/*! 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 28 Consider the equilibrium $$ ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{lr} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{array} $$

Short Answer

Expert verified
The equilibrium constant \(K_p\) is \(3.91 \times 10^{-32}\).

Step by step solution

01

Write the Given Reactions and Their Equilibrium Constants

We have two separate reactions: \(2\mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_{c1} = 2.0\) and \(2\mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g) + \mathrm{O}_2(g)\) with \(K_{c2} = 2.1 \times 10^{30}\). Our goal is to find the equilibrium constant \(K_p\) for the new reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\).
02

Derive Target Reaction

To derive the target reaction, we need to manipulate the given reactions. If we reverse reaction 2, it becomes \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\). This changes the equilibrium constant to \(K_{c2}' = \frac{1}{2.1 \times 10^{30}} = 4.76 \times 10^{-31}\).
03

Combine Reactions

Combine the reversed reaction 2 and reaction 1:- \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\) with \(K_{c2}' = 4.76 \times 10^{-31}\)- \(2\mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_{c1} = 2.0\)The sum gives \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_c = K_{c1} \times K_{c2}'\).
04

Calculate \(K_c\) for Target Reaction

Calculate \(K_c\) for the target reaction using the formula:\[ K_c = K_{c1} \times K_{c2}' = 2.0 \times 4.76 \times 10^{-31} = 9.52 \times 10^{-31} \]
05

Convert \(K_c\) to \(K_p\)

The relationship between \(K_c\) and \(K_p\) is given by \(K_p = K_c \times (RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas. For our reaction, \(\Delta n = (2 - (1+1+1)) = -1\).Using \(R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}\) and \(T = 298 \text{ K}\), we find:\[ K_p = 9.52 \times 10^{-31} \times (0.0821 \times 298)^{-1} = 9.52 \times 10^{-31} \times 0.041 \]\[ K_p = 3.91 \times 10^{-32} \]
06

Conclusion

The equilibrium constant \(K_p\) for the reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) at 298 K is \(3.91 \times 10^{-32}\).

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Key Concepts

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

Chemical Equilibrium
In chemistry, chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of the reactants and products remain constant over time. Chemical equilibrium is crucial because it helps to predict the behavior of reactions in different conditions.
The equilibrium constant, represented as either \( K_c \) for concentrations or \( K_p \) for partial pressures, quantitatively describes the ratio of product and reactant concentrations at equilibrium. This is determined by the nature of the reaction and temperature.
For example, in our target reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g) \), we need to find the equilibrium constant to understand how much \( \mathrm{NOBr} \) is formed and how much \( \mathrm{N}_2, \mathrm{O}_2, \) and \( \mathrm{Br}_2 \) remain.
Reaction Manipulation
Reaction manipulation is a technique used to modify chemical equations to derive a desired target equation. In solving equilibrium problems, we often manipulate given reactions to find information about a related reaction.
There are several common operations used in reaction manipulation:
  • Reversing a reaction, which involves swapping the reactants and products. When this is done, the equilibrium constant becomes the reciprocal of the original.
  • Combining reactions, adding two or more reactions to get a net reaction. The equilibrium constant for the net reaction is the product of the constants for the individual steps.
  • Multiplying a reaction by a coefficient scales the equilibrium constant exponentially.

In our example, reversing and combining given reactions allowed us to derive the reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g) \) and find its \( K_c \) value, which is crucial for further calculations.
Thermodynamics
Thermodynamics plays a vital role in understanding chemical equilibrium, especially when predicting how a reaction shifts with changes in temperature and pressure. The first and second laws of thermodynamics lay the foundation for calculating equilibrium constants.
One of the key insights is the relationship between \( K_c \) and \( K_p \), which represents concentrations and pressures, respectively. This relationship is given by: \[ K_p = K_c \times (RT)^{\Delta n} \]where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas from reactants to products.
In our reaction, \( \Delta n = -1 \) because we have fewer moles of gas on the product side than the reactant side. Using the values of \( R \) and \( T \), we can convert between \( K_c \) and \( K_p \), aiding in our comprehension of how pressure affects the reaction outcome.
This understanding is invaluable in many chemical processes, from industrial synthesis to biological systems.

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Most popular questions from this chapter

Consider the reaction $$ \begin{array}{l} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons \\ 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ} \end{array} $$ Does each of the following increase, decrease, or leave unchanged the yield of \(\mathrm{NO}\) at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathbf{c})\) decrease \(\left[\mathrm{O}_{2}\right] ;(\mathbf{d})\) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$ 2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\operatorname{Br}_{2}(g) $$ An equilibrium mixture in a 5.00-L vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) of \(\mathrm{Br}_{2}\). (a) Calculate \(K_{c}\). (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of \(\mathrm{NOBr}\) ?

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=\) \(8.59 \times 10^{-4}\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(50.7 \mathrm{kPa}\) \(\mathrm{PCl}_{3}, 50.7 \mathrm{kPa} \mathrm{Cl}_{2}\), and \(20.3 \mathrm{kPa} \mathrm{PCl}_{5}\) at this temperature. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. \((\mathbf{c})\) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

Consider the hypothetical reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons\) \(2 \mathrm{C}(g),\) for which \(K_{c}=0.25\) at a certain temperature. \(\mathrm{A}\) 1.00-L reaction vessel is loaded with \(1.00 \mathrm{~mol}\) of compound C, which is allowed to reach equilibrium. Let the variable \(x\) represent the number of mol/L of compound A present at equilibrium. (a) In terms of \(x\), what are the equilibrium concentrations of compounds \(\mathrm{B}\) and \(\mathrm{C}\) ? (b) What limits must be placed on the value of \(x\) so that all concentrations are positive? (c) By putting the equilibrium concentrations (in terms of \(x\) ) into the equilibriumconstant expression, derive an equation that can be solved for \(x .(\mathbf{d})\) The equation from part \((\mathrm{c})\) is a cubic equation (one that has the form \(\left.a x^{3}+b x^{2}+c x+d=0\right)\). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of \(x\) that you specified in part (b). The point at which the cubic equation crosses the \(x\) -axis is the solution. \((\mathbf{e})\) From the plot in part (d), estimate the equilibrium concentrations of \(\mathrm{A}, \mathrm{B},\) and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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