/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Write the equilibrium constant e... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+\) \(\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
The equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(s) \leftrightarrows \mathrm{CaO}(s)+ \mathrm{CO}_{2}(g)\) is given by: \[K = [\mathrm{CO}_{2}]\]

Step by step solution

01

Identify the gas-phase species

In this reaction, we have only one gas-phase species that is \(\mathrm{CO}_{2}(g)\). So, we will include only this in our equilibrium constant expression.
02

Write the equilibrium constant expression

The equilibrium constant expression K is given by: \[K = \frac{[\mathrm{CO}_{2}]^{1}}{1}\] Here, we have used the concentration of \(\mathrm{CO}_{2}(g)\) raised to the power of its stoichiometric coefficient 1 (assuming one mole each of reactants and products) in the numerator. We don't include \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{CaO}(s)\) in the expression, as their concentrations do not change during the reaction. So, the equilibrium constant expression for the given reaction is: \[K = [\mathrm{CO}_{2}]\] The equilibrium constant K will give information about the concentration of \(\mathrm{CO}_{2}\) at equilibrium for different temperatures.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that describes the state of a reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentration of reactants and products over time. It is a dynamic equilibrium, meaning both reactions continue to occur, but because they’re at the same rate, the overall concentrations remain constant.

Crucially, equilibrium does not mean that the reactants and products are present in equal amounts. Instead, it indicates that their ratios are stable. The position of equilibrium is represented by the equilibrium constant \( K \), which is a ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients. In the exercise with the decomposition of \(\mathrm{CaCO}_{3}(s)\), the equilibrium constant expression helps us understand how much \(\mathrm{CO}_{2}(g)\) will be present at equilibrium at a given temperature.

To experimentally determine \( K \), one would typically measure the concentrations of the gaseous products and reactants at equilibrium. Then, these values are plugged into the equilibrium expression to solve for the equilibrium constant. This constant is crucial in predicting how the system will react to changes in conditions, such as changes in concentration, pressure, or temperature, often illustrated through Le Chatelier's Principle.
Gaseous Reactions and Equilibrium
Gaseous reactions such as the decomposition of \(\mathrm{CaCO}_{3}(s)\) to form \(\mathrm{CaO}(s)\) and \(\mathrm{CO}_{2}(g)\) are particularly interesting in the context of equilibrium. The behavior of gases under changing conditions is governed by the ideal gas law, which relates the pressure, volume, temperature, and moles of a gas. At equilibrium, the partial pressure of the gaseous products and reactants is used in the equilibrium expression.

For a reaction involving gases, the equilibrium constant can be expressed in terms of partial pressures (usually designated as \( K_p \)) or concentrations (designated as \( K_c \)). In the exercise, only \(\mathrm{CO}_{2}(g)\) is a gas, and its concentration is used in the equilibrium expression. In real-world applications, this information is vital because it can predict how changing the pressure (by changing the volume or the amount of gas) affects the system's position of equilibrium, and consequently, the yields of a chemical process.
Stoichiometry and the Equilibrium Constant
Stoichiometry is the branch of chemistry that deals with the quantitative relationship between reactants and products in a chemical reaction. It is essential for calculating the equilibrium constant because the stoichiometric coefficients from the balanced chemical equation are used as exponents in the equilibrium expression. In our example with the reaction of \(\mathrm{CaCO}_{3}(s)\), the stoichiometric coefficient for \(\mathrm{CO}_{2}(g)\) is 1, reflecting the mole ratio in which \(\mathrm{CO}_{2}(g)\) is produced.

When writing the equilibrium constant expression, solid and liquid reactants and products (like \(\mathrm{CaCO}_{3}(s)\) and \(\mathrm{CaO}(s)\) in our case) are typically omitted because their concentrations are constant and do not affect the equilibrium. Only species in the gaseous or aqueous states are included. Stoichiometry is also employed when using the equilibrium constant to calculate unknown concentrations at equilibrium. If one value is known, stoichiometry allows for the calculation of the others, keeping in mind the mole ratios established by the balanced chemical equation.

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

What does a value of \(K_{\text {eq }}\) greater than \(10^{3}\) imply? Prove that your answer is correct by using the general expression \(K_{\mathrm{eq}}=[\) Products \(] /[\) Reactants \(]\).

At \(25^{\circ} \mathrm{C}\), the solubility of iron(III) hydroxide in water is \(4.49 \times 10^{-10} \mathrm{M}\). (a) What is the solubility in grams per liter? (b) What is the molar equilibrium concentration of each ion? (c) How many grams of iron(III) hydroxide could you dissolve in a 20,000-gallon swimming pool?

Using the definition of equilibrium, show how \(k_{\mathrm{f}} / k_{\mathrm{r}}\) for the one-step reaction \(\mathrm{R} \rightleftarrows \mathrm{P}\) is equal to the ratio \([\mathrm{P}] /[\mathrm{R}]\).

Consider the gas-phase reaction: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftarrows 2 \mathrm{NH}_{3}(g)\) Suppose it is at equilibrium inside of a closed vessel. Next, the volume of the closed vessel is suddenly decreased, increasing the overall pressure inside the flask. According to Le Châtelier's principle, which way will the reaction shift, and why? Then explain why the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) would not shift when exposed to the same type of disturbance.

When the reaction \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)\) is run at \(2000^{\circ} \mathrm{C}\), appreciable amounts of reactants and product are present at equilibrium. (a) A sealed 2.00-L container at \(2000{ }^{\circ} \mathrm{C}\) is filled with \(1.00\) mole of \(\mathrm{NO}(g)\) and nothing else. At that moment, which reaction is faster, forward or reverse? Justify your answer. (b) At equilibrium, the concentration of \(\mathrm{NO}(g)\) is \(0.0683 \mathrm{M}\) and the concentration of \(\mathrm{N}_{2}(g)\) is \(0.2159 \mathrm{M}\). What is the value of \(K_{\mathrm{eq}}\) at \(2000^{\circ} \mathrm{C} ?\)

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.