Chegg Apply Le Chatelier Principle for a Temperature Increase Using Molecular Art

Chapter 13. Fundamental Equilibrium Concepts

thirteen.3 Shifting Equilibria: Le Châtelier'southward Principle

Learning Objectives

By the stop of this department, you will be able to:

• Describe the ways in which an equilibrium system can exist stressed
• Predict the response of a stressed equilibrium using Le Châtelier's principle

As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction caliber (Q) is equal to the equilibrium constant (Thousand). We next address what happens when a organisation at equilibrium is disturbed so that Q is no longer equal to K. If a system at equilibrium is subjected to a perturbance or stress (such equally a alter in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of Q volition no longer equal the value of K. To re-establish equilibrium, the system will either shift toward the products (if Q < K) or the reactants (if Q > K) until Q returns to the same value equally K.

This process is described by Le Châtelier's principle: When a chemical system at equilibrium is disturbed, it returns to equilibrium past counteracting the disturbance. Equally described in the previous paragraph, the disturbance causes a alter in Q; the reaction will shift to re-found Q = Thousand.

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 have a mixture of reactants and products that accept not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we tin can compare the values of Q and K for the organization to predict the changes.

Effect of Alter in Concentration on Equilibrium

A chemical organisation at equilibrium tin exist temporarily shifted out of equilibrium by adding or removing i or more than of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the arrangement to equilibrium.

The stress on the system in Effigy 1 is the reduction of the equilibrium concentration of SCN⁻ (lowering the concentration of one of the reactants would cause Q to exist larger than M). As a consequence, Le Châtelier's principle leads the states to predict that the concentration of Fe(SCN)²⁺ should decrease, increasing the concentration of SCN⁻ function way back to its original concentration, and increasing the concentration of Iron³⁺ in a higher place its initial equilibrium concentration.

Figure i. (a) The test tube contains 0.1 M Atomic number 26³⁺. (b) Thiocyanate ion has been added to solution in (a), forming the ruby-red Fe(SCN)²⁺ ion. Fe³⁺(aq) + SCN⁻(aq) ⇌ Fe(SCN)²⁺(aq). (c) Silver nitrate has been added to the solution in (b), precipitating some of the SCN⁻ as the white solid AgSCN. Ag⁺(aq) + SCN⁻(aq) ⇌ AgSCN(south). The decrease in the SCN⁻ concentration shifts the start equilibrium in the solution to the left, decreasing the concentration (and lightening colour) of the Fe(SCN)^two+. (credit: modification of work by Mark Ott)

The consequence of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction:

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

The numeric values for this example take been determined experimentally. A mixture of gases at 400 °C with [H_two] = [I_two] = 0.221 M and [Hi] = one.563 M is at equilibrium; for this mixture, Q_c = M_c = 50.0. If H₂ is introduced into the organization and then quickly that its concentration doubles earlier it begins to react (new [H_two] = 0.442 M), the reaction will shift so that a new equilibrium is reached, at which [H₂] = 0.374 M, [I_ii] = 0.153 Grand, and [HI] = 1.692 M. This gives:

[latex]Q_c = \frac{[\text{HI}]^ii}{[\text{H}_2][\text{I}_2]} = \frac{(1.692)^2}{(0.374)(0.153)} = l.0 = K_c[/latex]

We have stressed this system by introducing boosted H₂. The stress is relieved when the reaction shifts to the right, using upward some (but not all) of the backlog H₂, reducing the corporeality of uncombined I_two, and forming boosted How-do-you-do.

Event of Modify in Pressure on Equilibrium

Sometimes we can change the position of equilibrium past changing the pressure of a organisation. Withal, changes in force per unit area have a measurable event only in systems in which gases are involved, and then but when the chemical reaction produces a modify in the total number of gas molecules in the arrangement. An easy style to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure level (as well as related factors like volume), information technology is important to recollect that equilibrium constants are divers with regard to concentration (for K_c ) or partial force per unit area (for K_P ). Some changes to total force per unit area, like adding an inert gas that is not part of the equilibrium, will modify the total force per unit area but not the fractional pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium.

Check out this link to see a dramatic visual demonstration of how equilibrium changes with force per unit area changes.

As nosotros increment the pressure level of a gaseous system 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 ane or more than of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the full number of molecules per unit of book will be favored because this relieves the stress. The contrary reaction would be favored past a subtract in pressure.

Consider what happens when we increase the pressure on a organization in which NO, O₂, and NO_two are at equilibrium:

[latex]2\text{NO}(thou)\;+\;\text{O}_2(g)\;{\rightleftharpoons}\;2\text{NO}_2(yard)[/latex]

The germination of additional amounts of NO₂ decreases the total number of molecules in the system because each fourth dimension 2 molecules of NO_two course, a total of iii molecules of NO and O_ii are consumed. This reduces the total pressure exerted past the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a subtract in the pressure on the system favors decomposition of NO₂ into NO and O₂, which tends to restore the force per unit area.

Now consider this reaction:

[latex]\text{Due north}_2(1000)\;+\;\text{O}_2(grand)\;{\rightleftharpoons}\;2\text{NO}(thou)[/latex]

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

Effect of Change in Temperature on Equilibrium

Changing concentration or pressure perturbs an equilibrium considering the reaction caliber is shifted away from the equilibrium value. Irresolute the temperature of a system at equilibrium has a dissimilar issue: A change in temperature really changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Châtelier's principle.

When hydrogen reacts with gaseous iodine, rut is evolved.

[latex]\text{H}_2(grand)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{HI}(chiliad)\;\;\;\;\;\;\;{\Delta}H = -ix.4\;\text{kJ\;(exothermic)}[/latex]

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

[latex]\text{H}_2(1000)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{Hullo}(chiliad)\;+\;\text{heat}[/latex]

Increasing the temperature of the reaction increases the internal free energy of the organisation. Thus, increasing the temperature has the effect of increasing the amount of one 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 I_two and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the product of heat, and favors the formation 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 HI has increased and the concentrations of H₂ and I₂ decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.five at 357 °C to fifty.0 at 400 °C.

Temperature affects the equilibrium betwixt NO₂ and N_twoO₄ in this reaction

[latex]\text{N}_2\text{O}_4(yard)\;{\rightleftharpoons}\;2\text{NO}_2(thousand)\;\;\;\;\;\;\;{\Delta}H = 57.20\;\text{kJ}[/latex]

The positive ΔH value tells u.s. that the reaction is endothermic and could exist written

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

At higher temperatures, the gas mixture has a deep brown colour, indicative of a significant corporeality of brown NO₂ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing free energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless N₂O₄ increases, and the concentration of dark-brown NO₂ decreases, causing the brown color to fade.

This interactive blitheness allows yous to apply Le Châtelier's principle to predict the effects of changes in concentration, pressure, and temperature on reactant and product concentrations.

Catalysts Practice Not Affect Equilibrium

As we learned during our study of kinetics, a catalyst tin speed upwards the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more chop-chop (by speeding upward 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 catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation

[latex]\text{Due north}_2(one thousand)\;+\;iii\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(thousand)[/latex]

A large quantity of ammonia is manufactured by this reaction. Each twelvemonth, ammonia is among the top x chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the U.s. each twelvemonth.

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

Fritz Haber

In the early 20th century, High german chemist Fritz Haber (Figure 2) developed a practical process for converting diatomic nitrogen, which cannot be used by plants as a nutrient, to ammonia, a form of nitrogen that is easiest for plants to blot.

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

The availability of nitrogen is a potent limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (North₂) is nutritionally unavailable due the tremendous stability of the nitrogen-nitrogen triple bond. For plants to use atmospheric nitrogen, the nitrogen must exist converted to a more than bioavailable form (this conversion is called nitrogen fixation).

Haber was born in Breslau, Prussia (shortly Wroclaw, Poland) in December 1868. He went on to study chemical science and, while at the University of Karlsruhe, he adult what would subsequently be known equally the Haber process: the catalytic germination of ammonia from hydrogen and atmospheric nitrogen nether high 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, equally information technology immune the product 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 constructed fertilizer production has increased the number of humans that arable state can support from 1.nine persons per hectare in 1908 to 4.3 in 2008.

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

In addition to his work in ammonia production, Haber is besides remembered by history every bit one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his part in these developments, Haber said, "During peace fourth dimension a scientist belongs to the Globe, but during war fourth dimension he belongs to his state."^[1] Haber defended the use of gas warfare against accusations that it was inhumane, maxim that death was death, by whatever ways it was inflicted. He stands every bit an instance of the ethical dilemmas that confront scientists in times of state of war and the double-edged nature of the sword of science.

Like Haber, the products made from ammonia can exist multifaceted. In addition to their value for agriculture, nitrogen compounds tin can also be used to achieve destructive ends. Ammonium nitrate has too been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the assail on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995.

It has long been known that nitrogen and hydrogen react to grade ammonia. Nevertheless, it became possible to manufacture ammonia in useful quantities by the reaction of nitrogen and hydrogen simply in the early on 20th century afterwards the factors that influence its equilibrium were understood.

To be practical, an industrial process must give a large yield of production relatively rapidly. Ane fashion to increase the yield of ammonia is to increase the pressure on the system in which Due north₂, H₂, and NH_three are at equilibrium or are coming to equilibrium.

[latex]\text{N}_2(g)\;+\;iii\text{H}_2(thousand)\;{\rightleftharpoons}\;ii\text{NH}_3(g)[/latex]

The germination of additional amounts of ammonia reduces the total pressure exerted by the arrangement and somewhat reduces the stress of the increased pressure.

Although increasing the force per unit area of a mixture of Due north₂, H_two, and NH₃ will increase the yield of ammonia, at low temperatures, the charge per unit of formation of ammonia is boring. At room temperature, for instance, the reaction is so slow that if we prepared a mixture of N_two and H₂, no detectable corporeality of ammonia would form during our lifetime. The formation of ammonia from hydrogen and nitrogen is an exothermic process:

[latex]\text{North}_2(thousand)\;+\;3\text{H}_2(g)\;{\longrightarrow}\;2\text{NH}_3(thousand)\;\;\;\;\;\;\;{\Delta}H = -92.two\;\text{kJ}[/latex]

Thus, increasing the temperature to increment the rate lowers the yield. If nosotros lower the temperature to shift the equilibrium to favor the germination of more than ammonia, equilibrium is reached more than slowly because of the large subtract of reaction charge per unit with decreasing temperature.

Part of the rate of formation lost by operating at lower temperatures can be recovered by using a catalyst. The cyberspace effect of the goad on the reaction is to crusade equilibrium to exist reached more than chop-chop.

In the commercial production of ammonia, weather condition of about 500 °C, 150–900 atm, and the presence of a catalyst are used to give the best compromise among rate, yield, and the toll of the equipment necessary to produce and incorporate loftier-pressure gases at loftier temperatures (Figure iii).

Effigy 3. Commercial production of ammonia requires heavy equipment to handle the high temperatures and pressures required. This schematic outlines the design of an ammonia plant.

Fundamental Concepts and Summary

Systems at equilibrium can be disturbed by 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 unlike on the reactant and product sides of the reaction. The system'southward response to these disturbances is described by Le Châtelier's principle: The system volition answer in a fashion that counteracts the disturbance. Not all changes to the system consequence in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions only does not modify the equilibrium, and changing pressure level or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side.

Disturbance	Observed Change as Equilibrium is Restored	Direction of Shift	Effect on K
reactant added	added reactant is partially consumed	toward products	none
production added	added product is partially consumed	toward reactants	none
subtract in book/increase in gas pressure	pressure level decreases	toward side with fewer moles of gas	none
increase in book/decrease in gas pressure	force per unit area increases	toward side with more moles of gas	none
temperature increase	estrus is captivated	toward products for endothermic, toward reactants for exothermic	changes
temperature decrease	heat is given off	toward reactants for endothermic, toward products for exothermic	changes
Table 2. Effects of Disturbances of Equilibrium and M

Chemistry Stop of Chapter Exercises

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

   Nether what conditions volition decomposition in a closed container proceed to completion so that no CaCO_iii remains?

2. Explain how to recognize the weather under which changes in pressure would touch on systems at equilibrium.
3. What property of a reaction 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 colour of the solution in part (b) of Effigy 1 if a small amount of NaOH were added and Iron(OH)_iii precipitated? Explain your answer.
5. The post-obit reaction occurs when a burner on a gas stove is lit:
   [latex]\text{CH}_4(grand)\;+\;2\text{O}_2(g)\;{\rightleftharpoons}\;\text{CO}_2(g)\;+\;2\text{H}_2\text{O}(thou)[/latex]

   Is an equilibrium among CH₄, O₂, CO_ii, and H₂O established under these conditions? Explain your answer.

6. A necessary pace in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO₃, from sulfur dioxide, SO_ii, and oxygen, O₂, shown here. At loftier temperatures, the rate of germination of Then_iii is college, only the equilibrium corporeality (concentration or fractional pressure) of And then₃ is lower than it would exist at lower temperatures.
   [latex]2\text{SO}_2(g)\;+\;\text{O}_2(one thousand)\;{\longrightarrow}\;2\text{SO}_3(1000)[/latex]

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

   (b) Is the reaction endothermic or exothermic?

7. Suggest 4 ways in which the concentration of hydrazine, North₂H₄, could be increased in an equilibrium described by the following equation:
   [latex]\text{North}_2(one thousand)\;+\;2\text{H}_2(g)\;{\rightleftharpoons}\;\text{N}_2\text{H}_4(g)\;\;\;\;\;\;\;{\Delta}H = 95\;\text{kJ}[/latex]
8. Advise four ways in which the concentration of PH_three could be increased in an equilibrium described by the post-obit equation:
   [latex]\text{P}_4(g)\;+\;half-dozen\text{H}_2(g)\;{\rightleftharpoons}\;4\text{PH}_3(yard)\;\;\;\;\;\;\;{\Delta}H = 110.5\;\text{kJ}[/latex]
9. How will an increase in temperature affect each of the post-obit equilibria? How will a subtract in the volume of the reaction vessel bear on each?

   (a) [latex]2\text{NH}_3(thousand)\;{\rightleftharpoons}\;\text{North}_2(1000)\;+\;3\text{H}_2(g)\;\;\;\;\;\;\;{\Delta}H = 92\;\text{kJ}[/latex]

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

   (c) [latex]2\text{O}_3(k)\;{\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(s)\;\;\;\;\;\;\;{\Delta}H = -176\;\text{kJ}[/latex]

10. How will an increase in temperature bear on each of the following equilibria? How volition a decrease in the volume of the reaction vessel affect each?

    (a) [latex]2\text{H}_2\text{O}(k)\;{\rightleftharpoons}\;2\text{H}_2(k)\;+\;\text{O}_2(1000)\;\;\;\;\;\;\;{\Delta}H = 484\;\text{kJ}[/latex]

    (b) [latex]\text{N}_2(m)\;+\;3\text{H}_2(g)\;{\rightleftharpoons}\;2\text{NH}_3(g)\;{\Delta}H = -92.two\;\text{kJ}[/latex]

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

    (d) [latex]\text{H}_2(1000)\;+\;\text{I}_2(s)\;{\rightleftharpoons}\;2\text{HI}(g)\;\;\;\;\;\;\;{\Delta}H = 53\;\text{kJ}[/latex]

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

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

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

    (b) What volition happen to the concentrations of H_two, CO, and CH_iiiOH at equilibrium if more H_ii is added?

    (c) What will happen to the concentrations of H₂, CO, and CH₃OH at equilibrium if CO is removed?

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

    (e) What will happen to the concentrations of H_two, CO, and CH₃OH at equilibrium if the temperature of the system 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 (K_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 volition happen to the concentrations of N₂, O₂, and NO at equilibrium if more O₂ is added?

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

    (d) What will happen to the concentrations of Due north₂, O_ii, and NO at equilibrium if NO is added?

    (e) What will happen to the concentrations of Northward₂, O_two, and NO at equilibrium if the force per unit area on the system is increased past reducing the volume of the reaction vessel?

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

    (g) What volition happen to the concentrations of Due north₂, O₂, and NO at equilibrium if a catalyst is added?

13. Water gas, a mixture of H₂ and CO, is an important industrial fuel produced past 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}(south)\;+\;\text{H}_2\text{O}(m)\;{\rightleftharpoons}\;\text{CO}(g)\;+\;\text{H}_2(g)\;\;\;\;\;\;\;{\Delta}H = 131.30\;\text{kJ}[/latex]

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

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

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

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

14. Pure fe metal can be produced by the reduction of iron(III) oxide with hydrogen gas.

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

    [latex]\text{Fe}_2\text{O}_3(s)\;+\;iii\text{H}_2(g)\;{\rightleftharpoons}\;2\text{Fe}(due south)\;+\;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 production at equilibrium if more than Fe is added?

    (c) What volition 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 product at equilibrium if H₂ is added?

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

    (f) What will happen to the concentration of each reactant and production 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 h2o?

    (a) Addition of NaOH

    (b) Addition of HCl

    (c) Addition of NH₄Cl

16. Acetic acrid is a weak acid that reacts with water co-ordinate 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]

    Will whatever of the following increase the pct of acetic acid that reacts and produces [latex]\text{CH}_3\text{CO}_2^{\;\;-}[/latex] ion?

    (a) Addition of HCl

    (b) Improver of NaOH

    (c) Improver of NaCH₃CO_two

17. Suggest ii 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}(l)\;{\rightleftharpoons}\;\text{H}_2\text{O}(1000)\;\;\;\;\;\;\;{\Delta}H = 41\;\text{kJ}[/latex]
19. Additional solid argent sulfate, a slightly soluble solid, is added to a solution of argent ion and sulfate ion at equilibrium with solid argent sulfate.
    [latex]two\text{Ag}^{+}(aq)\;+\;\text{And so}_4^{\;\;2-}(aq)\;{\rightleftharpoons}\;\text{Ag}_2\text{SO}_4(s)[/latex]

    Which of the post-obit will occur?

    (a) Ag⁺ or [latex]\text{And so}_4^{\;\;2-}[/latex] concentrations will not alter.

    (b) The added silver sulfate will deliquesce.

    (c) Boosted silver sulfate volition form and precipitate from solution as Ag⁺ ions and [latex]\text{And then}_4^{\;\;ii-}[/latex] ions combine.

    (d) The Ag⁺ ion concentration volition increase 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 constant 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 by counteracting the disturbance

position of equilibrium

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

stress

change to a reaction'due south conditions that may cause a shift in the equilibrium

Solutions

Answers to Chemical science Finish of Chapter Exercises

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

three. The modify 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 exist thought of as a reactant. Additional heat would shift an exothermic reaction back 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, information technology is not at equilibrium. Because the system is not bars, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding temper.

seven. Add N_ii; add H_ii; decrease the container volume; heat the mixture.

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

xi. (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), [H_two] increases, [CO] decreases, [CH_threeOH] decreases; (d), [H₂] increases, [CO] increases, [CH₃OH] increases; (e), [H_two] increases, [CO] increases, [CH₃OH] decreases; (f), no changes.

13. (a) [latex]K_c = \frac{[\text{CO}][\text{H}_2]}{[\text{H}_2\text{O}]}[/latex]; (b) [H_iiO] no modify, [CO] no modify, [H₂] no change; (c) [H₂O] decreases, [CO] decreases, [H_two] decreases; (d) [H₂O] increases, [CO] increases, [H₂] decreases; (f) [H₂O] decreases, [CO] increases, [H_ii] increases. In (b), (c), (d), and (e), the mass of carbon will alter, simply its concentration (activity) volition non alter.

15. But (b)

17. Add NaCl or another salt that produces Cl⁻ to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(southward).

19. (a)

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

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