By What Factor Will The Rate Of The Reaction Change If The Ph Decreases From 5.50 To 2.00?

Affiliate 13. Fundamental Equilibrium Concepts

13.3 Shifting Equilibria: Le Châtelier'south Principle

Learning Objectives

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

• Describe the ways in which an equilibrium arrangement tin be stressed
• Predict the response of a stressed equilibrium using Le Châtelier'due south principle

As we saw in the previous section, reactions keep in both directions (reactants go to products and products become to reactants). We can tell a reaction is at equilibrium if the reaction quotient (Q) is equal to the equilibrium constant (Yard). We next address what happens when a system at equilibrium is disturbed so that Q is no longer equal to 1000. If a system at equilibrium is subjected to a perturbance or stress (such as 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 system will either shift toward the products (if Q < K) or the reactants (if Q > Chiliad) until Q returns to the same value equally K.

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

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 have not notwithstanding reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a instance, we tin compare the values of Q and K for the organisation to predict the changes.

Event of Change in Concentration on Equilibrium

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

The stress on the system in Figure i is the reduction of the equilibrium concentration of SCN⁻ (lowering the concentration of one of the reactants would cause Q to be larger than Yard). Every bit a consequence, Le Châtelier'southward principle leads us to predict that the concentration of Fe(SCN)²⁺ should decrease, increasing the concentration of SCN⁻ part way back to its original concentration, and increasing the concentration of Feⁱⁱⁱ⁺ above its initial equilibrium concentration.

Figure i. (a) The exam tube contains 0.i M Fe³⁺. (b) Thiocyanate ion has been added to solution in (a), forming the red Fe(SCN)²⁺ ion. Fe^three+(aq) + SCN⁻(aq) ⇌ Iron(SCN)^two+(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 Atomic number 26(SCN)ⁱⁱ⁺. (credit: modification of work past Mark Ott)

The effect of a alter in concentration on a system at equilibrium is illustrated farther by the equilibrium of this chemical reaction:

[latex]\text{H}_2(m)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;2\text{Hi}(yard)\;\;\;\;\;\;\;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 [H₂] = [I_two] = 0.221 Thou and [HI] = ane.563 M is at equilibrium; for this mixture, Q_c = 1000_c = 50.0. If H₂ is introduced into the arrangement and so chop-chop that its concentration doubles before it begins to react (new [H₂] = 0.442 One thousand), the reaction will shift and then that a new equilibrium is reached, at which [H₂] = 0.374 M, [I_ii] = 0.153 M, and [Hello] = i.692 1000. This gives:

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

We have stressed this arrangement by introducing boosted H_two. The stress is relieved when the reaction shifts to the right, using up some (but non all) of the excess H₂, reducing the amount of uncombined I_two, and forming additional HI.

Outcome of Change in Pressure on Equilibrium

Sometimes we tin can alter the position of equilibrium by irresolute the pressure level of a system. However, changes in pressure have a measurable effect just in systems in which gases are involved, and so only when the chemical reaction produces a change in the total number of gas molecules in the system. An like shooting fish in a barrel way to recognize such a system is to wait for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (also equally related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for 1000_c ) or fractional pressure level (for 1000_P ). Some changes to total force per unit area, like adding an inert gas that is non part of the equilibrium, will change the total pressure but non the fractional pressures of the gases in the equilibrium constant expression. Thus, add-on of a gas not involved in the equilibrium volition not perturb the equilibrium.

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

Equally nosotros increment the pressure of a gaseous system at equilibrium, either by decreasing the volume of the organization or by calculation more of i of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Châtelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of book will be favored because this relieves the stress. The reverse reaction would exist favored by a decrease in pressure.

Consider what happens when we increase the pressure level on a organisation in which NO, O₂, and NO₂ are at equilibrium:

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

The germination of additional amounts of NO_ii decreases the total number of molecules in the system because each time two molecules of NO_two form, a total of 3 molecules of NO and O₂ are consumed. This reduces the total pressure exerted by the organisation and reduces, but does non completely salvage, the stress of the increased force per unit area. On the other paw, a decrease in the pressure on the system favors decomposition of NO₂ into NO and O_ii, which tends to restore the pressure.

Now consider this reaction:

[latex]\text{N}_2(chiliad)\;+\;\text{O}_2(thousand)\;{\rightleftharpoons}\;two\text{NO}(k)[/latex]

Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide.

Event of Change in Temperature on Equilibrium

Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Irresolute the temperature of a system at equilibrium has a different outcome: A change in temperature actually changes the value of the equilibrium constant. Nonetheless, we can qualitatively predict the effect of the temperature change past treating it as a stress on the organisation and applying Le Châtelier's principle.

When hydrogen reacts with gaseous iodine, heat is evolved.

[latex]\text{H}_2(thousand)\;+\;\text{I}_2(g)\;{\rightleftharpoons}\;ii\text{Hi}(grand)\;\;\;\;\;\;\;{\Delta}H = -9.4\;\text{kJ\;(exothermic)}[/latex]

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

[latex]\text{H}_2(grand)\;+\;\text{I}_2(one thousand)\;{\rightleftharpoons}\;2\text{How-do-you-do}(g)\;+\;\text{rut}[/latex]

Increasing the temperature of the reaction increases the internal energy of the arrangement. Thus, increasing the temperature has the effect of increasing the corporeality of one of the products of this reaction. The reaction shifts to the left to save the stress, and there is an increment 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 nowadays, favors the production of estrus, and favors the formation of hydrogen iodide.

When we 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 constant: At the new equilibrium the concentration of How-do-you-do has increased and the concentrations of H_ii 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 between NO₂ and N_twoO_iv in this reaction

[latex]\text{Northward}_2\text{O}_4(g)\;{\rightleftharpoons}\;two\text{NO}_2(g)\;\;\;\;\;\;\;{\Delta}H = 57.20\;\text{kJ}[/latex]

The positive ΔH value tells united states of america that the reaction is endothermic and could be written

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

At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown NO₂ molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless N_iiO₄ increases, and the concentration of brown NO_ii decreases, causing the brown color to fade.

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

Catalysts Do Non 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 arrangement to reach equilibrium more quickly (by speeding up the forrard and reverse reactions), a goad has no consequence on the value of an equilibrium abiding nor on equilibrium concentrations.

The interplay of changes in concentration or force per unit area, temperature, and the lack of an influence of a catalyst on a chemic equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation

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

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

Ammonia plays a vital role in our global economic system. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Big 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 also used in the steel industry.

Fritz Haber

In the early 20th century, German chemist Fritz Haber (Figure 2) developed a practical procedure for converting diatomic nitrogen, which cannot exist used by plants every bit a nutrient, to ammonia, a grade of nitrogen that is easiest for plants to absorb.

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

The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting 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 be converted to a more than bioavailable course (this conversion is called nitrogen fixation).

Haber was born in Breslau, Prussia (soon Wroclaw, Poland) in December 1868. He went on to report chemical science and, while at the Academy of Karlsruhe, he developed what would subsequently be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this piece of work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements. The Haber procedure was a boon to agriculture, every bit it allowed the product of fertilizers to no longer be dependent on mined feed stocks such every bit 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 abundant state can back up from 1.nine persons per hectare in 1908 to iv.three in 2008.

Figure 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 artillery.

In add-on to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During Globe War I, he played a major function 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, merely 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 death, by whatsoever ways it was inflicted. He stands as 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 tin exist multifaceted. In improver to their value for agriculture, nitrogen compounds can too be used to reach destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack 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. However, it became possible to manufacture ammonia in useful quantities past the reaction of nitrogen and hydrogen but in the early on 20th century later the factors that influence its equilibrium were understood.

To be practical, an industrial process must give a large yield of product relatively rapidly. One mode to increment the yield of ammonia is to increment the pressure on the system in which N₂, H_two, and NH_three are at equilibrium or are coming to equilibrium.

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

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

Although increasing the force per unit area of a mixture of N₂, H₂, and NH₃ will increase the yield of ammonia, at low temperatures, the rate of formation of ammonia is irksome. At room temperature, for example, the reaction is and then slow that if we prepared a mixture of N_ii and H_ii, no detectable amount of ammonia would class 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(g)\;\;\;\;\;\;\;{\Delta}H = -92.2\;\text{kJ}[/latex]

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

Role of the rate of germination lost by operating at lower temperatures tin can exist recovered by using a catalyst. The net effect of the catalyst on the reaction is to cause equilibrium to be reached more than rapidly.

In the commercial production of ammonia, conditions 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 contain high-pressure gases at loftier temperatures (Figure 3).

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

Central Concepts and Summary

Systems at equilibrium can be disturbed past changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure level changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The organization'south response to these disturbances is described past Le Châtelier's principle: The organisation will reply in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not modify the equilibrium, and irresolute pressure or book 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 as Equilibrium is Restored	Management of Shift	Result on G
reactant added	added reactant is partially consumed	toward products	none
production added	added product is partially consumed	toward reactants	none
decrease in volume/increase in gas pressure	pressure decreases	toward side with fewer moles of gas	none
increase in book/decrease in gas force per unit area	pressure increases	toward side with more moles of gas	none
temperature increase	rut 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 K

Chemistry End of Chapter Exercises

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

   Under what weather will decomposition in a airtight container proceed to completion then that no CaCO_iii remains?

2. Explain how to recognize the conditions under which changes in pressure would bear on systems at equilibrium.
3. What holding of a reaction tin we use to predict the effect of a modify 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 Fe(OH)₃ precipitated? Explain your answer.
5. The following reaction occurs when a burner on a gas stove is lit:
   [latex]\text{CH}_4(g)\;+\;two\text{O}_2(yard)\;{\rightleftharpoons}\;\text{CO}_2(grand)\;+\;2\text{H}_2\text{O}(g)[/latex]

   Is an equilibrium among CH₄, O₂, CO₂, and H_iiO established under these conditions? Explain your answer.

6. A necessary step in the manufacture of sulfuric acrid is the formation of sulfur trioxide, Then_iii, from sulfur dioxide, So₂, and oxygen, O₂, shown here. At high temperatures, the charge per unit of germination of SO₃ is higher, but the equilibrium amount (concentration or partial pressure level) of Then₃ is lower than it would exist at lower temperatures.
   [latex]two\text{SO}_2(g)\;+\;\text{O}_2(g)\;{\longrightarrow}\;2\text{And so}_3(grand)[/latex]

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

   (b) Is the reaction endothermic or exothermic?

7. Suggest four means in which the concentration of hydrazine, Due north₂H₄, could be increased in an equilibrium described by the following equation:
   [latex]\text{N}_2(yard)\;+\;two\text{H}_2(g)\;{\rightleftharpoons}\;\text{Due north}_2\text{H}_4(g)\;\;\;\;\;\;\;{\Delta}H = 95\;\text{kJ}[/latex]
8. Suggest four ways in which the concentration of PH_iii could be increased in an equilibrium described by the post-obit equation:
   [latex]\text{P}_4(g)\;+\;half dozen\text{H}_2(yard)\;{\rightleftharpoons}\;iv\text{PH}_3(chiliad)\;\;\;\;\;\;\;{\Delta}H = 110.5\;\text{kJ}[/latex]
9. How volition an increase in temperature affect each of the following equilibria? How will a subtract in the book of the reaction vessel affect each?

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

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

   (c) [latex]ii\text{O}_3(g)\;{\rightleftharpoons}\;iii\text{O}_2(yard)\;\;\;\;\;\;\;{\Delta}H = -285\;\text{kJ}[/latex]

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

10. How will an increase in temperature impact each of the post-obit equilibria? How will a subtract in the volume of the reaction vessel impact each?

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

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

    (c) [latex]2\text{Br}(g)\;{\rightleftharpoons}\;\text{Br}_2(g)\;\;\;\;\;\;\;{\Delta}H = -224\;\text{kJ}[/latex]

    (d) [latex]\text{H}_2(g)\;+\;\text{I}_2(s)\;{\rightleftharpoons}\;two\text{Howdy}(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 because 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 peradventure replace gasoline, tin be prepared from water gas and hydrogen at high temperature and pressure level in the presence of a suitable catalyst.

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

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

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

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

    (d) What volition happen to the concentrations of H_ii, CO, and CH₃OH at equilibrium if CH_iiiOH is added?

    (e) What will happen to the concentrations of H_two, CO, and CH_threeOH at equilibrium if the temperature of the system is increased?

    (f) What will happen to the concentrations of H_ii, CO, and CH₃OH at equilibrium if more catalyst is added?

12. Nitrogen and oxygen react at high temperatures.

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

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

    (b) What will happen to the concentrations of Northward_ii, O₂, and NO at equilibrium if more than O_ii is added?

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

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

    (e) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if the force per unit area on the organization is increased by reducing the volume of the reaction vessel?

    (f) What will happen to the concentrations of N₂, O₂, and NO at equilibrium if the temperature of the system is increased?

    (thousand) 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_two and CO, is an important industrial fuel produced by the reaction of steam with carmine hot coke, essentially pure carbon.

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

    [latex]\text{C}(southward)\;+\;\text{H}_2\text{O}(g)\;{\rightleftharpoons}\;\text{CO}(g)\;+\;\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 C is added?

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

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

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

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

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

    [latex]\text{Fe}_2\text{O}_3(s)\;+\;3\text{H}_2(thousand)\;{\rightleftharpoons}\;2\text{Iron}(s)\;+\;three\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 Fe 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 product at equilibrium if H_two is added?

    (e) What will happen to the concentration of each reactant and product at equilibrium if the pressure on the system 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 according to this equation:
    [latex]\text{NH}_3(aq)\;+\;\text{H}_2\text{O}(l)\;{\rightleftharpoons}\;\text{NH}_4^{\;\;+}(aq)\;+\;\text{OH}^{-}(aq)[/latex]

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

    (a) Addition of NaOH

    (b) Addition of HCl

    (c) Add-on of NH_fourCl

16. Acetic acid 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]

    Will any of the following increase the percent of acetic acid 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 two means 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.nine\;\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}(k)\;\;\;\;\;\;\;{\Delta}H = 41\;\text{kJ}[/latex]
19. Boosted solid silver sulfate, a slightly soluble solid, is added to a solution of silverish ion and sulfate ion at equilibrium with solid silver sulfate.
    [latex]two\text{Ag}^{+}(aq)\;+\;\text{Then}_4^{\;\;2-}(aq)\;{\rightleftharpoons}\;\text{Ag}_2\text{SO}_4(s)[/latex]

    Which of the following will occur?

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

    (b) The added argent sulfate will dissolve.

    (c) Additional argent sulfate volition form and precipitate from solution equally Ag⁺ ions and [latex]\text{SO}_4^{\;\;ii-}[/latex] ions combine.

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

20. The amino acid alanine has 2 isomers, α-alanine and β-alanine. When equal masses of these two 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 partial pressures of components of a reaction at equilibrium (commonly used to describe atmospheric condition before a disturbance)

stress

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

Solutions

Answers to Chemical science Terminate of Affiliate Exercises

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

3. The alter in enthalpy may be used. If the reaction is exothermic, the heat produced can exist thought of as a product. If the reaction is endothermic the heat added tin exist thought of as a reactant. Additional rut 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 cause it to shift to the reactants' side.

5. No, it is not at equilibrium. Because the organization is not 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; oestrus the mixture.

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

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

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

15. But (b)

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

nineteen. (a)

---

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

Posted by: kylefractoggen.blogspot.com

Share this post