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 an equilibrium expression for each chemical equation. (a) \(2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g)\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) (c) \(\mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g)\)

Short Answer

Expert verified
Reaction (a): K = [CO2]^2 / [CO]^2 [O2], Reaction (b): K = [NO]^2 / [N2] [O2], Reaction (c): K = [SbCl3] [Cl2] / [SbCl5], Reaction (d): K = [COCl2] / [CO] [Cl2]

Step by step solution

01

Understand the Equilibrium Constant

The equilibrium constant, denoted as K, is a value that expresses the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation, when the reaction is at equilibrium.
02

Write the Equilibrium Expressions

For a general reaction aA + bB ⇌ cC + dD, the equilibrium expression is K = [C]^c [D]^d / [A]^a [B]^b. We'll use this format to write the equilibrium expressions for each reaction given.
03

Determine the Equilibrium Expression for Reaction (a)

For the reaction 2CO(g) + O2(g) ⇌ 2CO2(g), the equilibrium expression is K = [CO2]^2 / [CO]^2 [O2].
04

Determine the Equilibrium Expression for Reaction (b)

For the reaction N2(g) + O2(g) ⇌ 2NO(g), the equilibrium expression is K = [NO]^2 / [N2] [O2].
05

Determine the Equilibrium Expression for Reaction (c)

For the reaction SbCl5(g) ⇌ SbCl3(g) + Cl2(g), the equilibrium expression is K = [SbCl3] [Cl2] / [SbCl5].
06

Determine the Equilibrium Expression for Reaction (d)

For the reaction CO(g) + Cl2(g) ⇌ COCl2(g), the equilibrium expression is K = [COCl2] / [CO] [Cl2].

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 state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, meaning there is no net change in the concentrations of reactants and products over time. This dynamic balance does not imply that the reactants and products are in equal concentrations, but rather that their ratios remain constant at equilibrium.

At the molecular level, chemical equilibrium occurs when the frequency of reactant molecules reacting to form products is the same as the frequency of product molecules decomposing back into reactants. Importantly, even at equilibrium, the actual exchanges continue to happen; it is only the macroscopic concentrations that appear stable.

Understanding equilibrium is crucial because it helps predict the extent to which a chemical reaction will proceed, and it forms the basis for many applications in chemical engineering, biochemistry, and various other fields.
Equilibrium Constant
The equilibrium constant () provides a quantitative measure of the position of equilibrium in a chemical reaction. Represented by the symbol , its value is determined by the relative concentrations of the products and reactants at equilibrium, each raised to the power of their stoichiometric coefficients from the balanced chemical equation.

The formula to calculate the equilibrium constant for a generic chemical reaction is expressed as:\[\begin{equation}K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\end{equation}\]where and are the molar concentrations of the products and reactants, respectively, and , , , and are their corresponding coefficients in the balanced equation. If exceeds 1, the products are favored; if equals 1, neither reactants nor products are favored; and if is less than 1, the reactants are favored.

Chemists use this constant to predict the direction of the reaction and to calculate the concentrations of different components of the reaction at equilibrium.
Reaction Quotient
The reaction quotient (), denoted as , is similar to the equilibrium constant in that it is a ratio of the concentrations of the products to the reactants at any point in time, not just at equilibrium. It serves as a tool to predict the direction in which a reaction will proceed to achieve equilibrium.

The expression for is almost identical to that of , but it uses the initial concentrations of reactants and products, rather than those at equilibrium. If equals , the system is already at equilibrium. If is greater than , the system will shift to the left, increasing the concentration of reactants. Conversely, if is less than , the system will shift to the right, increasing the concentration of products until equilibrium is reached.
Le Chatelier's Principle
Le Chatelier's Principle posits that if a stress is applied to a system at chemical equilibrium, the system will adjust itself to minimize that stress and eventually establish a new equilibrium. Stresses include changes in concentration, temperature, or pressure.

For instance, if the concentration of a reactant is increased, the system responds by consuming the excess reactant to produce more products, shifting the equilibrium position to the right. If the temperature of an exothermic reaction is increased, the principle predicts a shift toward the reactants, as heat is a product. This principle is invaluable for predicting how changes will affect a chemical system at equilibrium, helping chemists control reactions to favor the formation of desired products.

Le Chatelier's Principle also explains why certain reactions are pressure-sensitive. In reactions involving gases, increasing the pressure by decreasing the volume will shift the equilibrium toward the side with fewer moles of gas, according to this principle.

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

Coal, which is primarily carbon, can be converted to natural gas, primarily \(\mathrm{CH}_{4}\), by this exothermic reaction. $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g) $$ If this reaction mixture is at equilibrium, predict the effect (shift right, shift left, or no effect) of these changes. (a) adding more \(C\) to the reaction mixture (b) adding more \(\mathrm{H}_{2}\) to the reaction mixture (c) raising the temperature of the reaction mixture (d) lowering the volume of the reaction mixture (e) adding a catalyst to the reaction mixture

What is the effect of decreasing the concentration of a reactant in a reaction mixture at equilibrium?

Consider the reaction. $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ An equilibrium mixture of this reaction at a certain temperature has \([\mathrm{CO}]=0.105 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.114 \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.185 \mathrm{M}\). What is the value of the equilibrium constant at this temperature?

Consider the reaction. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ A solution is made containing initial \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.020 \mathrm{M}\). At equilibrium, \(\left[\mathrm{Cl}_{2}\right]=1.2 \times 10^{-2} \mathrm{M}\). Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and \(\mathrm{SO}_{2}\).

The solubility of \(\mathrm{CaCrO}_{4}\) at \(25^{\circ} \mathrm{C}\) is \(4.15 \mathrm{~g} / \mathrm{L}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{CaCrO}_{4}\).

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