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For each of the following equations, give the expression for the equilibrium constant \(K_{c}\) $$ \begin{array}{l} \text { a } \mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \\ \text { b } \mathrm{C}(s)+2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+2 \mathrm{~N}_{2}(g) \\ \text { c } \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{NaHCO}_{3}(s) \\\ \text { d } \mathrm{Fe}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{3}(s) \end{array} $$

Short Answer

Expert verified
a) \( K_c = [\mathrm{NH}_3][\mathrm{HCl}] \), b) \( K_c = \frac{[\mathrm{CO}_2][\mathrm{N}_2]^2}{[\mathrm{N}_2\mathrm{O}]^2} \), c) \( K_c = \frac{1}{[\mathrm{CO}_2]} \), d) \( K_c = \frac{1}{[\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3} \).

Step by step solution

01

Analyze Reaction (a)

For the reaction \( \mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \), identify the gaseous products. The equilibrium constant \( K_c \) for reactions involving solids is expressed only in terms of the concentrations of gaseous or aqueous species. Therefore, \( K_c = [\mathrm{NH}_3][\mathrm{HCl}] \).
02

Analyze Reaction (b)

Consider the reaction \( \mathrm{C}(s) + 2 \mathrm{N}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g) + 2 \mathrm{N}_{2}(g) \). Omit the solid carbon from the equilibrium expression, and include only gaseous species. Thus, \( K_c = \frac{[\mathrm{CO}_2][\mathrm{N}_2]^2}{[\mathrm{N}_2\mathrm{O}]^2} \).
03

Analyze Reaction (c)

The reaction \( \mathrm{Na}_{2} \mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{NaHCO}_{3}(s) \) includes solids and a liquid, which are not included in the equilibrium expression. As a result, \( K_c = \frac{1}{[\mathrm{CO}_2]} \), since products are solid and their activity is considered to be 1.
04

Analyze Reaction (d)

For the aqueous reaction \( \mathrm{Fe}^{3+}(aq) + 3 \mathrm{OH}^{-}(aq) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{3}(s) \), exclude the solid product from the expression. Therefore, \( K_c = \frac{1}{[\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3} \).

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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 in a reversible reaction when the rates of the forward and reverse reactions are equal. This means that the concentrations of all reactants and products remain constant over time. It is important to understand that equilibrium is dynamic, not static, meaning the reactions continue to occur, but there is no net change in concentration.

In chemical equilibrium, the system will spontaneously shift to counteract any changes made to the system, as described by Le Chatelier’s Principle. If you add more reactant or remove product, the system will adjust to minimize the change by favoring the forward reaction. Conversely, adding product or removing reactant will shift the equilibrium to favor the reverse reaction.

Understanding chemical equilibrium is crucial for predicting how different factors like concentration, temperature, and pressure affect the reaction. By controlling these variables, chemists can manipulate the conditions to favor the desired reaction direction.
Equilibrium Expressions
Equilibrium expressions, or equilibrium constant expressions, are mathematical representations of a reaction at equilibrium. They allow chemists to determine the ratio of product concentrations to reactant concentrations.

For reactions where all reactants and products are gases or solutions, the equilibrium expression for the reaction \( aA + bB \rightleftharpoons cC + dD \) is given by \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).

Note that in equilibrium expressions, the concentrations of solids and pure liquids are omitted as their activities are constant and taken as 1. Only the concentrations of aqueous and gaseous species are included. This is why in reactions (a) through (d), only gaseous or aqueous species contribute to the equilibrium constant \( K_c \).
  • Solids and liquids are not included in the expression.
  • Only consider aqueous solutions and gases.
  • The coefficients of the reaction become the exponents in the expression.
Knowing how to construct equilibrium expressions allows us to predict the behavior of a system in equilibrium, especially how it will respond to changes.
Gaseous Reactions
Gaseous reactions are chemical reactions where the reactants and products are in the gas phase. An interesting aspect of such reactions is how they respond to changes in pressure and temperature.

In gaseous equilibria, changes in pressure can disturb the equilibrium. According to Le Chatelier's Principle, if you increase the pressure by decreasing the volume, the equilibrium will shift towards the side with the fewer gas molecules. Conversely, decreasing the pressure by increasing the volume favors the side with more gas molecules.
  • In reaction \((a)\), the effect of pressure change would alter the concentrations of \( \mathrm{NH}_3 \) and \( \mathrm{HCl} \).
  • In reaction \((b)\), increasing pressure could favor forming \( \mathrm{CO}_2 \) and \( \mathrm{N}_2 \) gases, as the forward reaction consumes more gas molecules than it produces.
Understanding how gaseous reactions behave can help us harness this knowledge in industrial applications, such as optimizing yields in chemical manufacturing processes by manipulating reaction conditions.

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Most popular questions from this chapter

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals 49 at \(230^{\circ} \mathrm{C}\). If \(0.400 \mathrm{~mol}\) each of phosphorus trichloride and chlorine are added to a 4.0-L reaction vessel, what is the equilibrium composition of the mixture at \(230^{\circ} \mathrm{C} ?\)

The major constituents of the atmosphere are nitrogen, \(\mathrm{N}_{2}\), and oxygen, \(\mathrm{O}_{2}\). Dry atmospheric air at a pressure of 1.000 atm has a partial pressure of \(\mathrm{N}_{2}\) of 0.781 atm and a partial pressure of \(\mathrm{O}_{2}\) of 0.209 atm. At high temperatures, \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) react to produce nitrogen monoxide (nitric oxide), NO. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose a sample of dry air is raised to \(2127^{\circ} \mathrm{C}\) in a hot flame and then rapidly cooled to fix the amount of \(\mathrm{NO}\) formed. If \(K_{p}\) for this reaction at this temperature is 0.0025 , how many grams of NO would be produced from \(100.0 \mathrm{~g}\) of dry air?

A chemist put 1.18 mol of substance \(A\) and 2.85 mol of substance \(\mathrm{B}\) into a 10.0 - \(\mathrm{L}\) flask, which she then closed. A and B react by the following equation: $$ \mathrm{A}(g)+2 \mathrm{~B}(g) \rightleftharpoons 3 \mathrm{C}(g)+\mathrm{D}(g) $$ She found that the equilibrium mixture at \(25^{\circ} \mathrm{C}\) contained \(0.376 \mathrm{~mol}\) of \(\mathrm{D}\). How many moles of \(\mathrm{B}\) are in the flask at equilibrium at \(25^{\circ} \mathrm{C} ?\) a. \(2.47 \mathrm{~mol}\) b. \(3.60 \mathrm{~mol}\) c. \(2.52 \mathrm{~mol}\) d. \(2.10 \mathrm{~mol}\) e. \(2.41 \mathrm{~mol}\)

A chemist wants to prepare phosgene, \(\mathrm{COCl}_{2}\), by the following reaction: $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ He places \(4.00 \mathrm{~g}\) of chlorine, \(\mathrm{Cl}_{2}\), and an equal molar amount of carbon monoxide, \(\mathrm{CO}\), into a 10.00 - \(\mathrm{L}\) reaction vessel at \(395^{\circ} \mathrm{C}\). After the reaction comes to equilibrium, he adds another \(4.00 \mathrm{~g}\) of chlorine to the vessel in order to push the reaction to the right to get more product. What is the partial pressure of phosgene when the reaction again comes to equilibrium? \(K_{c}=1.23 \times 10^{3}\).

Nitrogen monoxide, NO, reacts with bromine, \(\mathrm{Br}_{2}\), to give nitrosyl bromide, \(\mathrm{NOBr}\). $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ A sample of 0.0524 mol NO with 0.0262 mol \(\mathrm{Br}_{2}\) gives an equilibrium mixture containing \(0.0311 \mathrm{~mol} \mathrm{NOBr}\). What is the composition of the equilibrium mixture?

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