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A sample of solid ammonium chloride was placed in an evacuated container and then heated so that it decomposed to ammonia gas and hydrogen chloride gas. After heating, the total pressure in the container was found to be 4.4 atm. Calculate \(K_{\mathrm{p}}\) at this temperature for the decomposition reaction $$\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$$,

Short Answer

Expert verified
The short answer based on the step-by-step solution is as follows: For the decomposition reaction \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HCl}(g)\), the partial pressures of ammonia and hydrogen chloride gases are found to be 2.2 atm each. The expression for the equilibrium constant \(K_p\) is given by \(K_p = [\mathrm{NH}_3(g)] \cdot [\mathrm{HCl}(g)]\). Substituting the partial pressures of ammonia and hydrogen chloride, we find \(K_p = (2.2 \: \mathrm{atm}) \cdot (2.2 \: \mathrm{atm}) = 4.84 \: \mathrm{atm^2}\). Therefore, the equilibrium constant \(K_p\) for this decomposition reaction is \(4.84 \: \mathrm{atm^2}\).

Step by step solution

01

Write the balanced equation for the decomposition reaction.

We are given the balanced equation for the decomposition reaction as \[ \mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g) + \mathrm{HCl}(g) \]
02

Calculate the partial pressures of the gases involved.

The pressure in the container after the reaction reached equilibrium is given as 4.4 atm, which is the sum of the partial pressures of ammonia and hydrogen chloride gases. As one mole of ammonia is produced for each mole of hydrogen chloride, their partial pressures must be equal. Let's denote the partial pressure of ammonia and hydrogen chloride gases as \(x\). Then, we have \( 2x = 4.4 \: \mathrm{atm} \) Now solve for \(x\): \( x = \frac{4.4 \: \mathrm{atm}}{2} = 2.2 \: \mathrm{atm} \) Therefore, the partial pressures of ammonia and hydrogen chloride are 2.2 atm each.
03

Write the expression for the equilibrium constant \(K_p\).

The equilibrium constant \(K_p\) can be expressed as the product of the partial pressures of the product gases divided by the partial pressure of the reactant. However, since the reactant is a solid, it does not have a partial pressure and can be treated as 1. Thus, the expression for \(K_p\) can be written as \[ K_p = [\mathrm{NH}_3(g)] \cdot [\mathrm{HCl}(g)] \]
04

Use the partial pressures obtained in Step 2 to calculate \(K_p\).

Substitute the partial pressures of ammonia and hydrogen chloride (2.2 atm each) into the expression for \(K_p\), then calculate the value of \(K_p\) as follows: \[ K_p = (2.2 \: \mathrm{atm}) \cdot (2.2 \: \mathrm{atm}) = 4.84 \: \mathrm{atm^2} \] Thus, the equilibrium constant \(K_p\) for the given decomposition reaction is 4.84 atm\(^2\).

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Key Concepts

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

Chemical Equilibrium
Imagine a dance where both partners move in harmony with each other, dancing forwards and backwards with no one ever taking the lead. This dance is a metaphor for chemical equilibrium, an important concept in chemistry that occurs when the forward and reverse reactions in a chemical process happen at equal rates. I think of it as a perfectly balanced scale, with the reactants and products reaching a state where their concentrations remain constant over time, even if they are not equal.

Understanding equilibrium involves recognizing that it is a dynamic process—chemical reactions continue to occur, but since the rates are equal, there are no net changes in the concentrations of the reactants and products. This balanced state can be described by an equilibrium constant, symbolized as K for reactions based on concentration, or as Kp for reactions involving gases and their partial pressures. The exact value of the equilibrium constant at a given temperature provides insight into the position of equilibrium and tells us the relative amounts of products and reactants present in the mixture.
Partial Pressures
When gases mix, like in a balloon or our atmosphere, each one exerts a pressure as if it were alone in the container—this is what we refer to as partial pressure. It’s as if each gas carries a part of the total pressure burden. This pressure contributed by a single type of gas in a mixture is proportionate to its mole fraction—in simple terms, its share of the total number of gas particles present.

Diving deeper into this concept, Dalton’s Law of Partial Pressures comes into play, stating that the total pressure of a gas mixture is the sum of the partial pressures of each individual component. In a reaction where gases are produced or consumed, the partial pressures can tell you a lot about the reaction's progression and, as we already know, they're crucial in calculating the equilibrium constant Kp in terms of pressure. Always keep in mind the power of this law when unraveling the mysteries of gas-related equilibrium!
Decomposition Reaction
When you hear the word 'decomposition', think of breaking down or splitting up. In chemistry, a decomposition reaction is a process where a single compound breaks down into two or more simpler substances. This can happen through various means like heating, as seen with our friend ammonium chloride turning into ammonia and hydrogen chloride.

These reactions can be simple, just like our example, or become quite complex, leading to multiple products and intricate equilibrium scenarios. What's captivating about these reactions is how they reflect nature's tendency to break down complex structures into simpler ones. Remember, in decomposition, complex molecules are effectively 'undoing' their formation process, which serves as an essential cycle in chemical reactions and the broader ecosystem.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C},\) equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g):\).$$\mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g)$$.A 2.00 -g sample of XY (molar mass \(=165 \mathrm{g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at 0.967 atm. As XY begins to dissociate, the piston moves until 35.0 mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).

Consider the decomposition equilibrium for dinitrogen pentoxide: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2(g)+\mathrm{O}_{2}(g)$$.At a certain temperature and a total pressure of 1.00 atm, the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0,\) will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to 0.50\%? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of 10.0

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{K},\) where the following reactions occurred:$$\begin{array}{c}2 \mathrm{FeSO}_{4}(s) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{array}$$.After equilibrium was reached, the total pressure was 0.836 atm and the partial pressure of oxygen was 0.0275 atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

A 1.604 -g sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and 6.400 g oxygen gas are sealed into a 2.50 -L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is 0.326 atm, and the pressure of water vapor is 4.45 atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

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