New Day

Current Quiz

You previously generated a quiz from this deck.

Spaced Repetition

This mode helps you retain information by reviewing flashcards at optimal intervals, focusing on what needs the most reinforcement each day.

New Flashcards Per Day

🗂️
Flashcards

The standard way to view your flashcards without any extras

🧠
Exam Mode

Practice active recall and assess your answers with AI

⏳
Spaced Repetition

Memorize your flashcards using a scientifically proven method

🎯
Generate Quiz

Use AI to generate a quiz based on the same content

Equilibrium Constant Kp

A-Level AQA Chemistry

This pack of flashcards covers the principles of gas mixtures and chemical equilibrium, focusing on concepts such as partial pressure, mole fraction, calculating Kp, and the effects of temperature and pressure on equilibrium. It includes definitions, calculations, and relationships pertinent to gaseous reactions and equilibrium constants.

1 / 13

Combine

All flashcards from the current deck will be added to the deck selected from this list.

Export

This text is tab separated and can be imported into various flashcard apps.

"Define Kp in terms of pressure and mole fraction."	"Kp = (N2.P)(X3H2.P3)/(P4), where P is total pressure and X is mole fraction."
"How does increasing pressure affect the position of equilibrium in a reaction with more moles of gas on the product side?"	"Increasing pressure will shift the equilibrium to the right, increasing the mole fractions of products and decreasing the mole fractions of reactants."
"Explain the effect of pressure on the position of equilibrium and Kp in a reaction with fewer moles of gas on the product side."	"Increasing pressure will shift the equilibrium to the side with fewer moles of gas, and the value of Kp stays the same."
"What happens to the equilibrium position and Kp when the temperature is increased in an exothermic reaction?"	"The equilibrium shifts left, and the value of Kp gets smaller as there are fewer products."
"Describe the effect of changing conditions on the value of K or Kp."	"K and Kp only change with temperature. They do not change if pressure or concentration are altered. A catalyst also has no effect on K or Kp."
"Define mole fractions."	"Mole fractions are the ratio of moles of a component to the total moles in a mixture."
"Describe what partial pressure is in a gas mixture."	"Partial pressure is the pressure that a gas would have if it alone occupied the volume occupied by the whole mixture."
"Describe the relationship between total pressure and partial pressures in a gas mixture."	"In a gas mixture, the total pressure is equal to the sum of the partial pressures of all the gases present."
"How would you calculate the mole fraction of N2 in the given mixture?"	"To calculate the mole fraction of N2, divide the number of moles of N2 by the total number of moles of all gases in the mixture."
"Define the equilibrium constant, Kp."	"Kp is the equilibrium constant that relates the partial pressures of the reactants and products in a gaseous reaction at a specific temperature."
"Do solids, liquids, and aqueous substances contribute to the equilibrium constant expression, Kp?"	"No, only gases are included in the Kp expression. Solids, liquids, and aqueous substances are ignored."
"How do you calculate the partial pressure of a gas in a mixture?"	"To calculate the partial pressure of a gas, multiply its mole fraction by the total pressure of the mixture."
"Define mole fraction in the context of gas mixtures."	"Mole fraction is the ratio of the number of moles of a specific gas to the total number of moles of all gases in a mixture."

Click to copy

Image

Front

Define Kp in terms of pressure and mole fraction.

Back

Kp = (N2.P)(X3H2.P3)/(P4), where P is total pressure and X is mole fraction.

Image

Print Flashcards

[{"id":979838,"front":"Define Kp in terms of pressure and mole fraction.","back":"Kp = (N2.P)(X3H2.P3)/(P4), where P is total pressure and X is mole fraction.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":0,"explanation":"Kp is a thermodynamic equilibrium constant that relates the partial pressures of reactants and products in a chemical reaction. It is defined as the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients, divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients. In this case, Kp is defined in terms of the total pressure (P) and the mole fractions (X) of the reactants and products. The equation provided allows for the calculation of Kp using the partial pressures and mole fractions.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979837,"front":"How does increasing pressure affect the position of equilibrium in a reaction with more moles of gas on the product side?","back":"Increasing pressure will shift the equilibrium to the right, increasing the mole fractions of products and decreasing the mole fractions of reactants.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":1,"explanation":"Increasing pressure will shift the equilibrium towards the side with fewer moles of gas to counteract the change. This is in accordance with Le Chatelier's principle, which states that a system at equilibrium will adjust to minimize the effect of any imposed change, in this case, the increase in pressure.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979835,"front":"Explain the effect of pressure on the position of equilibrium and Kp in a reaction with fewer moles of gas on the product side.","back":"Increasing pressure will shift the equilibrium to the side with fewer moles of gas, and the value of Kp stays the same.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":2,"explanation":"When pressure is increased in a reaction with fewer moles of gas on the product side, the equilibrium shifts to the side with fewer gas moles to alleviate the stress. This occurs without altering the Kp value, as it's determined by the temperature and composition of the reaction.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979832,"front":"What happens to the equilibrium position and Kp when the temperature is increased in an exothermic reaction?","back":"The equilibrium shifts left, and the value of Kp gets smaller as there are fewer products.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":3,"explanation":"When temperature rises in an exothermic reaction, the system shifts to counteract the increase, favoring the reactants. This shift to the left reduces the amount of products present, leading to a decrease in Kp since it's a ratio of products to reactants.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979831,"front":"Describe the effect of changing conditions on the value of K or Kp.","back":"K and Kp only change with temperature. They do not change if pressure or concentration are altered. A catalyst also has no effect on K or Kp.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":4,"explanation":"K and Kp are equilibrium constants that represent the extent of a chemical reaction at equilibrium. They are only affected by changes in temperature, not by alterations in pressure, concentration, or the presence of a catalyst. This is due to the fundamental relationship between temperature and the equilibrium constant, as described by the Arrhenius equation.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979826,"front":"Define mole fractions.","back":"Mole fractions are the ratio of moles of a component to the total moles in a mixture.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":6,"explanation":"Mole fractions are crucial in understanding the composition of mixtures, especially in chemistry and thermodynamics. They help in calculating properties like vapor pressure and concentration, allowing for precise predictions in reactions and solutions. This concept is fundamental for stoichiometry and phase equilibria.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979813,"front":"Describe what partial pressure is in a gas mixture.","back":"Partial pressure is the pressure that a gas would have if it alone occupied the volume occupied by the whole mixture.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":8,"explanation":"Partial pressure is a concept used in gas mixtures to describe the individual pressure exerted by each gas component as if it solely occupied the total volume. This allows for the calculation of each gas's contribution to the overall pressure in the mixture.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979820,"front":"Describe the relationship between total pressure and partial pressures in a gas mixture.","back":"In a gas mixture, the total pressure is equal to the sum of the partial pressures of all the gases present.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":10,"explanation":"In a gas mixture, each gas exerts pressure independently, contributing to the total pressure. This relationship is described by Dalton's Law of Partial Pressures, stating that the total pressure is the sum of the pressures each gas would exert if it alone occupied the volume.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979819,"front":"How would you calculate the mole fraction of N2 in the given mixture?","back":"To calculate the mole fraction of N2, divide the number of moles of N2 by the total number of moles of all gases in the mixture.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":11,"explanation":"Calculating mole fraction is essential in understanding the composition of gas mixtures. It provides insight into the relative abundance of each component, which is crucial for applications in chemistry, engineering, and environmental science. This ratio helps predict behavior in reactions and physical properties.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979818,"front":"Define the equilibrium constant, Kp.","back":"Kp is the equilibrium constant that relates the partial pressures of the reactants and products in a gaseous reaction at a specific temperature.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":12,"explanation":"Kp is crucial for understanding chemical equilibria in gaseous systems. It allows chemists to predict the direction of a reaction and the extent to which reactants convert to products. The value of Kp varies with temperature, reflecting the dynamic nature of chemical reactions and their dependence on environmental conditions.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979816,"front":"Do solids, liquids, and aqueous substances contribute to the equilibrium constant expression, Kp?","back":"No, only gases are included in the Kp expression. Solids, liquids, and aqueous substances are ignored.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":14,"explanation":"In equilibrium constant expressions, only gaseous reactants and products are included because their concentrations can change significantly with pressure. Solids and liquids have constant concentrations in a given system, making their contributions negligible in the context of equilibrium calculations.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979815,"front":"How do you calculate the partial pressure of a gas in a mixture?","back":"To calculate the partial pressure of a gas, multiply its mole fraction by the total pressure of the mixture.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":15,"explanation":"The partial pressure of a gas in a mixture is determined by considering the proportion of that gas in the overall mixture. This calculation method is based on Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas present.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}},{"id":979814,"front":"Define mole fraction in the context of gas mixtures.","back":"Mole fraction is the ratio of the number of moles of a specific gas to the total number of moles of all gases in a mixture.","front_language":"English","back_language":"English","front_image":"","back_image":"","order":16,"explanation":"Mole fraction is crucial in understanding gas behavior, particularly in calculations involving partial pressures and reactions. It helps in determining the contribution of each gas to the overall properties of the mixture, such as density and reactivity, which are essential in fields like chemistry and environmental science.","deleted_at":null,"flashcard_deck_id":"m9Yyg","original_id":null,"examMode":{"text":"","correct":"","incorrect":"","info":""}}]

Options

Arrange Cards

Chronological

Random

Card Direction

Front First

Back First

Share this deck

Click to copy

Delete this deck

Are you sure you want to delete this deck? This cannot be undone.

Equilibrium Constant Kp Flashcards

Sign Up

New Day

Current Quiz

Spaced Repetition

New Flashcards Per Day

🗂️
Flashcards

🧠
Exam Mode

⏳
Spaced Repetition

🎯
Generate Quiz

Equilibrium Constant Kp

A-Level AQA Chemistry

Combine

Export

Front

Back

Print Flashcards

Print Type

Font Size

Rows Per Page

Options

Arrange Cards

Card Direction

Share this deck

Delete this deck

Sign Up

Equilibrium Constant Kp Flashcards

Sign Up

New Day

Current Quiz

Spaced Repetition

New Flashcards Per Day

🗂️ Flashcards

🧠 Exam Mode

⏳ Spaced Repetition

🎯 Generate Quiz

Equilibrium Constant Kp

A-Level AQA Chemistry

Combine

Export

Front

Back

Print Flashcards

Print Type

Font Size

Rows Per Page

Options

Arrange Cards

Card Direction

Share this deck

Delete this deck

Sign Up

🗂️
Flashcards

🧠
Exam Mode

⏳
Spaced Repetition

🎯
Generate Quiz