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Decomposition of ammonium dichromate is shown in the designated series of photos. In a closed container this process reaches an equilibrium state. Write a balanced chemical equation for the equilibrium reaction. How is the equilibrium affected if (a) more ammonium dichromate is added to the equilibrium system? (b) more water vapor is added? (c) more chromium(III) oxide is added?

Short Answer

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(a) Shifts right; (b) Shifts left; (c) No effect.

Step by step solution

01

Write the decomposition reaction

The decomposition reaction of ammonium dichromate is shown by the equation: \[ \text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7 (s) \rightarrow \text{Cr}_2\text{O}_3 (s) + \text{N}_2 (g) + 4\text{H}_2\text{O} (g) \] This represents the ammonium dichromate breaking down into chromium(III) oxide, nitrogen gas, and water vapor.
02

Consider the equilibrium state

In the closed container, the reaction establishes an equilibrium between the reactants and products. Therefore, the equilibrium reaction would be: \[ \text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7 (s) \leftrightarrow \text{Cr}_2\text{O}_3 (s) + \text{N}_2 (g) + 4\text{H}_2\text{O} (g) \] This equation is balanced as written.
03

Analyze the effect of adding ammonium dichromate

Adding more ammonium dichromate (\(\text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7\)) to the system increases the concentration of reactants. According to Le Chatelier's principle, the equilibrium will shift to the right, favoring the formation of products \(\text{Cr}_2\text{O}_3\), \(\text{N}_2\), and \(\text{H}_2\text{O}\).
04

Analyze the effect of adding water vapor

Adding more water vapor (\(\text{H}_2\text{O}\)) will increase the concentration of gaseous products. According to Le Chatelier's principle, the equilibrium shifts to the left to counteract the change, favoring the decomposition of \(\text{Cr}_2\text{O}_3\), \(\text{N}_2\), and \(\text{H}_2\text{O}\) back to ammonium dichromate.
05

Analyze the effect of adding chromium(III) oxide

Adding more chromium(III) oxide \(\text{Cr}_2\text{O}_3\) does not affect the equilibrium because \(\text{Cr}_2\text{O}_3\) is a solid, and changes in the amount of a pure solid do not affect the position of equilibrium in a reaction.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in the study of chemical equilibria. It helps predict how a change in conditions affects a chemical equilibrium. When a system at equilibrium experiences a change in concentration, temperature, or pressure, it will adjust to minimize that change and achieve a new equilibrium state.
  • If a reactant is added, the equilibrium shifts to the right, favoring product formation.
  • If a product is added, the equilibrium shifts to the left, favoring reactant formation.
  • Changes in concentration of solids or liquids typically do not affect equilibrium, whereas changes in gases do.
In the case of ammonium dichromate decomposition, adding more ( ext{NH}_4 ext{)}_2 ext{Cr}_2 ext{O}_7 ext{)} will shift the equilibrium towards the formation of more products, while adding more water vapor shifts it the opposite way. This principle provides a powerful tool for predicting the direction of the shift based on various stressors.
Ammonium Dichromate Decomposition
Ammonium dichromate, ( ext{NH}_4 ext{)}_2 ext{Cr}_2 ext{O}_7, is an interesting compound due to its vivid decomposition reaction, which can be demonstrated through a striking volcanic-like eruption. During decomposition, it breaks into chromium(III) oxide ( ext{Cr}_2 ext{O}_3 ), nitrogen gas ( ext{N}_2 ), and water vapor ( 4 ext{H}_2 ext{O} ). This decomposition reaction is often used as a dramatic demonstration of chemical change.
In a closed container, the reaction reaches a dynamic equilibrium, meaning the rate of decomposition of ammonium dichromate equals the rate of formation from its decomposition products. Observing this process helps to understand the principles of equilibrium and reaction dynamics in a controlled environment. The reaction also serves as an excellent illustration for Le Chatelier's principle in action, particularly when observing shifts in equilibrium due to changes in reactant or product concentrations.
Balanced Chemical Equation
A balanced chemical equation represents the transmutation of reactants into products while maintaining the law of conservation of mass. This means the number of atoms of each element must remain constant through the reaction. For the decomposition of ammonium dichromate, the equation is:\[ \text{(NH}_4\text{)}_2\text{Cr}_2\text{O}_7\ (s) \leftrightarrow \text{Cr}_2\text{O}_3\ (s) + \text{N}_2\ (g) + 4\text{H}_2\text{O}\ (g) \]This balanced equation shows the breakdown of one mole of solid ammonium dichromate into one mole of chromium(III) oxide, one mole of nitrogen gas, and four moles of water vapor.
Properly balancing chemical equations is crucial in chemistry as it ensures stoichiometric relationships, allowing for predictions of amounts of reactants needed or products formed. Such representations are not only vital for understanding chemical reactions but also for qualitative analysis and calculations in real-world applications.

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Most popular questions from this chapter

If an equilibrium is product-favored, is its equilibrium constant large or small with respect to \(1 ?\) Explain.

Hydrogen, bromine, and \(\mathrm{HBr}\) in the gas phase are in equilibrium in a container of fixed volume. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) \quad \Delta_{r} H^{\circ}=-103.7 \mathrm{~kJ} / \mathrm{mol}\) How will each of these changes affect the indicated quantities? Write "increase," "decrease," or "no change." \begin{tabular}{l} \hline Change & {\(\left[\mathrm{Br}_{2}\right]\)} & {\([\mathrm{HBr}]\)} & \(K_{c}\) & \(K_{\mathrm{p}}\) \\ \hline Some \(\mathrm{H}_{2}\) is added to the \\ container. \\ The temperature of the gases \\ in the container is increased. \\ The pressure of \(\mathrm{HBr}\) is \\ increased. \end{tabular}

Carbon dioxide reacts with carbon to give carbon monoxide according to the equation $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ At \(700 .{ }^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) sealed flask at equilibrium contains $$ 0.10 \mathrm{~mol} \mathrm{CO}, 0.20 \mathrm{~mol} \mathrm{CO}_{2}, \text { and } 0.40 \mathrm{~mol} \mathrm{C} . \text { Calculate } $$ the equilibrium constant \(K_{\mathrm{P}}\) for this reaction at the specified temperature.

Consider the decomposition of ammonium hydrogen sulfide: $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) $$ In a sealed flask at \(25^{\circ} \mathrm{C}\) are \(10.0 \mathrm{~g} \mathrm{NH}_{4} \mathrm{HS},\) ammonia with a partial pressure of \(0.692 \mathrm{~atm}\), and \(\mathrm{H}_{2} \mathrm{~S}\) with a partial pressure of \(0.0532 \mathrm{~atm}\). When equilibrium is established, it is found that the partial pressure of ammonia has increased by \(12.4 \% .\) Calculate \(K_{\mathrm{P}}\) for the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) at \(25^{\circ} \mathrm{C}\).

The value of \(K_{\mathrm{c}}\) is \(3.7 \times 10^{-23}\) at \(25^{\circ} \mathrm{C}\) for $$ \mathrm{C}(\text { graphite })+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}) $$ Describe what will happen if \(3.5 \mathrm{~mol} \mathrm{CO}\) and \(3.5 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are mixed in a 1.5-L sealed graphite container with a suitable catalyst so that the reaction rate is rapid at this temperature.

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