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

Writing Equilibrium Constant Expressions Write equilibrium constant expressions for the following reactions. For gases use either pressures or concentrations. (a) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\) (b) \(\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g})\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g})\) (d) \(\mathrm{NiO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Ni}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\)

Short Answer

Expert verified
(a) \( K_c = \frac{[\mathrm{H}_2\mathrm{O}]^2[\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}_2]^2} \); (b) \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}][\mathrm{O}_2]^{1/2}} \); (c) \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \); (d) \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}]} \).

Step by step solution

01

Understanding Equilibrium Constant Expressions

Equilibrium constant expressions are written based on the products and reactants involved in a balanced chemical equation. They are denoted by \( K \) and are determined either by concentrations \( K_c \) or partial pressures \( K_p \). Solid substances do not appear in these expressions.
02

Reaction (a) Equilibrium Constant Expression

For the reaction \( 2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \), the equilibrium constant expression using concentrations \( K_c \) is given by:\[ K_c = \frac{[\mathrm{H}_2\mathrm{O}]^2[\mathrm{O}_2]}{[\mathrm{H}_2\mathrm{O}_2]^2} \]
03

Reaction (b) Equilibrium Constant Expression

For the reaction \( \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}_{2}(\mathrm{g}) \), the equilibrium constant expression using concentrations \( K_c \) is:\[ K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}][\mathrm{O}_2]^{1/2}} \]
04

Reaction (c) Equilibrium Constant Expression

For the reaction \( \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) \), since \( \mathrm{C}(\mathrm{s}) \) is a solid, it does not appear in the expression. Therefore, the equilibrium constant expression is:\[ K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \]
05

Reaction (d) Equilibrium Constant Expression

For the reaction \( \mathrm{NiO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Ni}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \), both \( \mathrm{NiO}(\mathrm{s}) \) and \( \mathrm{Ni}(\mathrm{s}) \) are solids, so they do not appear in the expression. The expression is:\[ K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}]} \]

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 concentrations of reactants and products no longer change with time. This may make it sound like the reaction has stopped, but that's not the case. The forward reaction, where reactants form products, occurs at the same rate as the reverse reaction, where products break down back into reactants. It's like two teams in a tug of war where neither team is winning.
The equilibrium constant, represented by \( K \), is a number that shows the ratio of product concentrations to reactant concentrations at equilibrium. This constant gives us useful insights into the position of equilibrium—whether the reaction favors products (much larger \( K \)) or reactants (much smaller \( K \)).
Remember, we only consider gases and aqueous solutions in these expressions, never solids or liquids. This is crucial when writing the equilibrium expressions.
Reaction Quotients
The reaction quotient, \( Q \), is very similar to the equilibrium constant, \( K \), but it can be calculated at any point in a reaction, not just at equilibrium. It helps in determining which direction a reaction will proceed to achieve equilibrium. If \( Q \) is different from \( K \), it tells us that the reaction is either moving forward or backward to reach equilibrium.
  • If \( Q < K \), the reaction proceeds in the forward direction, creating more products.
  • If \( Q > K \), the reaction goes in reverse, forming more reactants.
  • If \( Q = K \), the reaction is already at equilibrium.
Calculating \( Q \) involves the same setup as \( K \), using the current concentrations or pressures of the substances involved. This tool is particularly useful in industrial and laboratory settings to adjust conditions favorably.
Gaseous Reactions
Gaseous reactions are chemical reactions that involve gases as either reactants or products. These types of reactions can be particularly interesting because they involve changes in volume and pressure. Variables like pressure or concentration of gases can shift the position of equilibrium. This is described by Le Chatelier's principle, which states that if a change is made to a system at equilibrium, the system adjusts in a way that counters that change.
In equilibrium constant expressions for gaseous reactions, pressures can be used instead of concentrations. This can be noted in the form of \( K_p \) where the pressures of gases are used instead of concentrations, denoted as \( K_c \). The conversion between these two is given by the equation:\[ K_p = K_c (RT)^{\Delta n} \]where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in number of moles of gas. Understanding the dynamics of gaseous reactions helps chemists optimize conditions for industrial processes.
Reaction Balancing
Reaction balancing is an essential initial step before you can write an equilibrium constant expression. Every chemical equation must adhere to the law of conservation of mass, meaning the same number and types of atoms should appear on both sides of the equation. So, if you have 10 hydrogen atoms as reactants, you should also have 10 among the products.
Balanced reactions provide the correct stoichiometric coefficients, which are those numbers in front of chemical species in a reaction equation. These coefficients are important as they influence the exponents used in the equilibrium constant expressions. For example, in the reaction \( aA + bB \rightleftarrows cC + dD \), the equilibrium expression is \( K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).
Taking the time to correctly balance a reaction simplifies the process of creating accurate equilibrium constant expressions.

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

Consider the following equilibria involving \(\mathrm{SO}_{2}(\mathrm{g})\) and their corresponding equilibrium constants. \(\mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{SO}_{3}(\mathrm{g}) \quad K_{1}\) \(2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \quad K_{2}\) Which of the following expressions relates \(K_{1}\) to \(K_{2} ?\) (a) \(K_{2}=K_{1}^{2}\) (d) \(K_{2}=1 / K_{1}\) (b) \(K_{2}^{2}=K_{1}\) (e) \(K_{2}=1 / K_{1}^{2}\) (c) \(K_{2}=K_{1}\)

Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide if the salt is heated to a sufficiently high temperature. $$ \mathrm{NH}_{4} \mathrm{I}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{HI}(\mathrm{g}) $$ Some ammonium iodide is placed in a flask, which is then heated to \(400^{\circ} \mathrm{C}\). If the total pressure in the flask when equilibrium has been achieved is \(705 \mathrm{mm} \mathrm{Hg},\) what is the value of \(K_{\mathrm{p}}\) (when partial pressures are in atmospheres)?

Limestone decomposes at high temperatures. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(1000^{\circ} \mathrm{C}, K_{\mathrm{p}}=3.87 .\) If pure \(\mathrm{CaCO}_{3}\) is placed in a \(5.00-\mathrm{I}\) flask and heated to \(1000^{\circ} \mathrm{C},\) what quantity of \(\mathrm{CaCO}_{3}\) must decompose to achieve the equilibrium pressure of \(\mathrm{CO}_{2} ?\)

The equilibrium constant, \(K\), for the reaction $$ 2 \mathrm{NOCl}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ is \(3.9 \times 10^{-3}\) at \(300^{\circ} \mathrm{C} .\) A mixture contains the gases at the following concentrations: \([\mathrm{NOCl}]=5.0 \times 10^{-8} \mathrm{mol} / \mathrm{L}\) \([\mathrm{NO}]=2.5 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{Cl}_{2}\right]=2.0 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) Is the reaction at equilibrium at \(300^{\circ} \mathrm{C}\) ? If not, in which direction does the reaction proceed to come to equilibrium?

A A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas with a pressure of \(1.00 \mathrm{atm}\) is placed in a flask. When equilibrium is achieved, \(20.0 \%\) of the \(\mathrm{N}_{2} \mathrm{O}_{4}\) has been converted to \(\mathrm{NO}_{2}\) gas. (a) Calculate \(K_{\mathrm{p}}\) (b) If the original pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is 0.10 atm, what is the percent dissociation of the gas? Is the result in agree. ment with Le Chatelier's principle?

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