{-2} \]Calculate the numerical value:\[ 0.08314 \times 673.15 = 55.98211 \]\[ K_p = 2.00 \times (55.98211)^{-2} \approx 2.00 \times 0.0003189 \approx 0.0006378 \]Therefore, \(K_p \approx 0.000638 \).

Key Concepts

Kc vs Kp

In chemical equilibrium, we often deal with two types of equilibrium constants: the concentration constant, denoted as \(K_c\), and the pressure constant, denoted as \(K_p\). These constants help us understand the extent of a reaction at equilibrium, but they are used in different contexts.

• \(K_c\) is used when dealing with concentrations of substances in a reaction, typically in moles per liter (mol/L). It applies to reactions where the components are in a liquid or gaseous state, but concentration measurement is preferred.
• \(K_p\) is used for gas-phase reactions where pressures are more convenient to measure than concentrations. It is expressed in units of pressure such as atm or bar.

The relationship between \(K_c\) and \(K_p\) is derived from the ideal gas law and is given by the equation: \[ K_p = K_c (RT)^{\Delta n} \]Where:

• \(R\) is the ideal gas constant.
• \(T\) is the temperature in Kelvin.
• \(\Delta n\) is the change in moles of gas, determined by subtracting the number of moles of reactants from moles of products.

This equation shows how \(K_c\) can be converted to \(K_p\) when you know the temperature and the change in the number of moles.

Ideal Gas Constant

The ideal gas constant \(R\) is a crucial element in many gas-related equations, including the conversion between \(K_c\) and \(K_p\). It's value can vary depending on the units used for pressure and volume.

• For the exercise at hand, using units of bar and liter, \(R\) is 0.08314 \(\frac{L \, bar}{K \, mol}\).
• In other unit systems, such as if pressure is measured in atm, \(R\) would be 0.08206 \(\frac{L \, atm}{K \, mol}\).

The ideal gas constant helps relate the conditions of pressure, volume, and temperature in a system, which is essential in applying gas laws effectively. Its role in chemical equilibrium equations illustrates how it links different states and types of data (concentration versus pressure) harmoniously.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of the reactants and products remain constant over time.
Key features of chemical equilibrium include:

• The system must be closed with no exchange of matter with the surroundings.
• It can be approached from either direction; starting with reactants or products will eventually lead to equilibrium.
• The position of equilibrium can be affected by changes in temperature, pressure, and concentration, as dictated by Le Chatelier's principle.

In the context of our exercise, understanding equilibrium is essential to discerning why specific constants like \(K_c\) and \(K_p\) are used and how they define the behavior of the reaction \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\). This balanced equation helps to determine how concentrations and pressures shift until they reach equilibrium proportions.

Gas Laws in Chemistry

Gas laws describe the behavior of gases in different conditions and form the basis of relating pressure, volume, temperature, and moles through the ideal gas law and other similar laws.

• Boyle's law considers the inverse relationship between pressure and volume when temperature is constant.
• Charles' law shows the direct tie between volume and temperature with constant pressure.
• Avogadro's law connects the volume of gas with the amount of substance (in moles), maintaining constant temperature and pressure.

For most reactions that involve gases, especially those in equilibrium like the synthesis of ammonia (\(\text{NH}_3\)), these laws underpin the link between \(K_c\) and \(K_p\) by explaining how pressure and volume changes are interrelated to concentrations. These relationships are pivotal when determining conditions at equilibrium in gas-phase reactions.