Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved
Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. Copyright © Cengage Learning. All rights reserved

Chemical Equilibrium When compounds react, they eventually form a mixture of products and unreacted reactants, in a dynamic equilibrium. A dynamic equilibrium consists of a forward reaction, in which substances react to give products, and a reverse reaction, in which products react to give the original reactants. Chemical equilibrium is the state reached by a reaction mixture when the rates of the forward and reverse reactions have become equal. 2

Equilibrium is a state in which there are no observable changes as time goes by.
Chemical equilibrium is achieved when: the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant Physical equilibrium H2O (l) H2O (g) Chemical equilibrium N2O4 (g) 2NO2 (g)

Chemical Equilibrium Much like water in a U-shaped tube, there is constant mixing back and forth through the lower portion of the tube. “reactants” “products” It’s as if the forward and reverse “reactions” were occurring at the same rate. The system appears to static (stationary) when, in reality, it is dynamic (in constant motion). 2

N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4

Chemical Equilibrium The Haber process for producing ammonia from N2 and H2 does not go to completion: It establishes an equilibrium state where all three species are present. 2

A Problem to Consider Applying Stoichiometry to an Equilibrium Mixture. Suppose we place mol N2 and mol H2 in a reaction vessel at 450 oC and 10.0 atmospheres of pressure. The reaction is What is the composition of the equilibrium mixture if it contains mol NH3? 2

A Problem to Consider Using the information given, set up the following table. Initial 1.000 3.000 Change -x -3x +2x Equilibrium x x 2x = mol The equilibrium amount of NH3 was given as mol. Therefore, 2x = mol NH3 (x = mol). 2

A Problem to Consider Using the information given, set up the following table. Initial 1.000 3.000 Change -x -3x +2x Equilibrium x x 2x = mol Equilibrium amount of N2 = = mol N2 Equilibrium amount of H2 = (3 x 0.040) = mol H2 Equilibrium amount of NH3 = 2x = mol NH3 2

CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) H2(g) + CO2(g) You add more H2O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. The concentrations of each product will increase, the concentration of CO will decrease, and the concentration of water will be higher than the original equilibrium concentration, but lower than the initial total amount. Students may have many different answers (hydrogen goes up, but carbon dioxide in unchanged, etc.) Let them talk about this for a while – do not go over the answer until each group of students has come up with an explanation. This question also sets up LeChâtelier’s principle for later. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider an equilibrium mixture in a closed vessel reacting according to the equation: H2O(g) + CO(g) H2(g) + CO2(g) You add more H2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. This is the opposite scenario of the previous slide. The concentrations of water and CO will increase. The concentration of carbon dioxide decreases and the concentration of hydrogen will be higher than the original equilibrium concentration, but lower than the initial total amount. Copyright © Cengage Learning. All rights reserved

The Equilibrium Constant
Every reversible system has its own “position of equilibrium” under any given set of conditions. The ratio of products produced to unreacted reactants for any given reversible reaction remains constant under constant conditions of pressure and temperature. The numerical value of this ratio is called the equilibrium constant for the given reaction. 2

The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. For the general equation above, the equilibrium-constant expression would be: 2

The equilibrium-constant expression for a reaction is obtained by multiplying the concentrations of products, dividing by the concentrations of reactants, and raising each concentration to a power equal to its coefficient in the balanced chemical equation. The molar concentration of a substance is denoted by writing its formula in square brackets. 2

Examples Write equilibrium expressions for the following equilibria:
1. 2HI(aq) H2(g) + I2(g) 2. BaCl2(s) Ba2+(aq) + 2Cl-(aq) 3. CaCO3(s) CaCO(s) + CO2(g)

The equilibrium constant, Kc, is the value obtained for the equilibrium-constant expression when equilibrium concentrations are substituted. A large Kc indicates large concentrations of products at equilibrium. A small Kc indicates large concentrations of unreacted reactants at equilibrium. 2

Interpreting Equilibrium Constants
The numerical value of the equilibrium constant tells us the extent to which reactants have converted to products. 1. Keq greater than 1 x 102. Large value of Keq: numerator > denominator. At equilibrium mostly product present.

Which is larger, numerator or denominator? Ans: Denominator.
Keq less than 1 x 10-2. Which is larger, numerator or denominator? Ans: Denominator. At equilibrium which is present in the larger quantity? Reactants or Products? Ans: Reactants. Keq between 1 x 10-2 and 1 x 102 Equilibrium mixture contains significant concentration of both reactants and products.

constant

N2O4 (g) NO2 (g) K = [NO2]2 [N2O4] = 4.63 x 10-3 aA + bB cC + dD K = [C]c[D]d [A]a[B]b Law of Mass Action

Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1
Lie to the right Favor products K << 1 Lie to the left Favor reactants

The law of mass action states that the value of the equilibrium constant expression Kc is constant for a particular reaction at a given temperature, whatever equilibrium concentrations are substituted. Consider the equilibrium established in the Haber process. 2

The equilibrium-constant expression would be Note that the stoichiometric coefficients in the balanced equation have become the powers to which the concentrations are raised. 2

Equilibrium: A Kinetics Argument
If the forward and reverse reaction rates in a system at equilibrium are equal, then it follows that their rate laws would be equal. Consider the decomposition of N2O4, dinitrogen tetroxide. However, once some NO2 is produced it can recombine to form N2O4. 2

kf kr Call the decomposition of N2O4 the forward reaction and the formation of N2O4 the reverse reaction. These are elementary reactions, and you can immediately write the rate law for each. Here kf and kr represent the forward and reverse rate constants. 2

kf kr Ultimately, this reaction reaches an equilibrium state where the rates of the forward and reverse reactions are equal. Therefore, 2

kf kr Combining the constants you can identify the equilibrium constant, Kc, as the ratio of the forward and reverse rate constants. 2

Obtaining Equilibrium Constants for Reactions
Equilibrium concentrations for a reaction must be obtained experimentally and then substituted into the equilibrium-constant expression in order to calculate Kc. 2

Consider the reaction below Suppose we started with initial concentrations of CO and H2 of M and M, respectively. 2

Consider the reaction below When the system finally settled into equilibrium we determined the equilibrium concentrations to be as follows. Reactants Products [CO] = M [H2] = M [CH4] = M [H2O] = M 2

Consider the reaction below The equilibrium-constant expression for this reaction is: 2

Consider the reaction below If we substitute the equilibrium concentrations, we obtain: 2

Consider the reaction below Regardless of the initial concentrations (whether they be reactants or products), the law of mass action dictates that the reaction will always settle into an equilibrium where the equilibrium-constant expression equals Kc. 2

Consider the reaction below As an example, let’s repeat the previous experiment, only this time starting with initial concentrations of products: [CH4]initial = M and [H2O]initial = M 2

Consider the reaction below We find that these initial concentrations result in the following equilibrium concentrations. Reactants Products [CO] = M [H2] = M [CH4] = M [H2O] = M 2

Consider the reaction below Substituting these values into the equilibrium-constant expression, we obtain the same result. Whether we start with reactants or products, the system establishes the same ratio. 2

The Equilibrium Constant, Kp
In discussing gas-phase equilibria, it is often more convenient to express concentrations in terms of partial pressures rather than molarities. It can be seen from the ideal gas equation that the partial pressure of a gas is proportional to its molarity. 2

If we express a gas-phase equilibria in terms of partial pressures, we obtain Kp. Consider the reaction below. The equilibrium-constant expression in terms of partial pressures becomes: 2

In general, the numerical value of Kp differs from that of Kc. From the relationship n/V=P/RT, we can show that where Dn is the sum of the moles of gaseous products in a reaction minus the sum of the moles of gaseous reactants. 2

A Problem to Consider Consider the reaction
Kc for the reaction is 2.8 x 102 at 1000 oC. Calculate Kp for the reaction at this temperature. 2

A Problem to Consider Consider the reaction We know that
From the equation we see that Dn = -1. We can simply substitute the given reaction temperature and the value of R ( L.atm/mol.K) to obtain Kp. 2

A Problem to Consider Consider the reaction Since 2

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.
N2O4 (g) NO2 (g) Kp = NO2 P2 N2O4 P Kc = [NO2]2 [N2O4] In most cases Kc  Kp aA (g) + bB (g) cC (g) + dD (g) Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)

Homogeneous Equilibrium
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq) [CH3COO-][H3O+] [CH3COOH][H2O] Kc = ‘ [H2O] = constant [CH3COO-][H3O+] [CH3COOH] = Kc [H2O] ‘ Kc = General practice not to include units for the equilibrium constant.

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g) COCl2 (g) [COCl2] [CO][Cl2] = 0.14 0.012 x 0.054 Kc = = 220 Kp = Kc(RT)Dn Dn = 1 – 2 = -1 R = T = = 347 K Kp = 220 x ( x 347)-1 = 7.7

The equilibrium constant Kp for the reaction
is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm? 2NO2 (g) NO (g) + O2 (g) 2 Kp = 2 PNO PO PNO PO 2 = Kp PNO PO 2 = 158 x (0.400)2/(0.270)2 = 347 atm

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO3 (s) CaO (s) + CO2 (g) Kc = ‘ [CaO][CO2] [CaCO3] [CaCO3] = constant [CaO] = constant Kc x ‘ [CaCO3] [CaO] Kc = [CO2] = Kp = PCO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

Heterogeneous Equilibria
Consider the reaction below. The equilibrium-constant expression contains terms for only those species in the homogeneous gas phase…H2O, CO, and H2. 2

does not depend on the amount of CaCO3 or CaO
CaCO3 (s) CaO (s) + CO2 (g) PCO 2 = Kp PCO 2 does not depend on the amount of CaCO3 or CaO

Consider the following equilibrium at 295 K:
The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction? NH4HS (s) NH3 (g) + H2S (g) Kp = P NH3 H2S P = x = Kp = Kc(RT)Dn Kc = Kp(RT)-Dn Dn = 2 – 0 = 2 T = 295 K Kc = x ( x 295)-2 = 1.20 x 10-4

‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] Kc A + B C + D Kc
C + D E + F [E][F] [A][B] Kc = A + B E + F Kc Kc = Kc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

‘ N2O4 (g) 2NO2 (g) 2NO2 (g) N2O4 (g) = 4.63 x 10-3 K = [NO2]2 [N2O4]
= 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

Equilibrium Constant for the Sum of Reactions
For example, nitrogen and oxygen can combine to form either NO(g) or N2O (g) according to the following equilibria. (1) Kc = 4.1 x 10-31 (2) Kc = 2.4 x 10-18 Kc = ? Using these two equations, we can obtain Kc for the formation of NO(g) from N2O(g): (3) 2

Equilibrium Constant for the Sum of Reactions
To combine equations (1) and (2) to obtain equation (3), we must first reverse equation (2). When we do we must also take the reciprocal of its Kc value. (1) Kc = 4.1 x 10-31 (2) Kc = (3) 2

Conclusions About the Equilibrium Expression
Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When the balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus Knew = (Koriginal)n. K values are usually written without units. Copyright © Cengage Learning. All rights reserved

K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one equilibrium constant, K. Equilibrium position is a set of equilibrium concentrations. Copyright © Cengage Learning. All rights reserved

Predicting the Direction of Reaction
How could we predict the direction in which a reaction at non-equilibrium conditions will shift to reestablish equilibrium? To answer this question, substitute the current concentrations into the reaction quotient expression and compare it to Kc. The reaction quotient, Qc, is an expression that has the same form as the equilibrium-constant expression but whose concentrations are not necessarily at equilibrium. 2

For the general reaction the Qc expresssion would be: 2

For the general reaction If Qc > Kc, the reaction will shift left…toward reactants. If Qc < Kc, the reaction will shift right… toward products. If Qc = Kc, then the reaction is at equilibrium. 2

A Problem to Consider Consider the following equilibrium.
A 50.0 L vessel contains 1.00 mol N2, 3.00 mol H2, and mol NH3. In which direction (toward reactants or toward products) will the system shift to reestablish equilibrium at 400 oC? Kc for the reaction at 400 oC is 2

A Problem to Consider First, calculate concentrations from moles of substances. 1.00 mol 50.0 L 3.00 mol 0.500 mol 2

A Problem to Consider First, calculate concentrations from moles of substances. M M M The Qc expression for the system would be: 2

A Problem to Consider First, calculate concentrations from moles of substances. M M M Substituting these concentrations into the reaction quotient gives: 2

A Problem to Consider First, calculate concentrations from moles of substances. M M M Because Qc = 23.1 is greater than Kc = 0.500, the reaction will go to the left (toward reactants) as it approaches equilibrium. 2

Calculating Equilibrium Concentrations
Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in the equilibrium mixture. 2

For example, consider the following equilibrium. Suppose a gaseous mixture contained 0.30 mol CO, 0.10 mol H2, mol H2O, and an unknown amount of CH4 per liter. What is the concentration of CH4 in this mixture? The equilibrium constant Kc equals 3.92. 2

First, calculate concentrations from moles of substances. 0.30 mol 1.0 L 0.10 mol 0.020 mol ?? 2

First, calculate concentrations from moles of substances. ?? 0.30 M 0.10 M 0.020 M The equilibrium-constant expression is: 2

First, calculate concentrations from moles of substances. ?? 0.30 M 0.10 M 0.020 M Substituting the known concentrations and the value of Kc gives: 2

First, calculate concentrations from moles of substances. ?? 0.30 M 0.10 M 0.020 M You can now solve for [CH4]. The concentration of CH4 in the mixture is mol/L. 2

Suppose we begin a reaction with known amounts of starting materials and want to calculate the quantities at equilibrium. 2

Consider the following equilibrium. Suppose you start with mol each of carbon monoxide and water in a 50.0 L container. Calculate the molarity of each substance in the equilibrium mixture at 1000 oC. Kc for the reaction is 0.58 at 1000 oC. 2

First, calculate the initial molarities of CO and H2O. 1.000 mol 50.0 L 2

First, calculate the initial molarities of CO and H2O. M 0 M The starting concentrations of the products are 0. We must now set up a table of concentrations (starting, change, and equilibrium expressions in x). 2

Let x be the moles per liter of product formed. Initial 0.0200 Change -x +x Equilibrium x x The equilibrium-constant expression is: 2

Solving for x. Initial 0.0200 Change -x +x Equilibrium x x Substituting the values for equilibrium concentrations, we get: 2

Solving for x. Initial 0.0200 Change -x +x Equilibrium x x Or: 2

Solving for x. Initial 0.0200 Change -x +x Equilibrium x x Taking the square root of both sides we get: 2

Solving for x. Initial 0.0200 Change -x +x Equilibrium x x Rearranging to solve for x gives: 2

Solving for equilibrium concentrations. Initial 0.0200 Change -x +x Equilibrium x x If you substitute for x in the last line of the table you obtain the following equilibrium concentrations. M CO M CO2 M H2O M H2 2

The preceding example illustrates the three steps in solving for equilibrium concentrations. Set up a table of concentrations (starting, change, and equilibrium expressions in x). Substitute the expressions in x for the equilibrium concentrations into the equilibrium-constant equation. Solve the equilibrium-constant equation for the values of the equilibrium concentrations. 2

In some cases it is necessary to solve a quadratic equation to obtain equilibrium concentrations. The next example illustrates how to solve such an equation. 2

ICE At 12800C the equilibrium constant (Kc) for the reaction
Is 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br2 (g) Br (g) Let x be the change in concentration of Br2 Br2 (g) Br (g) Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) x x [Br]2 [Br2] Kc = Kc = ( x)2 x = 1.1 x 10-3 Solve for x

Kc = ( x)2 x = 1.1 x 10-3 4x x = – x 4x x = 0 -b ± b2 – 4ac  2a x = ax2 + bx + c =0 x = x = Br2 (g) Br (g) Initial (M) Change (M) Equilibrium (M) 0.063 0.012 -x +2x x x At equilibrium, [Br] = x = M or M At equilibrium, [Br2] = – x = M

Writing Equilibrium Constant Expressions
Review: Writing Equilibrium Constant Expressions The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. The equilibrium constant is a dimensionless quantity. In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. If a reaction can be expressed as a sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.

Le Chatelier’s Principle
Obtaining the maximum amount of product from a reaction depends on the proper set of reaction conditions. Le Chatelier’s principle states that when a system in a chemical equilibrium is disturbed by a change of temperature, pressure, or concentration, the equilibrium will shift in a way that tends to counteract this change. 2

Removing Products or Adding Reactants
Let’s refer back to the illustration of the U-tube in the first section of this chapter. “reactants” “products” It’s a simple concept to see that if we were to remove products (analogous to dipping water out of the right side of the tube) the reaction would shift to the right until equilibrium was reestablished. 2

Removing Products or Adding Reactants
Let’s refer back to the illustration of the U-tube in the first section of this chapter. “reactants” “products” Likewise, if more reactant is added (analogous to pouring more water in the left side of the tube) the reaction would again shift to the right until equilibrium is reestablished. 2

Le Châtelier’s Principle
Changes in Concentration continued aA + bB cC + dD Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

Effects of Pressure Change
A pressure change caused by changing the volume of the reaction vessel can affect the yield of products in a gaseous reaction only if the reaction involves a change in the total moles of gas present 2

If the products in a gaseous reaction contain fewer moles of gas than the reactants, it is logical that they would require less space. So, reducing the volume of the reaction vessel would, therefore, favor the products. Conversely, if the reactants require less volume (that is, fewer moles of gaseous reactant), then decreasing the volume of the reaction vessel would shift the equilibrium to the left (toward reactants). 2

Literally “squeezing” the reaction will cause a shift in the equilibrium toward the fewer moles of gas. It’s a simple step to see that reducing the pressure in the reaction vessel by increasing its volume would have the opposite effect. In the event that the number of moles of gaseous product equals the number of moles of gaseous reactant, vessel volume will have no effect on the position of the equilibrium. 2

Changes in Volume and Pressure A (g) + B (g) C (g) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

Effect of Temperature Change
Temperature has a significant effect on most reactions Reaction rates generally increase with an increase in temperature. Consequently, equilibrium is established sooner. In addition, the numerical value of the equilibrium constant Kc varies with temperature. 2

Let’s look at “heat” as if it were a product in exothermic reactions and a reactant in endothermic reactions. We see that increasing the temperature is analogous to adding more product (in the case of exothermic reactions) or adding more reactant (in the case of endothermic reactions). This ultimately has the same effect as if heat were a physical entity. 2

For example, consider the following generic exothermic reaction. Increasing temperature would be analogous to adding more product, causing the equilibrium to shift left. Since “heat” does not appear in the equilibrium-constant expression, this change would result in a smaller numerical value for Kc. 2

For an endothermic reaction, the opposite is true. Increasing temperature would be analogous to adding more reactant, causing the equilibrium to shift right. This change results in more product at equilibrium, amd a larger numerical value for Kc. 2

In summary: For an endothermic reaction (DH positive) the amounts of products are increased at equilibrium by an increase in temperature (Kc is larger at higher temperatures). For an exothermic reaction (DH is negative) the amounts of reactants are increased at equilibrium by an increase in temperature (Kc is smaller at higher temperatures). 2

Changes in Temperature Change Exothermic Rx Endothermic Rx Increase temperature K decreases K increases Decrease temperature K increases K decreases colder hotter

Effect of a Catalyst A catalyst is a substance that increases the rate of a reaction but is not consumed by it. It is important to understand that a catalyst has no effect on the equilibrium composition of a reaction mixture A catalyst merely speeds up the attainment of equilibrium. 2

Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner uncatalyzed catalyzed Catalyst lowers Ea for both forward and reverse reactions. Catalyst does not change equilibrium constant or shift equilibrium.

Change Equilibrium Constant Change Shift Equilibrium Concentration yes no Pressure yes no Volume yes no Temperature yes yes Catalyst no no

Worked Example 14.1

Worked Example 14.2

Worked Example 14.5

Worked Example 14.8

Worked Example 14.9

Worked Example 14.10b

Worked Example 14.11

Worked Example 14.12

Worked Example 14.13b

Copyright © Cengage Learning. All rights reserved

Similar presentations

Presentation on theme: "Copyright © Cengage Learning. All rights reserved"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Copyright © Cengage Learning. All rights reserved

Similar presentations

Presentation on theme: "Copyright © Cengage Learning. All rights reserved"— Presentation transcript:

Similar presentations

About project

Feedback