For the reaction below, \(K_{\mathrm{p}}=1.16\) at \(800 .{ }^{\circ} \mathrm{C}\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ If a 20.0-g sample of \(\mathrm{CaCO}_{3}\) is put into a \(10.0\) - \(\mathrm{L}\) container and heated to \(800 .{ }^{\circ} \mathrm{C}\), what percentage by mass of the \(\mathrm{CaCO}_{3}\) will react to reach equilibrium?

Short Answer

Expert verified
76.5% of the CaCO3 reacts to reach equilibrium at 800 °C.

Step by step solution

01

Write the Balanced Chemical Equation

We are given the balanced chemical equation: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) $$
02

Create an ICE Table

Next, we will set up the ICE table to keep track of the amount of each substance in moles in the reaction. | | CaCO3 | CaO | CO2 | |------------------|-----------------|--------------|--------------| | Initial (mol) | \(n_{CaCO3}\) | 0 | 0 | | Change (mol) | -x | +x | +x | | Equilibrium (mol)| \(n_{CaCO3}-x\) | x | x |
03

Calculate Moles of CaCO3

To find the initial moles of CaCO3, we'll use the given mass (20.0 g) and molecular weight of CaCO3 (100.09 g/mol): $$ n_{CaCO3} = \frac{20.0}{100.09} = 0.2\ \mathrm{mol} $$
04

Write the Expression for Kp

For the given reaction, the equilibrium constant Kp expression is given by: $$ K_{p} = \frac{P_{CO_2}}{1} $$ where \(P_{CO_2}\) is the partial pressure of CO2 at equilibrium. We know that the pressure is directly proportional to the moles of CO2 gas: $$ P_{CO_2} = \frac{n_{CO_2}}{V} \cdot R \cdot T $$ where V is the container volume (10.0 L), R is the ideal gas constant (0.0821 L·atm/K·mol), and T is the absolute temperature (800 °C = 1073 K).
05

Substitute CO2 Pressure and Solve for x

Using the known values and expressions, we get: $$ K_{p} = \frac{x}{0.2-x} $$ Substitute the CO2 pressure relation and Kp value into the equation and solve for x: $$ 1.16 = \frac{x}{(0.2-x) \cdot 10.0} \cdot 0.0821 \cdot 1073 $$ Now, we can solve for x: $$ x = 0.153\ \mathrm{mol} $$
06

Calculate the Percentage of CaCO3 That Reacts

The percentage by mass of CaCO3 that reacts is given by the ratio of the moles of reacted CaCO3 to the initial moles of CaCO3, multiplied by 100: $$ \frac{x}{n_{CaCO3}} \times 100 = \frac{0.153}{0.2} \times 100 = 76.5\% $$ Therefore, 76.5% of the CaCO3 reacts to reach equilibrium at 800 °C.

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.

Equilibrium Constant (Kp)
Understanding the equilibrium constant, represented as Kp, is essential for predicting the extent of a reaction when dealing with gases at a constant temperature. In the context of chemical equilibrium, Kp is a numerical value that provides insight into the relative quantities of reactants and products present when a reaction reaches its equilibrium state.

For a given reversible reaction involving gases, the Kp expression is formulated using the partial pressures of the gaseous components. It specifically applies to reactions where temperature remains unchanged because the equilibrium constant itself is sensitive to temperature changes. The key thing to remember is that Kp does not include solids or liquids, only gases.

The Kp for the reaction \[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_{2}(g) \]is given by \[ K_{\mathrm{p}} = \frac{{P_{\mathrm{CO}_2}}}{{1}} \]where \( P_{\mathrm{CO}_2} \) is the partial pressure of \( \mathrm{CO}_2 \) gas. In our problem, we were given that Kp equals 1.16 at 800°C. By using this constant, we were able to solve the problem and determine what percentage of calcium carbonate reacts in a closed system.
Le Chatelier's Principle
Le Chatelier's principle provides a predictive quality to chemical reactions when a system at equilibrium is subjected to changes such as concentration, pressure, volume, or temperature. It states that if an external stress is applied to a system in equilibrium, the system will adjust itself to minimize the impact of that disturbance.

This principle explains how shifts in equilibrium occur. For instance, if we increase the pressure of a system by decreasing the volume, a reaction that produces fewer moles of gas is favored. Oppositely, if the pressure is decreased, the reaction moves toward producing more gas.

Temperature changes are also significant. For exothermic reactions, an increase in temperature shifts the equilibrium towards the reactants, while for endothermic reactions, the equilibrium shifts towards the products. In the solution given, if we were to change the container size or the temperature, Le Chatelier's principle would guide us to anticipate how the equilibrium would shift. However, since we maintained constant conditions, the reaction's behavior was straightforward and predictable using Kp.
Partial Pressure
Partial pressure is a concept in chemistry that refers to the pressure a single gas in a mixture of gases would exert if it occupied the entire volume of the mixture at the same temperature. It plays a crucial role in calculating Kp since Kp is derived from the partial pressures of the gaseous reactants and products.

In our exercise, the partial pressure of \( \mathrm{CO}_2 \), \( P_{\mathrm{CO}_2} \), is a key element in determining the equilibrium state of the reaction. It is directly proportional to the number of moles of \( \mathrm{CO}_2 \) and can be calculated using the ideal gas law: \[ P_{\mathrm{CO}_2} = \frac{{n_{\mathrm{CO}_2}}}{{V}} \cdot R \cdot T \]With this relationship, we can link moles of gas to pressure, which enables us to use the equilibrium constant effectively. It's also worth noting that the partial pressure is independent of the presence of other gases, which allows for straightforward calculations in a mixture.
ICE Table Method
The ICE table method is a systematic way to organize and calculate changes in concentrations or pressures of reactants and products throughout a chemical reaction. ICE stands for Initial, Change, and Equilibrium, each representing a state of the reaction components.

Initially, you list the initial concentrations (or pressures) of reactants and products. 'Change' denotes the changes these components undergo as the reaction moves towards equilibrium. Lastly, 'Equilibrium' states the final concentrations once equilibrium is reached. By creating an ICE table, we easily visualize and solve for unknown variables in equilibrium reactions.

In the problem, we created an ICE table that helped us solve for the unknown value of x, which represented the change in moles of \( \mathrm{CaCO}_3 \). After setting up the table, we incorporated the given Kp value and partial pressure calculations to solve for the equilibrium concentrations, and in turn, determine the percentage of \( \mathrm{CaCO}_3 \) that reacted. The ICE table is an invaluable tool in equilibrium problems like these and greatly simplifies complex calculations.

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

At a particular temperature, \(K=2.0 \times 10^{-6}\) for the reaction $$ 2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) $$ If \(2.0 \mathrm{~mol} \mathrm{CO}_{2}\) is initially placed into a 5.0-L vessel, calculate the equilibrium concentrations of all species.

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\). After equilibrium is reached the total pressure is \(1.5\) atm and \(16 \%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g)\). a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C}\). b. If the volume of the cylinder is increased until the total pressure is \(1.0 \mathrm{~atm}\) (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g)\). c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{~atm}\) )?

Trans fats, or partially hydrogenated cooking oils, result from efforts of food manufacturers to increase the shelf life of many different forms of cooking oils. The free-flowing oils are treated with hydrogen, a catalyst, and high heat to change the structure of the oil, raising the melting point. The now solid oil is much more stable toward spoilage. A serious drawback to this process has been the realization that the trans form of the fat poses serious health risks when compared with the naturally occurring cis forms of the oils. The part of the hydrogenation process that results in the formation of trans fats can be summarized as follows: $$ \text { cis fat }(a q)+\mathrm{H}_{2}(a q) \rightleftharpoons \operatorname{trans} \operatorname{fat}(a q) $$ A typical value of \(K\) for this type of reaction is about \(5.0 .\) If we start with \(0.10 \mathrm{~mol} / \mathrm{L}\) each of cis fat and \(\mathrm{H}_{2}\), calculate the equilibrium concentrations.

In which direction will the position of the equilibrium $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ be shifted for each of the following changes? a. \(\mathrm{H}_{2}(g)\) is added. b. \(\mathrm{I}_{2}(g)\) is removed. c. \(\mathrm{HI}(g)\) is removed. d. In a rigid reaction container, some \(\operatorname{Ar}(g)\) is added. e. The volume of the container is doubled. f. The temperature is decreased (the reaction is exothermic).

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