/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 The equilibrium constant \(K_{\m... [FREE SOLUTION] | 91Ó°ÊÓ

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The equilibrium constant \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. \(P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}\) b. \(P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}\) c. \(P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}\)

Short Answer

Expert verified
For the given sets of conditions: a. The reaction will shift to the right. b. The reaction will shift to the left. c. The system is at equilibrium.

Step by step solution

01

Write the expression for Qp

The equation for Qp is given by: $$Q_p = \frac{P_{N_2} \times P_{O_2}}{P_{NO}^2}$$ Now, we will calculate the value of Qp for each set of conditions and compare it to the given equilibrium constant (Kp) to determine if the system is at equilibrium and, if not, in which direction the system will shift.
02

Calculate Qp for condition (a)

Substitute the given values for condition (a) into the Qp expression: $$Q_{p_{a}} = \frac{0.11 \times 2.0}{(0.012)^2} = 1520.8$$ #a. Compare Qp for condition (a) with Kp Since \(Q_{p_{a}} < K_p\) (1520.8 < 2400), the reaction will shift to the right (in favor of the products).
03

Calculate Qp for condition (b)

Substitute the given values for condition (b) into the Qp expression: $$Q_{p_{b}} = \frac{0.36 \times 0.67}{(0.0078)^2} = 3912.3$$ #b. Compare Qp for condition (b) with Kp Since \(Q_{p_{b}} > K_p\) (3912.3 > 2400), the reaction will shift to the left (in favor of the reactants).
04

Calculate Qp for condition (c)

Substitute the given values for condition (c) into the Qp expression: $$Q_{p_{c}} = \frac{0.51 \times 0.18}{(0.0062)^2} = 2396.7$$ #c. Compare Qp for condition (c) with Kp Since \(Q_{p_{c}} \approx K_p\) (2396.7 ≈ 2400), the system is at equilibrium for condition (c). To summarize, only condition (c) is at equilibrium. Condition (a) will shift to the right, and condition (b) will shift to the left.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, represented as \(K\), is crucial in understanding chemical equilibria. It represents the ratio of the concentrations of products to reactants at equilibrium. For reactions involving gases, this is often denoted as \(K_p\), which stands for the equilibrium constant in terms of partial pressures.
In the given reaction, the equilibrium expression for \(K_p\) is:\[K_p = \frac{P_{N_2} \times P_{O_2}}{P_{NO}^2}\]When a system is at equilibrium, the value of \(Q_p\) matches \(K_p\). Here, \(K_p\) is 2400, indicating the pressure conditions where the reactants and products are balanced. Using this value helps determine if the system is at equilibrium and, if not, in which direction it might adjust to reach equilibrium. If possible, calculating \(Q\) at different pressures using the same expression is an approach usually taken to compare with \(K_p\). This makes it easy to judge the state of equilibrium. If \(Q_p < K_p\), the reaction favors product formation. If \(Q_p > K_p\), reactant formation is favored.
Reaction Quotient
The reaction quotient, symbolized as \(Q\) or more specifically \(Q_p\) for gaseous reactions like the one presented, provides a snapshot of the current state of a reaction with respect to equilibrium. Just like \(K_p\), its expression is:\[Q_p = \frac{P_{N_2} \times P_{O_2}}{P_{NO}^2}\]Unlike \(K_p\), which is fixed at a given temperature, \(Q_p\) can vary depending on the current pressures of the system. By calculating \(Q_p\) and comparing it to \(K_p\) (the equilibrium constant), you can determine whether a reaction system is at equilibrium or predict its direction of shift.
  • If \(Q_p < K_p\), the reaction will shift to the right to produce more products.
  • If \(Q_p > K_p\), the reaction will shift to the left to produce more reactants.
  • If \(Q_p = K_p\), the system is at equilibrium, and there is no shift.
This comparison helps chemists understand whether a reaction is at equilibrium or, if not, which way it must proceed to reach equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept used to predict the behavior of a chemical system when it experiences a change in conditions such as concentration, temperature, or pressure. An understanding of Le Chatelier’s Principle is essential to manipulate reactions to achieve a desired product.In the context of the initial problem, once we calculate \(Q_p\), this principle can be used in conjunction with it to explain the direction of the reaction shift.
  • If a system's condition changes leading to \(Q_p > K_p\), Le Chatelier's principle predicts a shift to the left. This means the system will attempt to form more reactants to diminish the product pressures.
  • If \(Q_p < K_p\), the principle suggests a shift to the right, promoting the production of more products to reach new equilibrium.
  • When external pressure, volume, or concentration is changed to disrupt equilibrium, Le Chatelier’s Principle explains how reactions counteract the change.
Understanding and applying this principle is key for industries that aim to control chemical reactions more efficiently and economically.

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Most popular questions from this chapter

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. 2.0 moles of pure \(\mathrm{NOCl}\) in a 2.0 \(\mathrm{L}\) flask b. 1.0 mole of NOCl and 1.0 mole of NO in a 1.0 - flask c. 2.0 moles of \(\mathrm{NOCl}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) in a \(1.0-\mathrm{L}\) flask

At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ A \(1.00-\mathrm{L}\) flask containing 10.0 \(\mathrm{g} \mathrm{NaHCO}_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C} .\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

For the reaction $$2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)$$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium in a \(2.0-\) L container it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} \mathrm{M} .\) Calculate the moles of \(\mathrm{O}_{2}(g)\) present under these conditions.

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

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