As shown in Table \(15.2, K_{p}\) for the equilibrium $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ is \(4.51 \times 10^{-5}\) at \(450^{\circ} \mathrm{C}\). For each of the mixtures listed here, indicate whether the mixture is at equilibrium at \(450^{\circ} \mathrm{C}\). If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to achieve equilibrium. (a) \(98 \mathrm{~atm} \mathrm{NH}_{3}, 45 \mathrm{~atm} \mathrm{~N}_{2}, 55 \mathrm{~atm} \mathrm{H}_{2}\) (b) \(57 \mathrm{~atm} \mathrm{NH}_{3}, 143 \mathrm{~atm} \mathrm{~N}_{2},\) no \(\mathrm{H}_{2}\) (c) \(13 \mathrm{~atm} \mathrm{NH}_{2} 27 \mathrm{~atm} \mathrm{~N}_{2} 82 \mathrm{~atm} \mathrm{H}_{2}\)

Short Answer

Expert verified
(a) For the given mixture, \(Q_p = \frac{(98)^2}{(45)(55)^3}\). Comparing with \(K_p = 4.51 \times 10^{-5}\), the reaction will shift toward reactants. (b) Since there is no hydrogen present, the reaction cannot proceed and will shift toward reactants to produce more hydrogen gas. (c) For the given mixture, \(Q_p = \frac{(13)^2}{(27)(82)^3}\). Comparing with \(K_p = 4.51 \times 10^{-5}\), the reaction will shift toward products.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is: \[ \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \]
02

Identify the given equilibrium constant, \(K_p\)

We are given the \(K_p\) value, which is: \(K_p = 4.51 \times 10^{-5}\)
03

Calculate the reaction quotient, \(Q_p\), for each mixture

The reaction quotient, \(Q_p\), is defined as: \[ Q_p = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \] For each given mixture, plug in the partial pressures into the formula and calculate the value of \(Q_p\). (a) With a partial pressure of \(98 \, \text{atm}\) of ammonia, \(45 \, \text{atm}\) of nitrogen, and \(55\, \text{atm}\) of hydrogen, the reaction quotient is: \( Q_p = \frac{(98)^2}{(45)(55)^3} \) (b) With a partial pressure of \(57 \, \text{atm}\) of ammonia, \(143 \, \text{atm}\) of nitrogen, and no hydrogen, the reaction quotient is: \( Q_p = \frac{(57)^2}{(143)(0)^3} \) (c) With a partial pressure of \(13 \, \text{atm}\) of ammonia, \(27 \, \text{atm}\) of nitrogen, and \(82\, \text{atm}\) of hydrogen, the reaction quotient is: \( Q_p = \frac{(13)^2}{(27)(82)^3} \)
04

Compare \(Q_p\) and \(K_p\) for each mixture to determine the direction of the reaction

Compare the values of \(Q_p\) and \(K_p\) for each mixture: (a) If \(Q_p > K_p\), the reaction will shift toward reactants. If \(Q_p < K_p\), the reaction will shift toward products. If \(Q_p = K_p\), the system is at equilibrium. (b) Since there is no hydrogen present, the reaction cannot proceed. It will shift toward reactants to produce more hydrogen gas. (c) Compare the calculated \(Q_p\) value with \(K_p\). If \(Q_p > K_p\), the reaction will shift toward reactants. If \(Q_p < K_p\), the reaction will shift toward products. If \(Q_p = K_p\), the system is at equilibrium.

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Key Concepts

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

Equilibrium Constant (Kp)
The equilibrium constant, represented as Kp, is a fundamental concept in chemical equilibrium that provides a quantitative measure of the position of equilibrium in reactions involving gases. It is expressed in terms of partial pressures.

For a general reaction of the form:
\[aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\] the equilibrium constant, Kp, is given by the following expression: \[ K_{p} = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \] where PA, PB, PC, and PD are the partial pressures of the gases A, B, C, and D, respectively, and the lowercase letters represent the stoichiometric coefficients.

A Kp value that is very large suggests that, at equilibrium, there is a greater concentration of products compared to reactants. Conversely, a Kp value that is small indicates an equilibrium with greater concentrations of reactants. Understanding Kp helps us predict how changes in conditions, like pressure and temperature, affect the equilibrium position of a reaction.
Reaction Quotient (Qp)
The reaction quotient, Qp, helps us determine the direction in which a reaction mixture will shift in order to reach equilibrium. It is calculated using the same expression as the equilibrium constant but with the initial partial pressures of the reactants and products, not the equilibrium ones.

For the reaction equilibrium equation given earlier, the reaction quotient would be: \[ Q_{p} = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} \] When comparing Qp to Kp, if Qp is greater than Kp, the reaction will proceed in the direction of the reactants to reach equilibrium. If Qp is less than Kp, the reaction will proceed towards the products. If they are equal, the system is at equilibrium.

By calculating Qp for different mixtures and comparing it to Kp, students can predict which way a reaction will shift. This is particularly valuable in ensuring a reaction is on the right path to achieve the desired outcome.
Le Châtelier's Principle
Le Châtelier's principle is a key idea that predicts how a change in conditions, such as concentration, pressure, or temperature, will affect the position of equilibrium in a closed system. According to this principle, if a system at equilibrium is disturbed by a change in conditions, the system will adjust itself to counteract the effect of that disturbance, and establish a new equilibrium.

For example, if we increase the pressure on a reaction mixture involving gases, the system will respond by shifting the equilibrium position in the direction that produces fewer gas molecules. Similarly, adding more reactant to a system at equilibrium will result in a shift towards the production of more products until a new balance is achieved.

Le Châtelier's principle is essential for understanding how to control the yield of a reaction in industrial processes and for predicting the impact of environmental changes on reactions. This principle empowers students with the ability to hypothesize the behavior of a system when subjected to various perturbations, enabling them to grasp the dynamic nature of chemical equilibria.

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Most popular questions from this chapter

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\). A 7.5-L gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at \(450 \mathrm{~K}\). At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{~atm}\) \(P_{\mathrm{Cl}_{2}}=0.157 \mathrm{~atm},\) and \(P_{\mathrm{PCl}_{5}}=1.30 \mathrm{~atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at \(450 \mathrm{~K}\).

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of \(\mathrm{A}\) or the partial pressure of \(\mathrm{B}\) ? Explain.

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a 1: 1 ratio to form XY. A solution that is initially \(1.00 \mathrm{~m} M\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{~m} M\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free \(\mathrm{Y}\) remain. What is \(K_{c}\) for the reaction?

Consider the following equilibrium, for which \(K_{p}=0.0752\) at $$\begin{array}{l} 480{ }^{\circ} \mathrm{C}: \\ \quad 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \end{array}$$ (a) What is the value of \(K_{p}\) for the reaction \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?(\mathrm{c})\) What is the value of \(K_{c}\) for the reac- tion in part (b)?

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

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