/*! 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 55 Write the expression for the equ... [FREE SOLUTION] | 91Ó°ÊÓ

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Write the expression for the equilibrium constant \(K_{c}\) for each of the following equations: $$ \begin{array}{l} \text { a } \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \\ \text { b } \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \\ \text { c } \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{CO}_{2}(g) \\ \text { d } \mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-(a q)} \end{array} $$

Short Answer

Expert verified
a) \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \) ; b) \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}]} \) ; c) \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}} \) ; d) \( K_c = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 \).

Step by step solution

01

Understanding Equilibrium Constant Expression

The equilibrium constant \( K_c \) is defined for a chemical reaction in equilibrium in terms of the concentrations of the products and reactants involved in the reaction. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the \( K_c \) expression is \( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \). Note that pure solids and liquids do not appear in the equilibrium expression.
02

Write \( K_c \) for Reaction (a)

For the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), since \( \text{C} \) is a solid, it is not included in the \( K_c \) expression. Thus, \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \).
03

Write \( K_c \) for Reaction (b)

For the reaction \( \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \), both \( \text{FeO} \) and \( \text{Fe} \) are solids and are not included in the expression. Therefore, \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{CO}]} \).
04

Write \( K_c \) for Reaction (c)

The equation is \( \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{CO}_{2}(g) \). Both \( \text{Na}_2 \text{CO}_3 \) and \( \text{Na}_2 \text{SO}_4 \) are solids and omitted. Thus, \( K_c = \frac{[\mathrm{CO}_2]}{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}} \).
05

Write \( K_c \) for Reaction (d)

For the reaction \( \mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq)+2 \mathrm{I}^{-}(aq) \), \( \text{PbI}_2 \) is a solid and is not included. Thus, \( K_c = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2 \).

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to a state in a chemical reaction where the concentrations of all reactants and products remain constant over time. This occurs when the forward and reverse reactions happen at the same rate, meaning that there is no net change in the concentrations of products and reactants. Chemical equilibrium is dynamic, which means that the reactions are still occurring, but with no overall effect on the concentrations of the substances involved.

At equilibrium, the system is stable and any changes to conditions such as concentration, temperature, or pressure will result in a shift to restore a new balance. This concept is crucial for understanding how reactions can be controlled and predicted in laboratory and industrial settings.
Reactions in Equilibrium
Reactions in equilibrium involve both the forward and reverse reactions proceeding simultaneously. At this stage, specific conditions are met where the amounts of reactants and products reach a steady state.

In a general reversible reaction represented as \( aA + bB \rightleftharpoons cC + dD \), both sides of the equation are continuously converting into each other. The rate of the forward reaction equals the rate of the reverse reaction, leading to constant concentrations of \( A, B, C, \) and \( D \).

Understanding reactions in equilibrium allows chemists to predict and alter the outcome of a reaction by changing conditions like temperature and concentration. It's critical in processes like synthesis, where maximizing the yield of a desired product is important.
Le Chatelier's Principle
Le Chatelier's Principle provides an explanation of how a chemical equilibrium shifts when subjected to changes in concentration, pressure, or temperature. It states that if a system at equilibrium experiences a change in one of these conditions, the system will adjust in such a way as to counteract the change and a new equilibrium will be established.

For example, if the concentration of a reactant is increased, the equilibrium will shift to the right, producing more products to re-establish equilibrium. Conversely, if the temperature is increased in an exothermic reaction, the equilibrium shifts to favor the reactants, reducing the effect of the added heat.

Le Chatelier's Principle is an essential concept in chemistry, providing insight into the manipulation of reaction conditions to optimize the production of desired products.
Law of Mass Action
The law of mass action is a fundamental principle that explains how the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to the power of their respective coefficients in the balanced chemical equation.

This law is the basis for defining the equilibrium constant, \( K_c \). For a reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant is represented as \( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \).

Pure solids and liquids are not included in the \( K_c \) expression because their concentrations do not change. The law of mass action helps chemists calculate changes in concentrations and understand how shifts in reactions occur, laying the foundation for predicting the behavior of chemical systems under different conditions.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) in a closed system, ammonium hydrogen sulfide exists as the following equilibrium: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ a) When a sample of pure \(\mathrm{NH}_{4} \mathrm{HS}(s)\) is placed in an evacuated reaction vessel and allowed to come to equilibrium at \(25^{\circ} \mathrm{C}\), the total pressure is \(0.660 \mathrm{~atm}\). What is the value of \(K_{p} ?\) b) To this system, sufficient \(\mathrm{H}_{2} \mathrm{~S}(g)\) is injected until the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is three times that of the ammonia at equilibrium. What are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S} ?\) c) In a different experiment, \(0.750 \mathrm{~atm}\) of \(\mathrm{NH}_{3}\) and 0.500 atm of \(\mathrm{H}_{2} \mathrm{~S}\) are introduced into a 1.00 - \(\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). How many moles of \(\mathrm{NH}_{4} \mathrm{HS}\) are present when equilibrium is established?

Iodine, \(\mathrm{I}_{2}\), is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, \(700^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\). He obtained the \(K_{p}\) values 0.01106 and \(0.001745,\) but he couldn't remember which value went with what temperature. Can you please help him assign temperatures to the equilibrium constants? State your argument carefully.

Antimony(V) chloride, \(\mathrm{SbCl}_{5}\), decomposes on heating to give antimony(III) chloride, \(\mathrm{SbCl}_{3},\) and chlorine. $$ \mathrm{SbCl}_{5}(g) \rightleftharpoons \mathrm{SbCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ A closed \(3.50-\mathrm{L}\) vessel initially contains \(0.0125 \mathrm{~mol} \mathrm{SbCl}_{5}\) What is the total pressure at \(248^{\circ} \mathrm{C}\) when equilibrium is achieved? The value of \(K_{c}\) at \(248^{\circ} \mathrm{C}\) is \(2.50 \times 10^{-2}\).

The equilibrium constant \(K_{c}\) equals 0.0952 for the following reaction at \(227^{\circ} \mathrm{C}\). $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ What is the value of \(K_{p}\) at this temperature?

A 2.500 -mol sample of phosphorus pentachloride, \(\mathrm{PCl}_{5}\), decomposes at \(160^{\circ} \mathrm{C}\) and 1.00 atm to give \(0.338 \mathrm{~mol}\) of phosphorus trichloride, \(\mathrm{PCl}_{3}\), at equilibrium. $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ What is the composition of the final reaction mixture?

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