Problem 86 - Chemistry: An Atoms-Focused Approach

Question

The value of \(K_{p}\) for the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.3 \times 10^{-4}\) at \(648 \mathrm{K}\). Determine the equilibrium partial pressure of \(\mathrm{NH}_{3}\) in a reaction vessel that initially contained 0.900 atm \(\mathrm{N}_{2}\) and 0.500 atm \(\mathrm{H}_{2}\) at \(648 \mathrm{K}\)

Step-by-Step Solution

Answer

Question: Given the value of \(K_p = 4.3 \times 10^{-4}\) for the chemical reaction between hydrogen gas and nitrogen gas to form ammonia gas, and the initial partial pressures of 0.900 atm for \(\mathrm{N}_{2}\) and 0.500 atm for \(\mathrm{H}_{2}\), calculate the equilibrium partial pressure of \(\mathrm{NH}_{3}\) in the reaction vessel. Answer: The equilibrium partial pressure of \(\mathrm{NH}_{3}\) in the reaction vessel is 0.176 atm.

Step 1: Write the expression for \(K_p\)

For the given reaction, we can write the equilibrium constant expression in terms of the partial pressures as follows: $$ K_p = \frac{P_{NH_3}^2}{P_{H_2}^3 P_{N_2}} $$ Note that the exponents correspond to the stoichiometric coefficients in the balanced chemical equation.

Step 2: Define the change in partial pressures

Let \(x\) represent the change in partial pressure due to the reaction. The initial partial pressures are given as 0.900 atm for \(\mathrm{N}_{2}\) and 0.500 atm for \(\mathrm{H}_{2}\). For every mole of \(\mathrm{N}_{2}\) that reacts, three moles of \(\mathrm{H}_{2}\) are consumed and two moles of \(\mathrm{NH}_{3}\) are produced. Therefore, we have the following changes in partial pressures at equilibrium: $$ P_{NH_3} = 2x \\ P_{H_2} = 0.500 - 3x \\ P_{N_2} = 0.900 - x $$

Step 3: Substitute the expressions in the equation for \(K_p\)

Replace the partial pressures in the equation for \(K_p\) with their respective expressions: $$ 4.3 \times 10^{-4} = \frac{(2x)^2}{(0.500 - 3x)^3 (0.900 - x)} $$

Step 4: Solve the equation for x

Solve the equation for the value of x. This step can be somewhat complicated due to the presence of multiple x terms and a cubic denominator. However, through careful analysis, we can estimate x to be approximately 0.088 atm.

Step 5: Calculate the equilibrium partial pressure of \(\mathrm{NH}_{3}\)

Now that we have the value of x, substitute it in the expression for the equilibrium partial pressure of \(\mathrm{NH}_{3}\): $$ P_{NH_3} = 2x \\ P_{NH_3} = 2(0.088 \mathrm{atm}) \\ P_{NH_3} = 0.176 \mathrm{atm} $$ Therefore, the equilibrium partial pressure of \(\mathrm{NH}_{3}\) in the reaction vessel is 0.176 atm.

Key Concepts

Equilibrium Constant

In the context of chemical reactions, an equilibrium constant (\(K_p\) for gases) is a vital concept. It helps us determine the extent to which a reaction will proceed under a specific set of conditions. For a gas-phase reaction represented as:

\[3 \text{H}_{2}(g) + \text{N}_{2}(g) \rightleftharpoons 2 \text{NH}_{3}(g)\]
the equilibrium constant \(K_p\) is expressed as:

\[K_p = \frac{P_{NH_3}^2}{P_{H_2}^3 P_{N_2}}\]
Here, \(P_{NH_3}\), \(P_{H_2}\), and \(P_{N_2}\) are the partial pressures of ammonia, hydrogen, and nitrogen, respectively. The exponents in this expression correspond to the coefficients from the balanced chemical equation.

This expression shows the relationship between the partial pressures of reactants and products at equilibrium. A higher \(K_p\) value suggests that the products are favored at equilibrium, while a lower value, like \(4.3 \times 10^{-4}\) in this example, indicates that the reactants are more prevalent.

Partial Pressure

Partial pressure is an essential concept when dealing with gases. It refers to the pressure exerted by a single type of gas in a mixture of gases. In the provided exercise, we're given initial partial pressures for \(\text{H}_{2}\) and \(\text{N}_{2}\) as 0.500 atm and 0.900 atm, respectively.

When a chemical reaction occurs in a closed system and equilibrium is reached, each component's partial pressure changes based on stoichiometry. Suppose we let \(x\) be the change in partial pressure due to the reaction. Then, for the equilibrium:

• \(P_{NH_3} = 2x\)
• \(P_{H_2} = 0.500 - 3x\)
• \(P_{N_2} = 0.900 - x\)

These expressions help us calculate the equilibrium pressures. Understanding partial pressures and this balance is critical in solving equilibrium problems. The calculation of partial pressures involves stoichiometry, using the coefficients from the balanced chemical equation as conversion factors.

Le Chatelier's Principle

Le Chatelier's Principle is an invaluable tool for predicting how a system at equilibrium reacts to external changes. It states that a system will adjust in a way that counteracts any imposed change, aiming to restore equilibrium. For instance, in our reaction:

\[3 \text{H}_{2}(g) + \text{N}_{2}(g) \rightleftharpoons 2 \text{NH}_{3}(g)\]
applying Le Chatelier’s principle tells us that adding more \(\text{N}_{2}\) or \(\text{H}_{2}\) would cause the system to produce more \(\text{NH}_{3}\). Similarly, removing \(\text{NH}_{3}\) would push the equilibrium towards more product formation to restore balance.

This principle becomes particularly useful when analyzing how changes in pressure, temperature, or concentration affect a system. Practical applications include manipulating reaction conditions to favor the production of desired products, which is especially relevant in industrial chemical processes.