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 reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D} .\) Assume that both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse? Explain.

Short Answer

Expert verified
(a) At equilibrium, products (C and D) predominate due to the very large value of the equilibrium constant. (b) The forward reaction (A + B -> C + D) has the larger rate constant, as K = k_forward / k_reverse, and a large K indicates a larger k_forward compared to k_reverse.

Step by step solution

01

(a) Identifying the predominant species at equilibrium

The equilibrium constant (K) for a given reaction is defined as the ratio of the concentrations of products to reactants at equilibrium. Since we are given that the equilibrium constant is very large, it means that the concentration of products (C and D) is much higher than the concentration of reactants (A and B) at equilibrium. Therefore, the products (C and D) predominate at equilibrium.
02

(b) Understanding the relationship between equilibrium constant and rate constants

The equilibrium constant (K) is related to the rate constants of the forward (k_forward) and reverse (k_reverse) reactions by the following equation: K = k_forward / k_reverse. This equation tells us that if the equilibrium constant is very large, the forward rate constant must be much larger than the reverse rate constant since K is the ratio of k_forward to k_reverse.
03

(b) Determining which reaction has the larger rate constant

Based on our understanding of the relationship between the equilibrium constant and the rate constants, we can conclude that the forward reaction (A + B -> C + D) has the larger rate constant since K is very large. This means that the forward reaction is faster than the reverse reaction, ultimately leading to a higher concentration of products (C and D) at equilibrium.

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.

Rate Constants
The rate constants of a reaction are critical in understanding how quickly a reaction proceeds. Rate constants, denoted by the symbols \(k_{\text{forward}}\) and \(k_{\text{reverse}}\), represent the speed at which the forward and reverse reactions occur, respectively. These constants depend on various factors including temperature and the nature of the reactants involved.

In the context of equilibrium, the relationship between the rate constants and the equilibrium constant (\(K\)) is crucial. The equilibrium constant is given by the ratio of the forward and reverse rate constants:
  • \(K = \frac{k_{\text{forward}}}{k_{\text{reverse}}}\)
This equation shows that if the equilibrium constant is large, the forward rate constant is significantly larger than the reverse rate constant. This indicates a faster forward reaction, leading to a predominance of products at equilibrium. Understanding the size of these constants allows chemists to predict and manipulate reaction conditions for desired outcomes.
Reaction Rates
Reaction rates dictate how fast a reaction proceeds toward equilibrium. They are influenced by the concentration of reactants and the temperature at which the reaction occurs. For elementary reactions, the rate is directly proportional to the concentration of the reactants raised to the power of their stoichiometric coefficients.

The rate of reaction can be described by the rate law, which for a generic reaction \(A + B \rightarrow C + D\) is expressed as:
  • Rate = \(k \cdot [A][B]\)
The rate law shows that increasing the concentration of reactants will generally lead to an increase in the reaction rate. The temperature dependence of the rate is described by the Arrhenius equation, which states that the rate constant increases with temperature, thereby increasing the reaction rate. In equilibrium, the forward and reverse reactions occur at the same rate, maintaining a constant concentration of reactants and products. This dynamic state means that while the compositions remain constant, the individual molecules are continuously reacting.
Chemical Equilibrium
Chemical equilibrium occurs when the rates of the forward and reverse reactions in a closed system are equal, resulting in constant concentrations of reactants and products. It is a dynamic state where the reactions occur continuously but there is no net change in the concentration of species in the system.

The position of equilibrium is represented by the equilibrium constant \(K\), which quantitatively expresses the ratio of product to reactant concentrations. For a general reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is:
  • \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
A large \(K\) value indicates that products are favored at equilibrium, whereas a small \(K\) suggests reactants are favored.
Factors such as concentration, pressure, and temperature can shift the equilibrium position, a concept described by Le Chatelier's Principle. This principle states that if an external change is applied to a system at equilibrium, the system adjusts to counteract the effect and re-establish equilibrium. Understanding chemical equilibrium helps in predicting the concentrations of reactants and products at any given time, thus aiding in the control and optimization of industrial chemical processes.

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

Bioremediation is the use of microorganisms to degrade environmental pollutants. Many pollutants contain only carbon and hydrogen (oil being one example). The chemical reactions are complicated, but in general the microorganisms react the pollutant hydrocarbon with \(\mathrm{O}_{2}\) to produce \(\mathrm{CO}_{2}\) and other carbon-containing compounds that are incorporated into the organism's biomass. How would increasing levels of \(\mathrm{CO}_{2}\) in the environment affect the bioremediation reaction?

The equilibrium constant for the reaction $$\begin{array}{r} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \\ \text { is } K_{c}=1.3 \times 10^{-2} \text {at } 1000 \mathrm{~K} \text { . (a) At this tempera } \end{array}$$ (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calcu- $$\begin{array}{l} \text { late } K_{c} \text { for } 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \text { . } \\ K_{c} \text { for } \mathrm{NOBr}(g) \rightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \end{array}$$ (c) Calculate

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

For the equilibrium $$\mathrm{PH}_{3} \mathrm{BCl}_{3}(s) \rightleftharpoons \mathrm{PH}_{3}(g)+\mathrm{BCl}_{3}(g)$$ (b) After \(3.00 \mathrm{~g}\) of solid$$K_{p}=0.052 \text { at } 60^{\circ} \mathrm{C} \text { . (a) Calculate } K_{c}$$ \(\mathrm{PH}_{3} \mathrm{BCl}_{3}\) is added to a closed \(1.500-\mathrm{L}\) vessel at \(60^{\circ} \mathrm{C},\) the vessel is charged with \(0.0500 \mathrm{~g}\) of \(\mathrm{BCl}_{3}(g) .\) What is the equilibrium concentration of \(\mathrm{PH}_{3} ?\)

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

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