Problem 20 - Chemistry: Principles and Reactions

Question

Given the following data at \(25^{\circ} \mathrm{C}\), $$\begin{array}{cl}2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array}$$ calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The equilibrium constant (K) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\) is \(8 \times 10^{-14}\).

Step 1: Write down the required reaction

First, we need to determine the reaction for the formation of one mole of NOBr from its elements in the gaseous state. This reaction can be written as: $$\mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$

Step 2: Manipulate the given reactions to obtain the required reaction

We want to manipulate the given reactions in such a way that we can sum them up to get the required reaction. Our goal is to get the two reactions to look like the following: $$\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ and $$\mathrm{N}_{2}(g) + 2 \mathrm{NO}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ To do this, for the first given reaction, we simply need to multiply the reaction by \(\frac{1}{2}\), so: $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g)$$ becomes $$\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ The second given reaction doesn't need any manipulation, as it already is in the desired form.

Step 3: Manipulate the given equilibrium constants

When we change the stoichiometry of the reactions, we also need to change the equilibrium constants accordingly. For the first reaction, we have multiplied all the stoichiometric coefficients by \(\frac{1}{2}\), so we need to raise its equilibrium constant to the power of \(\frac{1}{2}\). For the first reaction, $$K_1 = 1 \times 10^{-30}$$ $$K_1^{'} = {K_1}^{\frac{1}{2}} = \left( 1 \times 10^{-30} \right)^{\frac{1}{2}} = 1 \times 10^{-15}$$ For the second reaction, we have not changed its stoichiometry, so its equilibrium constant remains the same: $$K_2 = 8 \times 10^1$$

Step 4: Combine the manipulated reactions and equilibrium constants

We now sum up the manipulated reactions to get the required reaction: $$\frac{1}{2} \mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ $$2 \mathrm{NO}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ $$\mathrm{N}_{2}(g) + \mathrm{O}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ Next, we multiply the manipulated equilibrium constants to obtain the equilibrium constant for the required reaction: $$K = K_1^{'} \times K_2 = (1 \times 10^{-15}) \times (8 \times 10^1) = 8 \times 10^{-14}$$ Hence, the equilibrium constant \(K\) for the formation of one mole of NOBr from its elements in the gaseous state at \(25^{\circ} \mathrm{C}\) is \(8 \times 10^{-14}\).

Key Concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol K, is fundamental in understanding chemical reactions that can occur in both forward and reverse directions, ultimately reaching a state of balance. In such a state, the concentrations of reactants and products remain constant over time. The equilibrium constant quantifies the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation.

In the context of the given problem, the equilibrium constant for each reaction was used to determine the overall equilibrium constant for the formation of one mole of NOBr from its elements. This involved manipulating known equilibrium constants based on adjustments to the reaction stoichiometry, illustrating the mathematical relationship between changes to the reaction equation and the corresponding equilibrium constant value. The final calculation, combining individual constants, gave a quantifiable measure of the reaction's tendency to proceed under specific conditions, reflecting the dynamic balance between all involved species.

Reaction Stoichiometry

In chemistry, reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. It is crucial for understanding how much of a substance is consumed or produced and for predicting the outcomes of reactions. The balanced chemical equation provides the stoichiometric coefficients that represent the molar ratios of each reactant and product.

When solving the problem of finding the equilibrium constant for a new reaction constructed from known reactions, we used the stoichiometry of the initial reactions to manipulate them into the desired form. We then adjusted the equilibrium constants according to the stoichiometric changes we made. In essence, reaction stoichiometry acts as a bridge that allows us to accurately connect different chemical equations and their equilibrium constants, showcasing the precise and calculable nature of chemical reactions.

Le Chatelier's Principle

Le Chatelier's Principle is a qualitative tool that helps predict how a system at equilibrium reacts to external changes such as concentration, temperature, and pressure. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. It’s a way for the system to re-establish balance.

For students seeking a deeper understanding, it's essential to recognize that while Le Chatelier's principle does not directly impact the mathematical calculation of equilibrium constants, it provides insight into the directional shifts of reactions when subjected to different stresses. Each change in conditions can lead to a predictable shift in the system, which is why mastering this principle is invaluable for both conceptual comprehension and practical application in chemistry.