Problem 23 - Chemistry The Central Science In Si Units

Question

The equilibrium constant for the reaction $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K} .(\mathbf{a})\) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\). (c) Calculate \(K_{c}\) for \(\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\).

Step-by-Step Solution

Answer

(a) Equilibrium favors NO and Br2. (b) \(K_c = 76.923\). (c) \(K_c = 8.771\).

Step 1: Analyzing the Equilibrium Constant

First, let's determine which side of the reaction the equilibrium favors based on the provided equilibrium constant, \(K_c = 1.3 \times 10^{-2}\). An equilibrium constant much smaller than 1 implies that, at equilibrium, the concentration of the reactants (\(\text{NO}\) and \(\text{Br}_2\)) is significantly higher than that of the products (\(\text{NOBr}\)). Therefore, this reaction at equilibrium favors the formation of reactants \(\text{NO}\) and \(\text{Br}_2\).

Step 2: Calculate Reverse Reaction Equilibrium Constant

The reverse reaction of \(2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\) is \(2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g)\). To find \(K_c\) for the reverse reaction, we take the reciprocal of the original \(K_c\):\[ K_{c, \text{reverse}} = \frac{1}{K_c} = \frac{1}{1.3 \times 10^{-2}} \approx 76.923 \]

Step 3: Calculate Half Reaction Equilibrium Constant

For the reaction \(\text{NOBr}(g) \rightleftharpoons \text{NO}(g) + \frac{1}{2} \text{Br}_2(g)\), this is effectively half the stoichiometry of the original reverse reaction. Since equilibrium constants for gas phase reactions are influenced by stoichiometry, if we take the square root of the \(K_c\) for the reaction \(2 \text{NOBr}(g) \rightleftharpoons 2 \text{NO}(g) + \text{Br}_2(g)\), we adjust it accordingly:\[ K_{c, \text{half}} = (K_{c, \text{reverse}})^{1/2} = (76.923)^{1/2} \approx 8.771 \]

Key Concepts

reaction stoichiometry

When dealing with chemical reactions, it is vital to understand the concept of reaction stoichiometry. Stoichiometry refers to the calculation of reactants and products in chemical reactions. It helps us interpret the quantitative relationships between different compounds involved in a reaction.

For example, consider the reaction: \[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\]In this reaction, the coefficients tell us that two moles of nitrogen monoxide (NO) react with one mole of bromine (Br₂) to produce two moles of nitrosyl bromide (NOBr). These coefficients are essential in converting moles of one substance to moles of another, allowing us to predict how much product will form or how much reactant is needed.

• Stoichiometry provides a balanced view of the reaction.
• Helps in calculating the equilibrium concentrations.
• Informs how to adjust reactants to achieve a desired yield of products.

Understanding stoichiometry is crucial as it ensures that our predictions of reaction outcomes are accurate.

chemical equilibrium

Chemical equilibrium describes the state of a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, concentrations of reactants and products remain constant over time, although they are not necessarily equal.

Let's evaluate the aforementioned reaction equilibrium:\[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\]The equilibrium constant, \(K_c\), for this reaction at 1000 K is given as \(1.3 \times 10^{-2}\). A small \(K_c\) value, much less than 1, indicates that the equilibrium position favors the reactants \(\text{NO}\) and \(\text{Br}_2\) rather than the product \(\text{NOBr}\). This tells us that, under these conditions, more reactant molecules are present at equilibrium compared to products.

• Equilibrium can be perturbed by changes in concentration, temperature, or pressure.
• Le Chatelier's Principle helps predict how the system will respond to these changes.
• The equilibrium constant remains the same for a given reaction at a specific temperature.

In summary, understanding chemical equilibrium helps in manipulating and optimizing reactions in chemical processes.

reversible reactions

A reversible reaction is one where the reactants form products, which can also revert back to reactants under certain conditions. The concept is crucial in chemical reactions as many processes do not go to completion but reach an equilibrium.

For instance, consider the reversible reaction:\[2 \text{NO}(g) + \text{Br}_2(g) \rightleftharpoons 2 \text{NOBr}(g)\]This reaction is reversible, meaning NO and Br₂ can react to form NOBr, and NOBr can decompose back into NO and Br₂. This nature of reversibility is why we observe chemical equilibrium, with the possibility of shifting the equilibrium position under different conditions.

• Reversible reactions are common in both industrial and laboratory chemical processes.
• They allow control over product yield by adjusting reaction conditions.
• Understanding reversibility is essential for optimizing reaction conditions and improving efficiency.

Ultimately, the concept of reversible reactions lays the groundwork for interpreting dynamic equilibria in chemical systems.