Question

For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ \(K_{\mathrm{p}}=3.5 \times 10^{4}\) at 1495 \(\mathrm{K} .\) What is the value of \(K_{\mathrm{p}}\) for the following reactions at 1495 \(\mathrm{K}\) ? a. \(\operatorname{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\) b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\)

Short answer

The equilibrium constants for the new reactions at 1495 K are: a. \( K_{p_{new}} = 1.69 \times 10^{-3} \) b. \( K_{p_{new}} = 2.86 \times 10^{-5} \) c. \( K_{p_{new}} = 1.87 \times 10^2 \)

Step-by-step solution

Identify the given equilibrium constant and temperature

We are given the initial reaction and its equilibrium constant: \[ \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\] \[ K_p = 3.5 \times 10^4 \] Temperature: \(1495 K\).

Apply the rules for manipulating equilibrium constants to find new constants

There are a few basic rules for manipulating equilibrium constants when changing a reaction: 1. If the reaction is reversed, then the equilibrium constant is the reciprocal of the original constant. 2. If the reaction is multiplied by a factor, then the equilibrium constant is raised to the power of that factor. Now, let's apply these rules to find the equilibrium constants for the new reactions. a. For the first reaction: \[ \mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Br}_{2}(g) \] Compare it to the original reaction. This reaction is the reverse of the original and is multiplied by a factor of 1/2. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = \frac{1}{K_p^{\frac{1}{2}}} \] \[ K_{p_{new}} = \frac{1}{(3.5 \times 10^4)^{\frac{1}{2}}} = 1.69 \times 10^{-3} \] b. For the second reaction: \[ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \] This reaction is the reverse of the original reaction. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = \frac{1}{K_p} \] \[ K_{p_{new}} = \frac{1}{3.5 \times 10^4} = 2.86 \times 10^{-5} \] c. For the third reaction: \[ \frac{1}{2} \mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g) \] This reaction is identical to the original reaction but multiplied by a factor of 1/2. Therefore, the equilibrium constant for this reaction is: \[ K_{p_{new}} = {K_p}^{\frac{1}{2}} \] \[ K_{p_{new}} = (3.5 \times 10^4)^{\frac{1}{2}} = 1.87 \times 10^2 \]

Present the results

The equilibrium constants for the new reactions at 1495 K are: a. \( K_{p_{new}} = 1.69 \times 10^{-3} \) b. \( K_{p_{new}} = 2.86 \times 10^{-5} \) c. \( K_{p_{new}} = 1.87 \times 10^2 \)

Key concepts

chemical reactions

Chemical reactions involve the transformation of one or more substances into different substances. This change is represented by a chemical equation. In the context of this exercise, the chemical equation is \(\mathrm{H}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{HBr}(g)\). Here, hydrogen gas and bromine gas react to form hydrogen bromide gas. These participating substances are called reactants and products, respectively.

• The arrow denotes an equilibrium state, meaning the reaction can proceed in both forward and reverse directions.
• Gases, as denoted by the \(g\) symbol, exhibit unique behaviors compared to solids and liquids, especially under high temperatures like 1495 K.
• Equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction.

Understanding chemical reactions is vital to study equilibrium constants, which quantify the balance between reactants and products at equilibrium.

reaction manipulation

Reaction manipulation refers to the changes made in a chemical reaction's representation to find new equilibrium constants. In the exercise, this involves reversing reactions or changing their stoichiometry by multiplying or dividing by factors.

• Reversing a reaction involves swapping reactants with products, affecting the equilibrium constant by taking its reciprocal.
• Adjusting the stoichiometry involves raising or lowering the equilibrium constant to the power of the factor used in the multiplication or division. For example, if you multiply the stoichiometry by 2, the equilibrium constant is squared.

These manipulations directly affect the mathematical expression of the equilibrium constant. They follow strict chemical rules ensuring that the reaction's thermodynamic properties remain consistent.

thermodynamics

Thermodynamics is the study of heat and energy transfer, and it relates closely to chemical reactions and their constants. It considers how energy is absorbed or released during reactions. The concept is pivotal in understanding why reactions at equilibrium maintain certain proportions of reactants and products.

• The equilibrium constant is a thermodynamic quantity that varies with temperature. At 1495 K in this example, it determines the specific balance.
• Energy changes determine if reactions are exothermic (releasing heat) or endothermic (absorbing heat).
• The direction and extent of a reaction under given conditions can be predicted using thermodynamic principles.

Thus, thermodynamics provides the framework for predicting reaction behaviors and their equilibrium states.

equilibrium principles

Equilibrium principles form the basis of understanding chemical reactions and how to work with equilibrium constants. At equilibrium, the concentrations of reactants and products remain steady over time but are not necessarily equal.

• The principle of dynamic equilibrium suggests that reactions continue to occur in both forward and reverse directions at equal rates.
• Le Châtelier's principle helps predict the effect of changes in conditions (like temperature or pressure) on the equilibrium position.
• The equilibrium constant \(K_p\), specific to gas-phase reactions, is a ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced equation.

These principles guide chemists in manipulating and predicting the behavior of reactions, especially under different conditions. Understanding equilibrium principles is crucial for students mastering chemical reactions and constants.