Which Will Cause The Value Of The Equilibrium Constant To Change?

Chapter 13. Central Equilibrium Concepts

xiii.3 Shifting Equilibria: Le Châtelier'due south Principle

Learning Objectives

Past the finish of this section, you lot will be able to:

• Draw the ways in which an equilibrium system tin be stressed
• Predict the response of a stressed equilibrium using Le Châtelier's principle

As nosotros saw in the previous section, reactions keep in both directions (reactants get to products and products go to reactants). We tin tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (Thou). Nosotros next address what happens when a arrangement at equilibrium is disturbed and then that Q is no longer equal to K. If a system at equilibrium is subjected to a perturbance or stress (such every bit a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of Q will no longer equal the value of K. To re-establish equilibrium, the organisation will either shift toward the products (if Q < Grand) or the reactants (if Q > Chiliad) until Q returns to the same value as K.

This process is described by Le Châtelier's principle: When a chemical organisation at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. Every bit described in the previous paragraph, the disturbance causes a change in Q; the reaction will shift to re-establish Q = 1000.

Predicting the Direction of a Reversible Reaction

Le Châtelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we accept a mixture of reactants and products that accept not yet reached equilibrium, the changes necessary to accomplish equilibrium may non be so obvious. In such a case, we can compare the values of Q and Yard for the system to predict the changes.

Effect of Modify in Concentration on Equilibrium

A chemical system at equilibrium can be temporarily shifted out of equilibrium by calculation or removing one or more of the reactants or products. The concentrations of both reactants and products and so undergo additional changes to return the system to equilibrium.

The stress on the organisation in Figure ane is the reduction of the equilibrium concentration of SCN⁻ (lowering the concentration of one of the reactants would cause Q to exist larger than K). As a event, Le Châtelier's principle leads united states of america to predict that the concentration of Iron(SCN)ⁱⁱ⁺ should decrease, increasing the concentration of SCN⁻ part way back to its original concentration, and increasing the concentration of Feⁱⁱⁱ⁺ to a higher place its initial equilibrium concentration.

Figure ane. (a) The examination tube contains 0.1 Yard Fe^three+. (b) Thiocyanate ion has been added to solution in (a), forming the cherry-red Fe(SCN)²⁺ ion. Atomic number 26³⁺(aq) + SCN⁻(aq) ⇌ Atomic number 26(SCN)²⁺(aq). (c) Argent nitrate has been added to the solution in (b), precipitating some of the SCN⁻ as the white solid AgSCN. Ag⁺(aq) + SCN⁻(aq) ⇌ AgSCN(s). The decrease in the SCN⁻ concentration shifts the first equilibrium in the solution to the left, decreasing the concentration (and lightening color) of the Fe(SCN)ⁱⁱ⁺. (credit: modification of work by Marker Ott)

The effect of a change in concentration on a system at equilibrium is illustrated further past the equilibrium of this chemic reaction:

[latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{Hello}(g)\;\;\;\;\;\;\;K_c = 50.0\;\text{at}\;400\;^{\circ}\text{C}[/latex]

The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with [H₂] = [I₂] = 0.221 M and [Hullo] = i.563 Thousand is at equilibrium; for this mixture, Q_c = K_c = l.0. If H₂ is introduced into the system then quickly that its concentration doubles before it begins to react (new [H₂] = 0.442 M), the reaction will shift so that a new equilibrium is reached, at which [H_ii] = 0.374 One thousand, [I₂] = 0.153 One thousand, and [How-do-you-do] = one.692 One thousand. This gives:

[latex]Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(1.692)^ii}{(0.374)(0.153)} = 50.0 = K_c[/latex]

We have stressed this organization by introducing additional H₂. The stress is relieved when the reaction shifts to the right, using upwards some (merely non all) of the backlog H₂, reducing the amount of uncombined I₂, and forming additional How-do-you-do.

Upshot of Modify in Pressure on Equilibrium

Sometimes nosotros tin change the position of equilibrium past changing the pressure of a system. Yet, changes in pressure have a measurable result just in systems in which gases are involved, and then just when the chemical reaction produces a change in the full number of gas molecules in the system. An piece of cake way to recognize such a organisation is to look for dissimilar numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (every bit well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for Thousand_c ) or partial pressure (for K_P ). Some changes to total pressure, similar adding an inert gas that is not office of the equilibrium, volition change the total pressure only non the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not adjy the equilibrium.

Bank check out this link to see a dramatic visual demonstration of how equilibrium changes with pressure changes.

As we increment the pressure of a gaseous organisation at equilibrium, either past decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of 1 or more of the components. In accord with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will exist favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure level.

Consider what happens when we increase the force per unit area on a arrangement in which NO, O₂, and NO_ii are at equilibrium:

[latex]2\text{NO}(thousand)\;+\;\text{O}_2(k)\;{\rightleftharpoons}\;ii\text{NO}_2(thousand)[/latex]

The formation of additional amounts of NO₂ decreases the total number of molecules in the system considering each time two molecules of NO₂ grade, a total of three molecules of NO and O₂ are consumed. This reduces the total pressure level exerted past the system and reduces, only does not completely save, the stress of the increased pressure. On the other manus, a decrease in the pressure level on the system favors decomposition of NO₂ into NO and O₂, which tends to restore the pressure.

At present consider this reaction:

[latex]\text{N}_2(g)\;+\;\text{O}_2(1000)\;{\rightleftharpoons}\;2\text{NO}(grand)[/latex]

Because there is no alter in the total number of molecules in the system during reaction, a alter in pressure does non favor either formation or decomposition of gaseous nitrogen monoxide.

Effect of Change in Temperature on Equilibrium

Changing concentration or force per unit area perturbs an equilibrium because the reaction caliber is shifted away from the equilibrium value. Irresolute the temperature of a organization at equilibrium has a unlike outcome: A change in temperature really changes the value of the equilibrium abiding. Still, nosotros can qualitatively predict the effect of the temperature change past treating it every bit a stress on the organization and applying Le Châtelier'south principle.

When hydrogen reacts with gaseous iodine, heat is evolved.

[latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{HI}(g)\;\;\;\;\;\;\;{\Delta}H = -9.four\;\text{kJ\;(exothermic)}[/latex]

Because this reaction is exothermic, nosotros can write it with heat as a product.

[latex]\text{H}_2(g)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{HI}(g)\;+\;\text{heat}[/latex]

Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the consequence of increasing the amount of ane of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H₂ and I₂ and a reduction in the concentration of Hullo. Lowering the temperature of this system reduces the amount of energy present, favors the product of heat, and favors the germination of hydrogen iodide.

When nosotros change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium abiding: At the new equilibrium the concentration of How-do-you-do has increased and the concentrations of H_two and I_two decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C.

Temperature affects the equilibrium betwixt NO_two and N_iiO₄ in this reaction

[latex]\text{N}_2\text{O}_4(grand)\;{\rightleftharpoons}\;2\text{NO}_2(g)\;\;\;\;\;\;\;{\Delta}H = 57.twenty\;\text{kJ}[/latex]

The positive ΔH value tells us that the reaction is endothermic and could be written

[latex]\text{heat}\;+\;\text{Northward}_2\text{O}_4(g)\;{\rightleftharpoons}\;2\text{NO}_2(g)[/latex]

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO₂ molecules. If, all the same, nosotros put a stress on the system past cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless Due north₂O₄ increases, and the concentration of brown NO₂ decreases, causing the brown color to fade.

This interactive animation allows you to employ Le Châtelier'southward principle to predict the effects of changes in concentration, pressure level, and temperature on reactant and product concentrations.

Catalysts Do Not Bear on Equilibrium

As nosotros learned during our study of kinetics, a catalyst tin speed upward the charge per unit of a reaction. Though this increase in reaction rate may cause a arrangement to attain equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations.

The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a goad on a chemic equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen co-ordinate to the equation

[latex]\text{N}_2(k)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(k)[/latex]

A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the superlative 10 chemicals, by mass, manufactured in the globe. Nearly ii billion pounds are manufactured in the United States each year.

Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an of import fertilizer for the growth of corn, cotton fiber, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel manufacture.

Fritz Haber

In the early 20th century, German chemist Fritz Haber (Figure ii) adult a practical process for converting diatomic nitrogen, which cannot exist used past plants as a nutrient, to ammonia, a class of nitrogen that is easiest for plants to absorb.

[latex]\text{North}_2(k)\;+\;3\text{H}_2(k)\;{\leftrightharpoons}\;2\text{NH}_3(g)[/latex]

The availability of nitrogen is a strong limiting factor to the growth of plants. Despite bookkeeping for 78% of air, diatomic nitrogen (N₂) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bail. For plants to use atmospheric nitrogen, the nitrogen must exist converted to a more bioavailable course (this conversion is called nitrogen fixation).

Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemical science and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen nether loftier temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land tin support from 1.ix persons per hectare in 1908 to 4.iii in 2008.

Effigy 2. The work of Nobel Prize recipient Fritz Haber revolutionized agricultural practices in the early 20th century. His work also affected wartime strategies, adding chemical weapons to the arms.

In addition to his piece of work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World State of war I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, "During peace time a scientist belongs to the Earth, but during war time he belongs to his country."^[1] Haber defended the use of gas warfare against accusations that information technology was inhumane, saying that death was decease, by whatsoever means it was inflicted. He stands equally an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science.

Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agronomics, nitrogen compounds tin can also exist used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was i of the components of the flop used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April nineteen, 1995.

It has long been known that nitrogen and hydrogen react to form ammonia. However, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen but in the early 20th century after the factors that influence its equilibrium were understood.

To be practical, an industrial procedure must give a large yield of production relatively quickly. One manner to increase the yield of ammonia is to increase the force per unit area on the system in which Northward₂, H_ii, and NH₃ are at equilibrium or are coming to equilibrium.

[latex]\text{North}_2(grand)\;+\;3\text{H}_2(thou)\;{\rightleftharpoons}\;2\text{NH}_3(g)[/latex]

The formation of additional amounts of ammonia reduces the total force per unit area exerted by the arrangement and somewhat reduces the stress of the increased pressure.

Although increasing the pressure of a mixture of Due north_ii, H_ii, and NH₃ will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is boring. At room temperature, for example, the reaction is so slow that if we prepared a mixture of N₂ and H_ii, no detectable amount of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:

[latex]\text{N}_2(g)\;+\;3\text{H}_2(g)\;{\longrightarrow}\;2\text{NH}_3(thousand)\;\;\;\;\;\;\;{\Delta}H = -92.2\;\text{kJ}[/latex]

Thus, increasing the temperature to increase the charge per unit lowers the yield. If we lower the temperature to shift the equilibrium to favor the formation of more ammonia, equilibrium is reached more than slowly because of the large decrease of reaction rate with decreasing temperature.

Part of the rate of formation lost past operating at lower temperatures tin can be recovered by using a catalyst. The net result of the goad on the reaction is to cause equilibrium to exist reached more than rapidly.

In the commercial production of ammonia, conditions of almost 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the price of the equipment necessary to produce and incorporate high-pressure level gases at high temperatures (Figure three).

Effigy three. Commercial production of ammonia requires heavy equipment to handle the loftier temperatures and pressures required. This schematic outlines the design of an ammonia institute.

Key Concepts and Summary

Systems at equilibrium tin can be disturbed past changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Châtelier's principle: The system will answer in a mode that counteracts the disturbance. Non all changes to the system event in a disturbance of the equilibrium. Calculation a catalyst affects the rates of the reactions but does non alter the equilibrium, and changing pressure or volume volition non significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and production side.

Disturbance	Observed Change every bit Equilibrium is Restored	Management of Shift	Effect on Chiliad
reactant added	added reactant is partially consumed	toward products	none
product added	added product is partially consumed	toward reactants	none
decrease in volume/increment in gas pressure	force per unit area decreases	toward side with fewer moles of gas	none
increase in volume/decrease in gas pressure	pressure increases	toward side with more moles of gas	none
temperature increase	estrus is absorbed	toward products for endothermic, toward reactants for exothermic	changes
temperature decrease	estrus is given off	toward reactants for endothermic, toward products for exothermic	changes
Table 2. Effects of Disturbances of Equilibrium and K

Chemistry End of Chapter Exercises

1. The post-obit equation represents a reversible decomposition:
   [latex]\text{CaCO}_3(southward)\;{\rightleftharpoons}\;\text{CaO}(s)\;+\;\text{CO}_2(k)[/latex]

   Under what atmospheric condition will decomposition in a closed container go along to completion then that no CaCO_iii remains?

2. Explain how to recognize the weather under which changes in pressure would touch systems at equilibrium.
3. What property of a reaction tin can we use to predict the effect of a change in temperature on the value of an equilibrium constant?
4. What would happen to the color of the solution in part (b) of Figure 1 if a small amount of NaOH were added and Atomic number 26(OH)₃ precipitated? Explain your respond.
5. The following reaction occurs when a burner on a gas stove is lit:
   [latex]\text{CH}_4(g)\;+\;2\text{O}_2(thousand)\;{\rightleftharpoons}\;\text{CO}_2(g)\;+\;two\text{H}_2\text{O}(g)[/latex]

   Is an equilibrium among CH₄, O_ii, CO_two, and H₂O established nether these conditions? Explicate your respond.

6. A necessary step in the industry of sulfuric acrid is the formation of sulfur trioxide, SO_three, from sulfur dioxide, So₂, and oxygen, O_two, shown here. At high temperatures, the charge per unit of germination of SO_iii is higher, but the equilibrium amount (concentration or partial force per unit area) of So₃ is lower than it would be at lower temperatures.
   [latex]two\text{SO}_2(k)\;+\;\text{O}_2(thou)\;{\longrightarrow}\;2\text{So}_3(g)[/latex]

   (a) Does the equilibrium constant for the reaction increase, decrease, or remain most the aforementioned as the temperature increases?

   (b) Is the reaction endothermic or exothermic?

7. Propose four means in which the concentration of hydrazine, N_twoH₄, could be increased in an equilibrium described by the following equation:
   [latex]\text{Due north}_2(g)\;+\;2\text{H}_2(chiliad)\;{\rightleftharpoons}\;\text{N}_2\text{H}_4(g)\;\;\;\;\;\;\;{\Delta}H = 95\;\text{kJ}[/latex]
8. Advise iv means in which the concentration of PH_three could be increased in an equilibrium described by the following equation:
   [latex]\text{P}_4(yard)\;+\;6\text{H}_2(g)\;{\rightleftharpoons}\;iv\text{PH}_3(g)\;\;\;\;\;\;\;{\Delta}H = 110.5\;\text{kJ}[/latex]
9. How will an increment in temperature affect each of the following equilibria? How will a decrease in the book of the reaction vessel bear on each?

   (a) [latex]ii\text{NH}_3(g)\;{\rightleftharpoons}\;\text{N}_2(grand)\;+\;3\text{H}_2(1000)\;\;\;\;\;\;\;{\Delta}H = 92\;\text{kJ}[/latex]

   (b) [latex]\text{N}_2(chiliad)\;+\;\text{O}_2(k)\;{\rightleftharpoons}\;two\text{NO}(g)\;\;\;\;\;\;\;{\Delta}H = 181\;\text{kJ}[/latex]

   (c) [latex]2\text{O}_3(g)\;{\rightleftharpoons}\;3\text{O}_2(thousand)\;\;\;\;\;\;\;{\Delta}H = -285\;\text{kJ}[/latex]

   (d) [latex]\text{CaO}(s)\;+\;\text{CO}_2(g)\;{\rightleftharpoons}\;\text{CaCO}_3(south)\;\;\;\;\;\;\;{\Delta}H = -176\;\text{kJ}[/latex]

10. How will an increment in temperature affect each of the following equilibria? How will a subtract in the volume of the reaction vessel affect each?

    (a) [latex]2\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;two\text{H}_2(grand)\;+\;\text{O}_2(one thousand)\;\;\;\;\;\;\;{\Delta}H = 484\;\text{kJ}[/latex]

    (b) [latex]\text{North}_2(g)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;two\text{NH}_3(g)\;{\Delta}H = -92.2\;\text{kJ}[/latex]

    (c) [latex]two\text{Br}(chiliad)\;{\rightleftharpoons}\;\text{Br}_2(g)\;\;\;\;\;\;\;{\Delta}H = -224\;\text{kJ}[/latex]

    (d) [latex]\text{H}_2(chiliad)\;+\;\text{I}_2(s)\;{\rightleftharpoons}\;ii\text{HI}(g)\;\;\;\;\;\;\;{\Delta}H = 53\;\text{kJ}[/latex]

11. Water gas is a one:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: [latex]\text{H}_2\text{O}(1000)\;+\;\text{C}(s)\;{\rightleftharpoons}\;\text{H}_2(g)\;+\;\text{CO}(thousand)[/latex]. Methanol, a liquid fuel that could peradventure replace gasoline, tin exist prepared from h2o gas and hydrogen at high temperature and pressure level in the presence of a suitable goad.

    (a) Write the expression for the equilibrium constant (One thousand_c ) for the reversible reaction

    [latex]2\text{H}_2(grand)\;+\;\text{CO}(g)\;{\rightleftharpoons}\;\text{CH}_3\text{OH}(g)\;\;\;\;\;\;\;{\Delta}H = -90.2\;\text{kJ}[/latex]

    (b) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if more H_two is added?

    (c) What volition happen to the concentrations of H_ii, CO, and CH_iiiOH at equilibrium if CO is removed?

    (d) What will happen to the concentrations of H_two, CO, and CH_threeOH at equilibrium if CH₃OH is added?

    (e) What will happen to the concentrations of H_ii, CO, and CH₃OH at equilibrium if the temperature of the arrangement is increased?

    (f) What will happen to the concentrations of H₂, CO, and CH_iiiOH at equilibrium if more than goad is added?

12. Nitrogen and oxygen react at high temperatures.

    (a) Write the expression for the equilibrium abiding (One thousand_c ) for the reversible reaction

    [latex]\text{N}_2(g)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{NO}(g)\;\;\;\;\;\;\;{\Delta}H = 181\;\text{kJ}[/latex]

    (b) What will happen to the concentrations of N₂, O_ii, and NO at equilibrium if more O_two is added?

    (c) What will happen to the concentrations of Northward₂, O₂, and NO at equilibrium if Due north₂ is removed?

    (d) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if NO is added?

    (e) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if the pressure on the arrangement is increased by reducing the volume of the reaction vessel?

    (f) What will happen to the concentrations of Northward₂, O₂, and NO at equilibrium if the temperature of the organization is increased?

    (g) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if a catalyst is added?

13. Water gas, a mixture of H_ii and CO, is an important industrial fuel produced by the reaction of steam with blood-red hot coke, substantially pure carbon.

    (a) Write the expression for the equilibrium constant for the reversible reaction

    [latex]\text{C}(s)\;+\;\text{H}_2\text{O}(thousand)\;{\rightleftharpoons}\;\text{CO}(grand)\;+\;\text{H}_2(g)\;\;\;\;\;\;\;{\Delta}H = 131.30\;\text{kJ}[/latex]

    (b) What will happen to the concentration of each reactant and product at equilibrium if more than C is added?

    (c) What will happen to the concentration of each reactant and production at equilibrium if H_iiO is removed?

    (d) What will happen to the concentration of each reactant and product at equilibrium if CO is added?

    (eastward) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the organization is increased?

14. Pure iron metallic can exist produced by the reduction of iron(III) oxide with hydrogen gas.

    (a) Write the expression for the equilibrium constant (Chiliad_c ) for the reversible reaction

    [latex]\text{Fe}_2\text{O}_3(s)\;+\;three\text{H}_2(g)\;{\rightleftharpoons}\;2\text{Fe}(s)\;+\;iii\text{H}_2\text{O}(g)\;\;\;\;\;\;\;{\Delta}H = 98.7\;\text{kJ}[/latex]

    (b) What will happen to the concentration of each reactant and product at equilibrium if more than Atomic number 26 is added?

    (c) What will happen to the concentration of each reactant and product at equilibrium if H₂O is removed?

    (d) What will happen to the concentration of each reactant and production at equilibrium if H₂ is added?

    (e) What volition happen to the concentration of each reactant and product at equilibrium if the pressure on the organisation is increased by reducing the book of the reaction vessel?

    (f) What will happen to the concentration of each reactant and product at equilibrium if the temperature of the system is increased?

15. Ammonia is a weak base that reacts with water co-ordinate to this equation:
    [latex]\text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

    Will whatsoever of the following increase the percent of ammonia that is converted to the ammonium ion in water?

    (a) Addition of NaOH

    (b) Add-on of HCl

    (c) Add-on of NH₄Cl

16. Acetic acrid is a weak acid that reacts with water according to this equation:
    [latex]\text{CH}_3\text{CO}_2\text{H}(aq)\;+\;\text{H}_2\text{O}(aq)\;{\rightleftharpoons}\;\text{H}_3\text{O}^{+}(aq)\;+\;\text{CH}_3\text{CO}_2^{\;\;-}(aq)[/latex]

    Volition any of the following increase the percent of acetic acrid that reacts and produces [latex]\text{CH}_3\text{CO}_2^{\;\;-}[/latex] ion?

    (a) Addition of HCl

    (b) Addition of NaOH

    (c) Addition of NaCH₃CO₂

17. Suggest 2 ways in which the equilibrium concentration of Ag⁺ can be reduced in a solution of Na⁺, Cl⁻, Ag⁺, and [latex]\text{NO}_3^{\;\;-}[/latex], in contact with solid AgCl.
    [latex]\text{Na}^{+}(aq)\;+\;\text{Cl}^{-}(aq)\;+\;\text{Ag}^{+}(aq)\;+\;\text{NO}_3^{\;\;-}(aq)\;{\rightleftharpoons}\;\text{AgCl}(s)\;+\;\text{Na}^{+}(aq)\;+\;\text{NO}_3^{\;\;-}(aq)[/latex]
    [latex]{\Delta}H = -65.9\;\text{kJ}[/latex]
18. How can the pressure of water vapor be increased in the following equilibrium?
    [latex]\text{H}_2\text{O}(50)\;{\rightleftharpoons}\;\text{H}_2\text{O}(g)\;\;\;\;\;\;\;{\Delta}H = 41\;\text{kJ}[/latex]
19. Additional solid argent sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate.
    [latex]2\text{Ag}^{+}(aq)\;+\;\text{SO}_4^{\;\;2-}(aq)\;{\rightleftharpoons}\;\text{Ag}_2\text{SO}_4(due south)[/latex]

    Which of the following will occur?

    (a) Ag⁺ or [latex]\text{SO}_4^{\;\;2-}[/latex] concentrations volition non change.

    (b) The added silver sulfate volition deliquesce.

    (c) Boosted silver sulfate will form and precipitate from solution equally Ag⁺ ions and [latex]\text{Then}_4^{\;\;2-}[/latex] ions combine.

    (d) The Ag⁺ ion concentration will increment and the [latex]\text{So}_4^{\;\;2-}[/latex] ion concentration volition subtract.

20. The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these ii compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium abiding for ionization [latex](\text{HX}\;{\rightleftharpoons}\;\text{H}^{+}\;+\;\text{X}^{-})[/latex]?

Glossary

Le Châtelier's principle

when a chemical system at equilibrium is disturbed, it returns to equilibrium past counteracting the disturbance

position of equilibrium

concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe weather before a disturbance)

stress

change to a reaction's weather condition that may cause a shift in the equilibrium

Solutions

Answers to Chemical science End of Chapter Exercises

1. The amount of CaCO₃ must be so small that [latex]P_{\text{CO}_2}[/latex] is less than K_P when the CaCO₃ has completely decomposed. In other words, the starting corporeality of CaCO₃ cannot completely generate the full [latex]P_{\text{CO}_2}[/latex] required for equilibrium.

iii. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added tin be thought of equally a reactant. Additional rut would shift an exothermic reaction dorsum to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would crusade it to shift to the reactants' side.

5. No, it is not at equilibrium. Because the system is non confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

seven. Add together N₂; add H₂; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift correct, ΔP increase = no upshot; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increment = shift correct.

eleven. (a) [latex]K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{H}_2]^two[\text{CO}]}[/latex]; (b) [H₂] increases, [CO] decreases, [CH₃OH] increases; (c), [H₂] increases, [CO] decreases, [CH₃OH] decreases; (d), [H₂] increases, [CO] increases, [CH_threeOH] increases; (e), [H₂] increases, [CO] increases, [CH_threeOH] decreases; (f), no changes.

xiii. (a) [latex]K_c = \frac{[\text{CO}][\text{H}_2]}{[\text{H}_2\text{O}]}[/latex]; (b) [H_twoO] no change, [CO] no modify, [H_ii] no modify; (c) [H₂O] decreases, [CO] decreases, [H_ii] decreases; (d) [H_iiO] increases, [CO] increases, [H₂] decreases; (f) [H_twoO] decreases, [CO] increases, [H₂] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

xv. Only (b)

17. Add NaCl or some other table salt that produces Cl⁻ to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

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Source: https://opentextbc.ca/chemistry/chapter/13-3-shifting-equilibria-le-chateliers-principle/

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