Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider the following reactions: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad\) and \(\quad \mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \rightleftharpoons 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

Short Answer

Expert verified
In summary, the two property differences between these reactions related to equilibrium are: 1. Differences in their equilibrium constant expressions due to the state of iodine (gaseous in the first reaction and solid in the second reaction). Specifically, \(K_1 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\) for the first reaction, and \(K_2 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]}\) for the second reaction. 2. Differences in the effect of pressure or volume changes due to the different participation of gaseous and solid species in the reactions, according to Le Chatelier's principle. The first reaction will be more sensitive to pressure and volume changes since all species involved are gases, while the second reaction will be less affected by these changes due to the presence of solid iodine.

Step by step solution

01

1. Reaction Quotient (Q) & Equilibrium Constant (K)

The reaction quotient (Q) is defined as the ratio of the concentrations of the products raised to their stoichiometric coefficients to the concentrations of the reactants raised to their stoichiometric coefficients. For the first reaction, the reaction quotient is given by: \[Q_1 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\] At equilibrium, the reaction quotient (Q) becomes the equilibrium constant (K). Therefore, the equilibrium constant for the first reaction is: \[K_1 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\] For the second reaction, the equilibrium expression is: \[Q_2 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]}\] Since the concentration of solid iodine (I2) does not affect the equilibrium, it is omitted from the equation. At equilibrium, the reaction quotient becomes the equilibrium constant: \[K_2 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]}\] Notice that the equilibrium constant expressions for the two reactions are different.
02

2. Le Chatelier's Principle

Le Chatelier's principle states that if a change in conditions is imposed on a system at equilibrium, the system will adjust itself in a way to counteract the change and restore equilibrium. For the first reaction, changes in pressure or volume will affect the equilibrium since all species involved are gases: \[\mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)\] If the pressure increases or the volume decreases, the system will shift towards the side with fewer moles of gas, which is the side with the products (HI). Conversely, if the pressure decreases or the volume increases, the system will shift towards the side with more moles of gas, which is the side with the reactants (H2 and I2). For the second reaction, changes in pressure or volume will have no effect on the position of solid iodine in the equilibrium: \[\mathrm{H}_2(g) + \mathrm{I}_2(s) \rightleftharpoons 2 \mathrm{HI}(g)\] Since the concentration of solid iodine (I2) does not change with pressure or volume, only the gaseous reactant and product will be affected. Therefore, Le Chatelier's principle will have a different impact on the second reaction compared to the first one. In summary, the two property differences between these reactions related to equilibrium are: 1. Differences in their equilibrium constant expressions due to the state of iodine (gaseous in the first reaction and solid in the second reaction). 2. Differences in the effect of pressure or volume changes due to the different participation of gaseous and solid species in the reactions, according to Le Chatelier's principle.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Quotient
The reaction quotient, often represented by the letter \( Q \), is a valuable tool for assessing the progress of a chemical reaction. It is calculated by taking the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients. For example, in the reaction \( \mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \), the reaction quotient \( Q_1 \) is expressed as:
  • \( Q_1 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \)
At any given moment in time during the reaction, this expression tells us the current state of the reaction. By comparing \( Q \) with the equilibrium constant \( K \), one can determine the direction in which the reaction will proceed to achieve equilibrium.
Notice, in the second reaction, iodine is solid, so it does not appear in \( Q_2 \). Solids have a constant concentration that does not change as the reaction progresses, so:
  • \( Q_2 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]} \)
This omission leads to a different interpretation for reactions involving solids, highlighting the importance of understanding the physical state of reactants and products in equilibrium calculations.
Equilibrium Constant
The equilibrium constant \( K \) is a fundamental aspect of chemical equilibrium, representing the state of balance in a chemical reaction where the forward and backward reactions occur at the same rate. It is defined based on the concentration of products and reactants at equilibrium similar to the reaction quotient but is unique to a reaction's temperature. For our reactions:
  • First Reaction: \( K_1 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]} \)
  • Second Reaction: \( K_2 = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]} \)
The difference between \( K_1 \) and \( K_2 \) lies in the presence of solid iodine in the second reaction, which does not influence the equilibrium expression. This reflects the critical rule in equilibrium calculations: only the concentrations of species in the gaseous or aqueous state are included.
Knowing \( K \) helps to predict the extent of a reaction; a large \( K \) means a reaction heavily favors products, while a small \( K \) signifies a reaction favoring reactants. It's key to remember \( K \) can change with temperature, but is constant for a given reaction at a given temperature.
Le Chatelier's Principle
Le Chatelier's Principle provides a system for predicting the effect of changes in concentration, pressure, or temperature on a chemical equilibrium. This principle states that if a system at equilibrium experiences a disturbance, it will adjust in a way that tends to counteract that change.
For the reaction \( \mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \), adding pressure by decreasing volume shifts the equilibrium toward fewer moles of gas—toward the products. Conversely, reducing pressure shifts it toward the reactants.
However, the second reaction, \( \mathrm{H}_2(g) + \mathrm{I}_2(s) \rightleftharpoons 2 \mathrm{HI}(g) \), behaves differently because solid iodine does not participate in pressure changes. This is crucial when predicting the behavior of reactions involving solids versus gases under changing conditions. Each type of reactant reacts differently to increases or decreases in pressure, reflecting the nuanced applications of Le Chatelier’s Principle.
Gaseous vs Solid Reactants
Understanding the distinction between gaseous and solid reactants in equilibrium is essential. Gases and solutions are dynamic participants in chemical reactions because their concentrations can change, influencing the equilibrium state. For gases, every molecule can potentially affect the equilibrium expression, as seen in:
  • \( \mathrm{H}_2(g) + \mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) \)
Solids, however, like iodine in the reaction \( \mathrm{H}_2(g) + \mathrm{I}_2(s) \rightleftharpoons 2 \mathrm{HI}(g) \), maintain a constant state under set conditions and do not appear in the equilibrium expression. This means changes to the system such as pressure shifts will impact the equilibrium involving gases but leave solid phases largely unaffected.
This characteristic is critical for formulating forecasts about how reactions proceed in different states, leading to more accurate predictions for chemical systems in real-world scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O}\) , and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O}\) , and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\) .

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. 1.0 \(\mathrm{g} \mathrm{H}_{2} \mathrm{O}\) and 2.0 \(\mathrm{g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a 1.0 -L flask. b. 1.0 mole of pure HOCl is placed in a 2.0 \(\mathrm{L}\) flask.

In a given experiment, 5.2 moles of pure NOCl were placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$ a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that is present at equilibrium. b. Calculate the equilibrium concentrations for all species.

The gas arsine, \(\mathrm{AsH}_{3},\) decomposes as follows: $$2 \mathrm{AsH}_{3}(g) \rightleftharpoons 2 \mathrm{As}(s)+3 \mathrm{H}_{2}(g)$$ In an experiment at a certain temperature, pure \(\mathrm{AsH}_{3}(g)\) was placed in an empty, rigid, sealed flask at a pressure of 392.0 torr. After 48 hours the pressure in the flask was observed to be constant at 488.0 torr. a. Calculate the equilibrium pressure of \(\mathrm{H}_{2}(g)\) b. Calculate \(K_{\mathrm{p}}\) for this reaction.

Hydrogen for use in ammonia production is produced by the reaction $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \frac{\text { Nicatalyst }}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free