Problem 27 - Chemistry

Question

At a given temperature, \(K=1.3 \times 10^{-2}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate values of \(K\) for the following reactions at this temperature. a. \(\frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\) b. \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) c. \(\mathrm{NH}_{3}(g) \rightleftharpoons \frac{1}{2} \mathrm{N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) d. \(2 \mathrm{N}_{2}(g)+6 \mathrm{H}_{2}(g) \rightleftharpoons 4 \mathrm{NH}_{3}(g)\)

Step-by-Step Solution

Answer

For the given reactions at this temperature, the equilibrium constants are: a. \(K_a \approx 0.114\) b. \(K_b \approx 76.9\) c. \(K_c \approx 8.67\) d. \(K_d \approx 1.69 \times 10^{-4}\)

Step 1: Recall the basic properties of manipulating equilibrium constants

When manipulating a chemical reaction to create a new reaction, the equilibrium constant is also affected according to: 1. If the reaction is reversed, take the reciprocal of the equilibrium constant. 2. If the whole equation is multiplied or divided by a constant, raise the equilibrium constant to the power of that constant. We'll use these properties for calculating the values of \(K\) for each of the given reactions.

Step 2: Calculate the equilibrium constant for reaction a

For reaction a, the initial reaction is divided by 2: $$\frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g) \rightleftharpoons NH_{3}(g)$$ Since the coefficients are multiplied by 1/2, we'll need to raise the initial equilibrium constant to the power of 1/2: $$K_a = \left(K\right)^\frac{1}{2} = \left(1.3 \times 10^{-2}\right)^\frac{1}{2} \approx 0.114$$

Step 3: Calculate the equilibrium constant for reaction b

For reaction b, the initial reaction is reversed: $$2 NH_{3}(g) \rightleftharpoons N_{2}(g) + 3 H_{2}(g)$$ Since the reaction is reversed, take the reciprocal of the initial equilibrium constant: $$K_b = \frac{1}{K} = \frac{1}{1.3 \times 10^{-2}} \approx 76.9$$

Step 4: Calculate the equilibrium constant for reaction c

For reaction c, the initial reaction is divided by 2 and then reversed: $$NH_{3}(g) \rightleftharpoons \frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g)$$ To find the equilibrium constant for this reaction, first raise the initial \(K\) to the power of 1/2, then take the reciprocal: $$K_c = \frac{1}{\left(K\right)^\frac{1}{2}} = \frac{1}{\left(1.3 \times 10^{-2}\right)^\frac{1}{2}} \approx 8.67$$

Step 5: Calculate the equilibrium constant for reaction d

For reaction d, the initial reaction is multiplied by 2: $$2 N_{2}(g) + 6 H_{2}(g) \rightleftharpoons 4 NH_{3}(g)$$ Since the coefficients are multiplied by 2, raise the initial equilibrium constant to the power of 2: $$K_d = \left(K\right)^2 = \left(1.3 \times 10^{-2}\right)^2 \approx 1.69 \times 10^{-4}$$ Finally, we have the equilibrium constants for all the given reactions: a. \(K_a \approx 0.114\) b. \(K_b \approx 76.9\) c. \(K_c \approx 8.67\) d. \(K_d \approx 1.69 \times 10^{-4}\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs in a chemical reaction when the rates of the forward and reverse reactions are equal. This means that the concentration of reactants and products remains constant over time. It doesn't mean the reactants and products are equal in concentration, but rather, their formation rates are balanced.
This concept is crucial in understanding how reactions reach a state where no net change is observed, even though molecular activities continue.
In a system at equilibrium, various factors like concentration, pressure, and temperature don't disturb the balance unless acted upon externally. This stability is represented by the equilibrium constant, symbolized as \( K \).
\( K \) provides insight into the ratio of product concentrations to reactant concentrations, raised to their respective stoichiometric coefficients, at equilibrium. Therefore, a larger \( K \) indicates a product-favored reaction, while a smaller \( K \) indicates a reactant-favored one.

Manipulating Equilibrium Constants

Understanding how to manipulate equilibrium constants is essential when dealing with chemical reactions. This manipulation involves changing the way a reaction is written to reflect different stoichiometries.

• If you reverse a chemical equation, you take the reciprocal of the equilibrium constant.
• If you multiply or divide the whole equation by a constant, you raise the equilibrium constant to the power of that constant.

These principles allow us to derive new equilibrium constants from known ones for the same reactions under different conditions. This manipulation doesn't change the nature of the reaction but only evaluates it under a new perspective.
Reactions can be scaled up or down in magnitude, but their intrinsic properties, captured by the equilibrium constant, adhere to precise mathematical relationships.

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a value that can be calculated in the same way as the equilibrium constant \( K \), but for non-equilibrium conditions.
However, while \( K \) is specific to equilibrium states, \( Q \) can describe the ratio of product to reactant concentrations (or pressures) at any point in time during the reaction. By comparing \( Q \) with \( K \), we can predict the direction in which a reaction will proceed to reach equilibrium.

• \( Q < K \): The reaction will move forward, producing more products.
• \( Q > K \): The reaction will move in reverse, producing more reactants.
• \( Q = K \): The reaction is at equilibrium.

Using \( Q \) helps us understand the current state of a reaction system and infer how to achieve equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle describes how a chemical system at equilibrium responds to external changes. When a system at equilibrium is disturbed by a change in concentration, temperature, or pressure, the principle predicts that the system will adjust to minimize that disturbance.
According to Le Chatelier's principle:

• If the concentration of a reactant or product is changed, the system will shift to counteract the change and restore equilibrium.
• If the pressure is changed, the system will shift towards the side with fewer moles of gas to counter the pressure change.
• If the temperature changes, the equilibrium will shift in the direction that absorbs the added heat (endothermic direction) or releases heat (exothermic direction).

This principle is a powerful tool for predicting how reactions will shift in response to changes and forms the basis of many chemical processes in industries and natural systems.