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 general equilibrium constant expression for each of the following: (a) \(3 \mathrm{~A} \rightleftarrows 2 \mathrm{C}\) (b) \(A+B \rightleftarrows 2 C\) (c) \(3 \mathrm{~A}+5 \mathrm{~B} \rightleftarrows \mathrm{C}+4 \mathrm{D}\)

Short Answer

Expert verified
(a) \( K = \frac{[C]^2}{[A]^3} \); (b) \( K = \frac{[C]^2}{[A][B]} \); (c) \( K = \frac{[C][D]^4}{[A]^3[B]^5} \)."

Step by step solution

01

Understand the Equilibrium Constant

The equilibrium constant, denoted as \(K\), represents the ratio of the concentration of products to the concentration of reactants at equilibrium. It's expressed as a function of the concentrations of the products and reactants, each raised to their respective coefficients from the balanced chemical equation.
02

Write the Equilibrium Constant for Reaction (a)

For the reaction \(3 \mathrm{~A} \rightleftarrows 2 \mathrm{C}\), the equilibrium constant expression is written as \[ K = \frac{[C]^2}{[A]^3} \]where \([C]\) and \([A]\) represent the concentrations of \(C\) and \(A\) at equilibrium.
03

Write the Equilibrium Constant for Reaction (b)

For the reaction \(A + B \rightleftarrows 2 C\), the equilibrium constant expression is written as \[ K = \frac{[C]^2}{[A][B]} \]where \([C]\), \([A]\), and \([B]\) are the equilibrium concentrations of \(C\), \(A\), and \(B\) respectively.
04

Write the Equilibrium Constant for Reaction (c)

For the reaction \(3 \mathrm{~A} + 5 \mathrm{~B} \rightleftarrows \mathrm{C} + 4 \mathrm{D}\), the equilibrium constant expression is \[ K = \frac{[C][D]^4}{[A]^3[B]^5} \]where \([C]\), \([D]\), \([A]\), and \([B]\) denote the equilibrium concentrations of \(C\), \(D\), \(A\), and \(B\) respectively.

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 is a fundamental concept in chemistry where the rate of the forward reaction equals the rate of the reverse reaction. This creates a state of balance where the concentrations of reactants and products remain constant over time. However, it's essential to note that this doesn't mean that reactants and products are in equal concentrations, only that their rates of formation are equal. In this balanced state, although the chemical reactions continue to occur, there is no net change in the concentration of reactants and products.

Achieving chemical equilibrium depends on several factors, including temperature, pressure (for gases), and the initial concentrations of reactants and products. Changes in these conditions can shift the equilibrium position, leading to either an increase in products or a return to more reactants, as described by Le Chatelier’s Principle. Understanding chemical equilibrium is crucial in predicting the outcomes of reactions and in industrial chemical processes where efficiency and yield are vital.
Concentration of Reactants and Products
The concentration of reactants and products plays a key role in defining the state of equilibrium in a chemical reaction. At equilibrium, although the concentrations of reactants ([A], [B]) and products ([C], [D]) aren't changing, they are not necessarily equal. Instead, they reach a specific ratio that is defined by the equilibrium constant expression (K). This expression is derived from the balanced chemical equation and indicates the relative amounts of each substance at equilibrium.

In the equation K = \( \frac{[C]^2}{[A][B]} \), for example, the squared concentration of C indicates that twice as much product is being produced compared to one unit of reactant being consumed. The square and other powers in the equilibrium constant expression reflect the stoichiometry of the balanced chemical equation. This serves as a guideline for how much of each species is present when the system is at balance. Understanding these concentrations helps chemists manipulate reactions to favor the production of desired products by adjusting the starting amounts or reaction conditions.
Balanced Chemical Equation
A balanced chemical equation is essential in any chemical reaction as it ensures that the law of conservation of mass is upheld. This means that the number of atoms of each element is the same on both sides of the equation, reflecting that mass and energy cannot be created or destroyed. In practical terms, balancing a chemical equation involves making sure that there are equal numbers of each type of atom on both sides of the equation.

For instance, in the equation 3A + 5B \( \rightleftarrows \) C + 4D, the coefficients tell us how many molecules of each species participate in the reaction. These coefficients are used in the equilibrium constant expression to represent the power to which each concentration is raised. Balancing equations is crucial because incorrect balancing leads to incorrect predictions of the quantities of products formed or reactants consumed. By understanding the balanced chemical equations fully, one can correctly set up the equilibrium constant expressions and effectively predict and manipulate the outcome of chemical reactions.

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

Although a clock with a swinging pendulum is not a chemical equilibrium system, explain how it represents a dynamic, reversible process.

Which of the following theoretical factors decreases the rate of a reaction? (a) decrease collision frequency (b) decrease collision energy (c) ineffective collision orientation

Cadmium sulfide dissociates slightly in an aqueous solution as follows: $$ \mathrm{CdS}(s) \rightleftarrows \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q) $$ Predict the direction of equilibrium shift for each of the following stresses: (a) increase \(\left[\mathrm{Cd}^{2+}\right]\) (b) increase \(\left[\mathrm{S}^{2-}\right]\) (c) decrease \(\left[\mathrm{Cd}^{2+}\right]\) (d) decrease \(\left[\mathrm{S}^{2-}\right]\) (e) add solid \(\mathrm{CdS}\) (f) add solid \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) (g) add solid \(\mathrm{NaNO}_{3}\) (h) add \(\mathrm{H}^{+}\)

Given the chemical equation for the ionization of water $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Predict the direction of equilibrium shift for each of the following stresses: (a) increase \(\left[\mathrm{H}^{+}\right]\) (b) decrease \(\left[\mathrm{OH}^{-}\right]\) (c) increase \(\mathrm{pH}\) (d) decrease \(\mathrm{pH}\)

Write the general equilibrium constant expression for each of the following: (a) \(2 \mathrm{~A} \rightleftarrows \mathrm{C}\) (b) \(A+2 B \rightleftarrows 3 C\) (c) \(2 \mathrm{~A}+3 \mathrm{~B} \rightleftarrows 4 \mathrm{C}+\mathrm{D}\)

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