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Consider the system \(4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-1530.4 \mathrm{~kJ}\) (a) How will the amount of ammonia at equilibrium be affected by 1\. removing \(\mathrm{O}_{2}(g) ?\) 2\. adding \(\mathrm{N}_{2}(g) ?\) 3\. adding water? 4\. expanding the container at constant pressure? 5\. increasing the temperature? (b) Which of the above factors will increase the value of \(K ?\) Which will decrease it?

Short Answer

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**Question**: Based on the given reaction, determine the effect of various factors on the equilibrium position and the equilibrium constant, K. **Short Answer**: 1. Removing O2 (g) from the system will increase the amount of ammonia at equilibrium and increase the value of K. 2. Adding N2 (g) to the system will decrease the amount of ammonia at equilibrium and decrease the value of K. 3. Adding water will not affect the amount of ammonia at equilibrium or the value of K. 4. Expanding the container at constant pressure will increase the amount of ammonia at equilibrium and increase the value of K. 5. Increasing the temperature will increase the amount of ammonia at equilibrium but will not determine the direction in which K changes in this case.

Step by step solution

01

Part a - Effect of removing O2 (g)

According to Le Chatelier's Principle, if a change is applied to a system in equilibrium, the system will adjust in such a way as to counteract that change and restore the equilibrium. In this case, removing O2 will cause the reaction to shift to the left, consuming N2 and H2O and producing NH3 and O2. Thus, the amount of ammonia at equilibrium will increase.
02

Part a - Effect of adding N2 (g)

Adding N2 to the system will cause the reaction to shift to the right, consuming NH3 and O2 and producing N2 and H2O. This will decrease the amount of ammonia at equilibrium.
03

Part a - Effect of adding water

Since water is a liquid and does not appear in the equilibrium constant expression, adding water to the system will have no effect on the equilibrium position. Therefore, the amount of ammonia at equilibrium will not change in this case.
04

Part a - Effect of expanding the container at constant pressure

Expanding the container lowers the concentration of all reactants and products. According to Le Chatelier's Principle, the reaction will shift towards the side with more moles of gaseous species to counteract this change. In this case, the left side (reactant side) has more moles of gaseous species, so the reaction will shift to the left. This will result in an increase in the amount of ammonia at equilibrium.
05

Part a - Effect of increasing the temperature

Since the reaction is exothermic (ΔH is negative), increasing the temperature will cause the reaction to shift towards the side that absorbs heat, which is the endothermic direction (where ΔH is positive), or to the left in this case. This will result in an increase in the amount of ammonia at equilibrium.
06

Part b - Factors that increase the value of K

The value of K will increase if the forward reaction is favored. In this case, removing O2 (g) and expanding the container at constant pressure will favor the formation of NH3 and cause K to increase.
07

Part b - Factors that decrease the value of K

The value of K will decrease if the reverse reaction is favored. In this case, adding N2 (g) will favor the consumption of NH3 and cause K to decrease. Note that adding water does not affect K, and changes in temperature will affect K, but we have not been asked to determine whether K increases or decreases with increasing temperature in this case.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction in a closed system. When a system reaches equilibrium, the concentrations of reactants and products remain constant over time, but this does not mean they are equal. It is a dynamic state where the concentrations of all reactants and products are steady, and the reactions still occur, but with no net change.

Coming to the example provided, when the system comprising of ammonia, oxygen, nitrogen, and water reaches equilibrium, the formation of nitrogen and water from the reaction of ammonia with oxygen balances with the backward reaction of nitrogen and water forming ammonia and oxygen. This is a classic example of a gaseous reaction equilibrium, where the behavior of the system can be predicted using Le Chatelier's Principle.
Equilibrium Constant (K)
The equilibrium constant, denoted as K, is a value that quantifies the ratio of the concentrations of products to reactants at equilibrium. Each equilibrium reaction has a unique K value at a given temperature. It is calculated using an expression where the concentrations of the products are raised to the power of their stoichiometric coefficients and divided by the similar product of reactant concentrations.

For the reaction in the exercise, the equilibrium constant expression would not include liquid water as it does not affect the concentration of dissolved species. Thus, changes in the amounts of gaseous reactants or products can shift the position of equilibrium, influencing the K value. As mentioned in the solution, removing oxygen or expanding the container shifts the equilibrium to the left, increasing the K value, since these actions favor the production of reactants. Conversely, adding nitrogen shifts the equilibrium to the right, favoring products, and this lowers the K value.
Effect of Temperature on Equilibrium
Temperature is a crucial factor in the behavior of equilibrium systems. According to Le Chatelier's Principle, when a system at equilibrium undergoes a temperature change, the equilibrium position will shift to counteract that change.

In the given exercise, the reaction is exothermic, which means it releases heat. Increasing the temperature adds heat to the system and the equilibrium will shift to favor the endothermic direction, or the reverse reaction, in an attempt to absorb the excess heat. This shift to the left will increase the concentration of reactants at equilibrium. It is also crucial to understand that, unlike other factors affecting equilibrium, changes in temperature actually change the value of K. In an exothermic reaction, increasing the temperature results in a decrease of K; for endothermic reactions, increasing the temperature would increase K.

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Most popular questions from this chapter

Sulfur oxychloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), decomposes to sulfur dioxide and chlorine gases. $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ At a certain temperature, the equilibrium partial pressures of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) are \(1.88 \mathrm{~atm}, 0.84 \mathrm{~atm},\) and \(0.27 \mathrm{~atm}\) respectively. (a) What is \(K\) at that temperature? (b) Enough \(\mathrm{Cl}_{2}\) condenses to reduce its partial pressure to 0.68 atm. What are the partial pressures of all gases when equilibrium is reestablished?

For the system $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ \(K\) is 26 at \(300^{\circ} \mathrm{C}\). In a \(10.0-\mathrm{L}\) flask at \(300^{\circ} \mathrm{C}\), a gaseous mixture consists of all three gases with the following partial pres- $$ \text { sures: } P_{\mathrm{PCl}_{5}}=0.026 \mathrm{~atm}, P_{\mathrm{PCl}_{3}}=0.65 \mathrm{~atm}, P_{\mathrm{Cl}_{2}}=0.33 \mathrm{~atm} $$ (a) Is the system at equilibrium? Explain. (b) If the system is not at equilibrium, in which direction will the system move to reach equilibrium?

At \(800^{\circ} \mathrm{C}, K=2.2 \times 10^{-4}\) for the following reaction $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ Calculate \(K\) at \(800^{\circ} \mathrm{C}\) for (a) the synthesis of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) from \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) gases. (b) the decomposition of one mole of \(\mathrm{H}_{2} \mathrm{~S}\) gas.

Given the following data at \(25^{\circ} \mathrm{C}\) $$ \begin{array}{ll} 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K=1 \times 10^{-30} \\ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K=8 \times 10^{1} \end{array} $$ Calculate \(K\) for the formation of one mole of NOBr from its elements in the gaseous state.

Given the following descriptions of reversible reactions, write a balanced equation (smallest whole-number coefficients) and the equilibrium constant expression for each. (a) Nickel metal reacts with carbon monoxide to form nickel tetracarbonyl \(\left(\mathrm{Ni}(\mathrm{CO})_{4}\right)\) gas. (b) Aqueous nitrous acid in equilibrium with hydrogen and nitrite ions. (c) Chlorine gas and bromide ions in equilibrium with liquid bromine and chloride ions.

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