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Write the expressions for the equilibrium constants \(K_{P}\) of these thermal decompositions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
(a) \( K_{P,a} = P_{CO_2} \cdot P_{H_2O} \); (b) \( K_{P,b} = (P_{SO_2})^2 \cdot P_{O_2} \).

Step by step solution

01

Identify the Gaseous Components

In both reactions, identify the species that are in the gas phase because these will contribute to the equilibrium constant in terms of pressure, \( K_P \). In reaction (a), \( \mathrm{CO}_{2}(g) \) and \( \mathrm{H}_{2} \mathrm{O}(g) \) are the gaseous components. In reaction (b), \( \mathrm{SO}_{2}(g) \) and \( \mathrm{O}_{2}(g) \) are the gases.
02

Write the Equilibrium Constant Expression for Reaction (a)

For the reaction \( 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{CO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(g) \), the equilibrium constant \( K_P \) is given by the partial pressures of the gaseous products, since solids are not included in \( K_P \). The expression is: \[ K_{P,a} = P_{CO_2} \cdot P_{H_2O} \] where \( P_{CO_2} \) and \( P_{H_2O} \) are the partial pressures of carbon dioxide and water vapor respectively.
03

Write the Equilibrium Constant Expression for Reaction (b)

For the reaction \( 2 \mathrm{CaSO}_{4}(s) \rightleftharpoons 2 \mathrm{CaO}(s) + 2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \), only the gaseous products contribute to \( K_P \). Therefore, the expression is: \[ K_{P,b} = (P_{SO_2})^2 \cdot P_{O_2} \] where \( P_{SO_2} \) and \( P_{O_2} \) are the partial pressures of sulfur dioxide and oxygen respectively.

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Key Concepts

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

K_P
The equilibrium constant, known as \(K_P\), is crucial when dealing with gases in chemical reactions. It helps in understanding the extent of a reaction under set conditions of pressure and temperature.
In the case of reactions involving gases, \(K_P\) is calculated using the partial pressures of the reactants and products in their gaseous forms.
This concept becomes particularly significant in thermal decomposition reactions, where solid reactants decompose to produce gases.
  • Solids and liquids do not appear in the \(K_P\) expression, which significantly simplifies the calculation.
  • Only gaseous components are considered, as they contribute to the system's pressure.
For instance, in a reaction that forms \(\mathrm{CO_{2}(g)}\) and \(\mathrm{H_{2}O(g)}\), \(K_P\) would be expressed using these gases' partial pressures. Formulating \(K_P\) requires knowledge of these pressures, providing insight into how product or reactant-favored the equilibrium state is.
thermal decomposition
Thermal decomposition refers to the process where a chemical compound breaks down into simpler compounds or elements when heated. This is a common type of reaction in which heat acts as the catalyst to drive the decomposition.
It often involves solid reactants decomposing into gaseous products.
  • The decomposition of certain compounds requires specific conditions such as high temperatures.
  • The formation of gaseous products can often be predicted based on the initial composition of the reacting materials.
An example is the decomposition of sodium bicarbonate (\(\mathrm{NaHCO_{3}}\)), which upon heating breaks down to form sodium carbonate (\(\mathrm{Na_{2}CO_{3}}\)), carbon dioxide (\(\mathrm{CO_{2}}\)) gas, and water vapor (\(\mathrm{H_{2}O(g)}\)).
In this type of reaction, measuring and expressing the equilibrium constant (\(K_P\)) is essential as it indicates how much product is formed under equilibrated conditions.
partial pressure
Partial pressure is an essential concept when discussing gases in a chemical reaction, especially when calculating \(K_P\).
Partial pressure refers to the pressure that a single gas in a mixture would exert if it alone occupied the entire volume. It essentially measures a gas's contribution to the total pressure of the system.
To calculate an equilibrium constant \(K_P\), it's important to determine the partial pressures of all gaseous reactants and products involved.
  • Each gas's partial pressure in a mixture can be found using the ideal gas law or by knowing its mole fraction in the mixture multiplied by total pressure.
  • Partial pressures are directly proportional to the concentration of gases, given constant temperature and volume.
In the example of the thermal decomposition reactions provided, the partial pressures of \(\mathrm{CO_{2}}\), \(\mathrm{H_{2}O}\), \(\mathrm{SO_{2}}\), and \(\mathrm{O_{2}}\) are extracted from total pressure to compute \(K_P\). Understanding partial pressures helps predict how changes in conditions (like temperature or volume) might affect the reaction's equilibrium position.
gaseous phase
The gaseous phase is one of the most dynamic environments for chemical reactions. It involves molecules in constant random motion, leading to frequent collisions and reactions.
In the context of equilibrium constants, only species in the gaseous phase contribute to \(K_P\).
This distinction is because gases expand to fill the volume of their container, creating pressure that directly influences equilibrium.
  • Gaseous reactants or products play a major role in the establishment and maintenance of a chemical reaction's equilibrium.
  • Their behavior can be described via the ideal gas law, providing a framework to predict outcomes.
When considering the reactions from the exercise, identifying gaseous products like \(\mathrm{CO_{2}}\), \(\mathrm{H_{2}O}\), \(\mathrm{SO_{2}}\), and \(\mathrm{O_{2}}\) allows us to accurately write \(K_P\).
Gases' unique properties mean that variations in pressure and volume can dramatically alter the ratio of reactants to products at equilibrium, making the gaseous phase crucial for understanding reaction dynamics.

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Most popular questions from this chapter

A quantity of 1.0 mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was introduced into an evacuated vessel and allowed to attain equilibrium at a certain temperature $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ The average molar mass of the reacting mixture was \(70.6 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of the gases. (b) Calculate \(K_{P}\) for the reaction if the total pressure was 1.2 atm. (c) What would be the mole fractions if the pressure were increased to 4.0 atm by reducing the volume at the same temperature?

When heated, ammonium carbamate decomposes as $$\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)$$ At a certain temperature the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{P}\) for the reaction.

One mole of \(\mathrm{N}_{2}\) and 3 moles of \(\mathrm{H}_{2}\) are placed in a flask at \(397^{\circ} \mathrm{C}\). Calculate the total pressure of the system at equilibrium if the mole fraction of \(\mathrm{NH}_{3}\) is found to be 0.21. The \(K_{P}\) for the reaction is \(4.31 \times 10^{-4}\).

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}:\) $$4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g)$$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a \(2.0-\mathrm{L}\) flask at \(1024^{\circ} \mathrm{C} .\) (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

Define homogeneous equilibrium and heterogeneous equilibrium. Give two examples of each.

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