/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Write equilibrium-constant expre... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write equilibrium-constant expressions \(K_{c}\) for each of the following reactions. a. \(\mathrm{N}_{2} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{NO}(g)\) b. \(2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\) c. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) d. \(\mathrm{PCl}_{3}(g)+3 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}(g)+3 \mathrm{HCl}(g)\)

Short Answer

Expert verified
a. \( K_c = \frac{[\mathrm{NO}_2][\mathrm{NO}]}{[\mathrm{N}_2\mathrm{O}_3]} \); b. \( K_c = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2\mathrm{S}]^2} \); c. \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{NO}]^2[\mathrm{O}_2]} \); d. \( K_c = \frac{[\mathrm{P}(\mathrm{NH}_2)_3][\mathrm{HCl}]^3}{[\mathrm{PCl}_3][\mathrm{NH}_3]^3} \).

Step by step solution

01

Identify the Reaction for Part (a)

The reaction given is \( \mathrm{N}_{2} \mathrm{O}_{3}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{NO}(g) \). Identify the products and reactants to write the equilibrium constant expression.
02

Write the Expression for Part (a)

The equilibrium constant expression \( K_{c} \) is given by the ratio of the concentrations of the products raised to their coefficients to the concentrations of the reactants raised to their coefficients: \[ K_{c} = \frac{[\mathrm{NO}_{2}][\mathrm{NO}]}{[\mathrm{N}_{2} \mathrm{O}_{3}]} \].
03

Identify the Reaction for Part (b)

The reaction given is \( 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \). Identify the products and reactants to construct the equilibrium constant expression.
04

Write the Expression for Part (b)

The equilibrium constant expression \( K_{c} \) is \[ K_{c} = \frac{[\mathrm{H}_{2}]^2[\mathrm{S}_{2}]}{[\mathrm{H}_{2} \mathrm{~S}]^2} \], using the stoichiometric coefficients as exponents.
05

Identify the Reaction for Part (c)

The reaction is \( 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \). Identify the products and reactants to formulate the equilibrium constant.
06

Write the Expression for Part (c)

The equilibrium constant expression \( K_{c} \) is \[ K_{c} = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]} \]. This accounts for the stoichiometric coefficients.
07

Identify the Reaction for Part (d)

The chemical equation provided is \( \mathrm{PCl}_{3}(g)+3 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}(g)+3 \mathrm{HCl}(g) \). Note the coefficients of each molecule.
08

Write the Expression for Part (d)

The equilibrium constant expression \( K_{c} \) for this reaction is \[ K_{c} = \frac{[\mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}][\mathrm{HCl}]^3}{[\mathrm{PCl}_{3}][\mathrm{NH}_{3}]^3} \], considering the stoichiometry.

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 91Ó°ÊÓ!

Key Concepts

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

Chemical Reactions
Chemical reactions involve the transformation of substances called reactants into different substances known as products. During this process, the atoms in the reactants are rearranged to form new chemical bonds, resulting in the products. Understanding these reactions is crucial as they underpin many processes in chemistry and are fundamental to the formation of new materials.
The general form of a chemical reaction is:
  • Reactants → Products
Here, reactants are the starting substances that go through a chemical change. Products are the substances formed as a result of the reaction. In a chemical equation, reactants are typically written on the left side, while products are written on the right side, separated by an arrow.
Balanced chemical equations are essential for representing reactions accurately. This means the number of each type of atom on the reactant side must equal the number of that atom on the product side, adhering to the law of conservation of mass. Balancing a reaction ensures that the mass and charge are equal on both sides of the equation.
Chemical Equilibrium
Chemical equilibrium occurs in reversible reactions where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, although the reactions are still proceeding. This balanced state is known as dynamic equilibrium.
The equilibrium constant, denoted as \( K_c \), is a crucial concept that helps in understanding chemical equilibrium. It quantifies the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation:
  • \( K_c = \frac{[Products]}{[Reactants]} \)
This expression varies depending on the reaction and provides a mathematical way to represent the position of equilibrium.
If \( K_c \) is large, the position of equilibrium favors the products; if small, it favors the reactants. However, it is important to remember that equilibrium constants do not provide information about reaction rate or kinetics, only the position of equilibrium.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It relies on the mole concept and balanced chemical equations to determine these relationships.
The coefficients in a balanced chemical equation represent the stoichiometric ratios of the reactants and products. This is essential for identifying the amounts needed or produced in a reaction:
  • For example, in the reaction \( 2 \, H_2 + O_2 \rightarrow 2 \, H_2O \), the coefficients (2, 1, and 2) dictate the ratio in which the substances react and form.
Understanding stoichiometry is important for calculating the equilibrium constant as it involves using these ratios to express the concentration terms correctly. When writing the equilibrium expression, these coefficients become the exponents in the equation, reflecting the stoichiometric quantities of each substance involved in the reaction.
Stoichiometry helps in predicting how much product can be formed from given reactants and is used extensively in chemical engineering, manufacturing, and laboratory settings.

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

Gaseous acetic acid molecules have a certain tendency to form dimers. (A dimer is a molecule formed by the association of two identical, simpler molecules.) The equilibrium constant \(K_{p}\) at \(25^{\circ} \mathrm{C}\) for this reaction is \(1.3 \times 10^{3}\). a. If the initial pressure of \(\mathrm{CH}_{3} \mathrm{COOH}\) monomer (the simpler molecule) is \(7.5 \times 10^{-3}\) atm, what are the pressures of monomer and dimer when the system comes to equilibrium? (The simpler quadratic equation is obtained by assuming that all of the acid molecules have dimerized and then some of it dissociates to monomer.) b. Why do acetic acid molecules dimerize? What type of structure would you draw for the dimer? C. As the temperature decreases, would you expect the percentage of dimer to increase or decrease? Why?

Ammonium hydrogen sulfide, \(\mathrm{NH}_{4} \mathrm{HS}\), is unstable at room temperature and decomposes: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ You have placed some solid ammonium hydrogen sulfide in a closed flask. Which of the following would produce less hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), which is a poisonous gas? a. Removing some \(\mathrm{NH}_{3}\) from the flask b. Adding some \(\mathrm{NH}_{3}\) to the flask c. Removing some of the \(\mathrm{NH}_{4} \mathrm{HS}\) d. Increasing the pressure in the flask by adding helium gas Explain each of your answers.

Methanol is prepared industrially from synthesis gas (CO and \(\mathrm{H}_{2}\) ). $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) ; \Delta H^{\circ}=-21.7 \mathrm{kcal} $$ Would the fraction of methanol obtained at equilibrium be increased by raising the temperature? Explain.

During the commercial preparation of sulfuric acid, sulfur dioxide reacts with oxygen in an exothermic reaction to produce sulfur trioxide. In this step, sulfur dioxide mixed with oxygen-enriched air passes into a reaction tower at about \(420^{\circ} \mathrm{C}\), where reaction occurs on a vanadium(V) oxide catalyst. Discuss the conditions used in this reaction in terms of its effect on the yield of sulfur trioxide. Are there any other conditions that you might explore in order to increase the yield of sulfur trioxide?

A 2.00-L vessel contains \(1.00 \mathrm{~mol} \mathrm{~N}_{2}, 1.00 \mathrm{~mol} \mathrm{H}_{2}\), and \(2.00 \mathrm{~mol} \mathrm{NH}_{3} .\) What is the direction of reaction (forward or reverse) needed to attain equilibrium at \(400^{\circ} \mathrm{C}\) ? The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(0.51\) at \(400^{\circ} \mathrm{C}\).

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.