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Consider the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}_{\text {thombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C},\) predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{atm}\) c. \(P_{\mathrm{NO}_{2}}=0.29 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm}\)

Short Answer

Expert verified
a. No shift in either direction, the reaction is already at equilibrium. b. The reaction will shift to the left, favoring the formation of NO2. c. The reaction will shift to the left, favoring the formation of NO2.

Step by step solution

01

a. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = P_{N_2O_4} = 1.0\;atm\). According to the balanced equation for Reaction 1, two moles of NO2 are being converted into one mole of N2O4. We can calculate the reaction quotient (Q) for Reaction 1: Reaction 1: \(Q=[N_2O_4]/[NO_2]^2\) Substitute the given pressures: \(Q=(1.0)/(1.0)^2=1\) Compare Q and K_c (K_c is also 1 for this Reaction at 25°C). Since \(Q=K_c\), the reaction is already at equilibrium, and there will be no shift in either direction.
02

b. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = 0.21\;atm\) and \(P_{N_2O_4} = 0.50\;atm\). Calculate the reaction quotient (Q) for Reaction 1: Substitute the given pressures: \(Q=(0.50)/(0.21)^2=11.30\) Compare Q and K_c. Since \(Q>K_c\), the reaction will shift to the left, favoring the formation of NO2.
03

c. Determine the direction for Reaction 1 with given pressures.

Here, the initial pressures are \(P_{NO_2} = 0.29\;atm\) and \(P_{N_2O_4} = 1.6\;atm\). Calculate the reaction quotient (Q) for Reaction 1: Substitute the given pressures: \(Q=(1.6)/(0.29)^2=18.97\) Compare Q and K_c. Since \(Q>K_c\), the reaction will shift to the left, favoring the formation of NO2. From the given exercise, only Reaction 1 had partial pressures to analyze. Thus, the directions for Reaction 1 have been predicted based on the partial pressures provided.

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Key Concepts

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

Reaction Quotient
The reaction quotient, denoted as \( Q \), is a valuable tool in chemical equilibrium. It helps us determine the current state of a reaction relative to its equilibrium state. By using the reaction quotient, we can assess whether a reaction is at equilibrium or if it will shift towards the reactants or products.
  • The expression for \( Q \) is similar to the equilibrium constant expression, \( K_c \). It uses the current concentrations or partial pressures of the chemical species involved in a reaction.
  • For a balanced chemical reaction like \( 2\, \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \), the expression for the reaction quotient is \( Q = \frac{P_{\text{N}_2\text{O}_4}}{P_{\text{NO}_2}^2} \).
By substituting the given partial pressures into the \( Q \) expression, we can compare \( Q \) to \( K_c \):
  • If \( Q = K_c \), the system is at equilibrium, meaning no shift is needed.
  • If \( Q < K_c \), the reaction will shift to the right, producing more products.
  • If \( Q > K_c \), the reaction will shift to the left, forming more reactants.
This process offers a thorough understanding of reactions under changing conditions, helping predict the direction to reach equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle provides insight into how a chemical system at equilibrium responds to external changes. It states that if you disturb a system at equilibrium, the reaction will shift in a direction that counteracts the disturbance, thereby restoring equilibrium conditions.
  • For shifts due to concentration changes, increasing the concentration or pressure of a reactant usually results in an equilibrium shift towards the products.
  • Conversely, increasing the concentration of a product may force the system to shift towards the reactants, opposing the change.
In the exercise, when the reaction quotient \( Q \) was greater than \( K_c \), Le Chatelier's Principle predicted a shift towards the reactants. This shift balances the excess concentration of products. Understanding this principle allows students to predict how different scenarios influence the equilibrium state. Furthermore, knowing how Le Chatelier's Principle applies in chemical reactions is essential for various applications, including industrial processes and natural phenomena.
Gas Phase Reactions
Gas phase reactions are chemical reactions that involve substances in their gaseous state. These reactions are significant because gases uniformly mix, allowing reactions to occur throughout the entire mixture. In the context of equilibrium, there are special considerations to keep in mind:
  • Partial pressures often replace concentration measures when discussing gas phase reactions in equilibrium. This is because the pressure of a gas is directly proportional to its concentration when temperature and volume are constant.
  • In an equilibrium setup involving gas phase reactions, changes in conditions such as pressure, temperature, or volume, can significantly impact the reaction's direction.
  • For ammonium reactions, the equilibrium constant, \( K_p \), is often referenced. This accounts for the pressures of the gases involved, offering a depiction of the system's equilibrium state based on pressure metrics.
Understanding gas phase reactions and their dependence on partial pressures is beneficial for grasping how reactions behave under different conditions, influencing industrial practices like the Haber process or even natural geological phenomena.

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Most popular questions from this chapter

The value of \(\Delta G^{\circ}\) for the reaction $$2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+13 \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(l)$$ is \(-5490 .\) kJ. Use this value and data from Appendix 4 to calculate the standard free energy of formation for \(\mathrm{C}_{4} \mathrm{H}_{10}(g).\)

Consider the reaction: $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=-92.50 \mathrm{kJ}\) Which of the following statements is(are) true? a. This is an endothermic reaction. b. \(\Delta S^{\circ}\) for this reaction is negative. c. If the temperature is increased, the ratio \(\frac{\mathrm{PCl}_{5}}{\mathrm{PCl}_{3}}\) will increase. d. \(\Delta G^{\circ}\) for this reaction has to be negative at all temperatures. e. When \(\Delta G^{\circ}\) for this reaction is negative, then \(K_{\mathrm{p}}\) is greater than 1.00.

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C},\) calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 79.)

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

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