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

Write the equilibrium expression for each of the following reactions. a. \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g)\)

Short Answer

Expert verified
a. Kc = \(\frac{[O_2]^3}{[O_3]^2}\) b. Kc = \(\frac{[CO_2][H_2O]^2}{[CH_4][O_2]^2}\) c. Kc = \(\frac{[C_2H_4Cl_2]}{[C_2H_4][Cl_2]}\)

Step by step solution

01

a. Write the equilibrium expression for the reaction 2 O₃(g) ⇌ 3 O₂(g)

The reaction is: 2 O₃(g) ⇌ 3 O₂(g) To write the equilibrium expression, we'll use Kc = [Products]/[Reactants]. Kc = \(\frac{[O_2]^3}{[O_3]^2}\)
02

b. Write the equilibrium expression for the reaction CH₄(g) + 2 O₂(g) ⇌ CO₂(g) + 2 H₂O(g)

The reaction is: CH₄(g) + 2 O₂(g) ⇌ CO₂(g) + 2 H₂O(g) To write the equilibrium expression, we'll use Kc = [Products]/[Reactants]. Kc = \(\frac{[CO_2][H_2O]^2}{[CH_4][O_2]^2}\)
03

c. Write the equilibrium expression for the reaction C₂H₄(g) + Cl₂(g) ⇌ C₂H₄Cl₂(g)

The reaction is: C₂H₄(g) + Cl₂(g) ⇌ C₂H₄Cl₂(g) To write the equilibrium expression, we'll use Kc = [Products]/[Reactants]. Kc = \(\frac{[C_2H_4Cl_2]}{[C_2H_4][Cl_2]}\)

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.

Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction reaches a point where the concentrations of reactants and products remain constant over time. This balance happens because the rate at which the reactants convert into products is equal to the rate at which products revert to reactants. At equilibrium, the system is stabilized, though reactions at the molecular level continue.

Equilibrium can be observed in closed systems, where no substances are added or removed. It is characterized by a particular ratio of product and reactant concentrations, known as the equilibrium constant (\( K_c \)). This constant is unique for each reaction and can be calculated by putting the concentrations of the products over the reactants, each raised to the power of their respective coefficients in the balanced equation.

Finding chemical equilibrium is crucial for understanding how conditions affect the yield of reactions, which can be manipulated by changing concentrations, temperature, or pressure. Knowing this helps in industrial applications, like maximizing product formation.
Reaction Rates
Reaction rates tell us how quickly or slowly a reaction occurs, indicating the speed at which reactants turn into products. Factors influencing these rates include concentration, temperature, surface area, and the presence of catalysts.

  • Concentration: Increasing the concentration of reactants usually increases the reaction rate because more particles are available to collide.
  • Temperature: Higher temperatures lead to faster particle movement, causing more frequent and energetic collisions, thus speeding up the reaction.
  • Surface Area: Smaller particle sizes increase surface area, offering more area for collisions.
  • Catalysts: These substances speed up a reaction without being consumed by lowering the activation energy needed for the reaction to proceed.
These factors are essential in controlling reactions in fields like pharmacology and manufacturing, ensuring processes are safe and efficient.

Balancing reaction rates is key in reaching equilibrium, as the rate of the forward reaction needs to match the reverse for equilibrium to be established.
Chemical Reactions
Chemical reactions involve the transformation of one or more substances into new substances. This process can be represented by chemical equations, which illustrate the reactants (starting materials) and products (results of the reaction).
  • Reactants: Substances that undergo change.
  • Products: Substances formed as a result of a reaction.
The coefficients in a balanced chemical equation indicate the proportion of molecules involved, which is crucial for calculations related to reaction stoichiometry and equilibrium expressions.

There are various types of reactions, including synthesis, decomposition, single replacement, and double replacement. Each type has unique characteristics and conditions under which they occur.

Understanding the nature of chemical reactions helps in predicting outcomes of chemical processes, allows for proper adjustments in experimental conditions, and ensures that desired products are obtained efficiently.

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

When a reaction system has reached chemical equilibrium, the concentrations of the reactants and products no longer change with time. Why does the amount of product no longer increase, even though large concentrations of the reactants may still be present?

The equilibrium constant for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ has the value \(9.2 \times 10^{-6}\) at a particular temperature. The system is analyzed at equilibrium, and it is found that the concentrations of \(\mathrm{NOCl}(g)\) and \(\mathrm{NO}(g)\) are \(0.44 \mathrm{M}\) and \(1.5 \times 10^{-3} M,\) respectively. What is the concentration of \(\mathrm{Cl}_{2}(g)\) in the equilibrium system under these conditions?

For the process $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ it is found that the equilibrium concentrations at a particular temperature are \(\left[\mathrm{H}_{2}\right]=1.4 M,\left[\mathrm{CO}_{2}\right]=1.3 M,[\mathrm{CO}]=0.71 M,\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.66 M .\) Calculate the equilibrium constant \(K\) for the reaction under these conditions.

In general terms, what does the equilibrium constant for a reaction represent? What is the algebraic form of the equilibrium constant for a typical reaction? What do square brackets indicate when we write an equilibrium constant?

\(K_{\text {sp }}\) for copper(II) hydroxide, \(\mathrm{Cu}(\mathrm{OH})_{2},\) has a value \(2.2 \times 10^{-20}\) at 25 " \(\mathrm{C}\). Calculate the solubility of copper(II) hydroxide in \(\mathrm{mol} / \mathrm{L}\) and \(\mathrm{g} / \mathrm{L}\) at \(25 \mathrm{C}\)

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.

Sign-up for free