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

May favor either products or reactants Equilibrium Is: Macroscopically static  Microscopically dynamic May favor either products or reactants If products are favored, the equilibrium position of the reaction lies far to the right If reactants are favored, the equilibrium position of the reaction lies far to the left Copyright © Cengage Learning. All rights reserved

Changes in Concentration N2(g) + 3H2(g) 2NH3(g) Copyright © Cengage Learning. All rights reserved

Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. Copyright © Cengage Learning. All rights reserved

The Changes with Time in the Rates of Forward and Reverse Reactions Copyright © Cengage Learning. All rights reserved

H2O(g) + CO(g) H2(g) + CO2(g) 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

H2O(g) + CO(g) H2(g) + CO2(g) 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

Factors Determining Equilibrium Position of a Reaction Initial concentrations Relative energies of reactants and products Relative degree of organization of reactants and products

Consider the following reaction at equilibrium: jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). l m [C] [D] K = [A] j [B] k Copyright © Cengage Learning. All rights reserved

Interactive Example 13.1 - Solution Applying the law of mass action gives Coefficient of NO2 Coefficient of H2O Coefficient of O2 Coefficient of NH3

Interactive Example 13.2 - Calculating the Values of K The following equilibrium concentrations were observed for the Haber process for synthesis of ammonia at 127°C:

Interactive Example 13.2 - Calculating the Values of K (Continued) Calculate the value of K at 127°C for this reaction Calculate the value of the equilibrium constant at 127°C for the following reaction: Calculate the value of the equilibrium constant at 127° C given by the following equation:

Interactive Example 13.2 - Solution (a) The balanced equation for the Haber process is Thus, Note that K is written without units

Interactive Example 13.2 - Solution (b) To determine the equilibrium expression for the dissociation of ammonia, the reaction is written in the reverse order This leads to the following expression:

Interactive Example 13.2 - Solution (c) Determine the equilibrium constant using the law of mass action Compare the above expression to the one obtained in solution (a)

Interactive Example 13.2 - Solution (c) (Continued) Thus,

Equilibrium Position versus Equilibrium Constant Refers to each set of equilibrium concentrations There can be infinite number of positions for a reaction Depends on initial concentrations Equilibrium constant One constant for a particular system at a particular temperature Remains unchanged Depends on the ratio of concentrations

Table 13.1 - Synthesis of Ammonia at Different Concentrations of Nitrogen and Hydrogen

Equilibrium Expression - Conclusions Consider the following reaction: The equilibrium expression is Reversing the original reaction results in a new expression

Equilibrium Expression - Conclusions (Continued) Multiplying the original reaction by the factor n gives The equilibrium expression becomes

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

Equilibrium position is a set of equilibrium concentrations. 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

K involves concentrations. Kp involves pressures. Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Equilibrium pressures at a certain temperature: Copyright © Cengage Learning. All rights reserved

Example N2(g) + 3H2(g) 2NH3(g) Copyright © Cengage Learning. All rights reserved

The Relationship Between K and Kp Kp = K(RT)Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = 0.08206 L·atm/mol·K T = temperature (in Kelvin) Copyright © Cengage Learning. All rights reserved

Interactive Example 13.5 - Calculating K from Kp Using the value of Kp obtained in example 13.4, calculate the value of K at 25°C for the following reaction:

Interactive Example 13.5 - Solution The value of Kp can be used to calculate K using the formula Kp = K(RT)Δn T = 25 + 273 = 298 K Δn = 2 – (2+1) = –1 Thus, Sum of product coefficients Sum of reactant coefficients Δn

Interactive Example 13.5 - Solution (Continued) Therefore,

Example N2(g) + 3H2(g) 2NH3(g) Using the value of Kp (3.9 × 104) from the previous example, calculate the value of K at 35°C. Copyright © Cengage Learning. All rights reserved

Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N2(g) + 3H2(g) 2NH3(g) HCN(aq) H+(aq) + CN-(aq) Copyright © Cengage Learning. All rights reserved

Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase: 2KClO3(s) 2KCl(s) + 3O2(g) 2H2O(l) 2H2(g) + O2(g) Copyright © Cengage Learning. All rights reserved

The concentrations of pure liquids and solids are constant. The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present. The concentrations of pure liquids and solids are constant. 2KClO3(s) 2KCl(s) + 3O2(g) Copyright © Cengage Learning. All rights reserved

Write the expressions for K and Kp for the following processes: Interactive Example 13.6 - Equilibrium Expressions for Heterogeneous Equilibria Write the expressions for K and Kp for the following processes: Solid phosphorus pentachloride decomposes to liquid phosphorus trichloride and chlorine gas Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate

Interactive Example 13.6 - Solution (a) The balanced equation for the reaction is The equilibrium expressions are In this case neither the pure solid PCl5 nor the pure liquid PCl3 is included in the equilibrium expressions

Interactive Example 13.6 - Solution (b) The balanced equation for the reaction is The equilibrium expressions are The solids are not included

Uses of the Equilibrium Constant Helps in predicting features of reactions, such as determining: Tendency (not speed) of a reaction to occur Whether a given set of concentrations represent an equilibrium condition Equilibrium position that will be achieved from a given set of initial concentrations

Uses of the Equilibrium Constant - Example (Continued 4) Let x be the number of molecules that need to disappear so that the system can reach equilibrium This table contains three columns. Each column has three rows below it. The column on the left is titled initial conditions. Row 1 reads 9 molecules of A, row 2 reads 12 molecules of B, row 3 reads 0 molecules of D, and row 4 reads 0 molecules of C. The column in the middle in untitled. Row 1 reads x molecules of A disappear, row 2 reads x molecules of B disappear, row 3 reads x molecules of D form, and row 4 reads x molecules of C form. The column on the right is titled equilibrium conditions. Row 1 reads 9 minus x molecules of A, row 2 reads 12 minus x molecules of B, row 3 reads x molecules of D, and row 4 reads x molecules of C.

The Extent of a Reaction A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right. Reaction goes essentially to completion. Copyright © Cengage Learning. All rights reserved

The Extent of a Reaction A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left. Reaction does not occur to any significant extent. Copyright © Cengage Learning. All rights reserved

Size of K and Time Required to Reach Equilibrium Not directly related Time required depends on the rate of the reaction Determined by the size of the activation energy Size of K is determined by thermodynamic factors Example - Energy difference between products and reactants Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! If the equilibrium lies to the right, the value for K is __________. large (or >1) If the equilibrium lies to the left, the value for K is ___________. small (or <1) Large (or >1); small (or < 1) Copyright © Cengage Learning. All rights reserved

Reaction Quotient, Q Used when all of the initial concentrations are nonzero. Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Copyright © Cengage Learning. All rights reserved

Reaction Quotient, Q Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left. Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right. Consuming reactants and forming products, to attain equilibrium. Copyright © Cengage Learning. All rights reserved

Interactive Example 13.7 - Using the Reaction Quotient For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0×10–2 Predict the direction in which the system will shift to reach equilibrium in the following case: [NH3]0 = 1.0×10–3 M [N2]0 = 1.0×10–5 M [H2]0 = 2.0×10–3 M

Interactive Example 13.7 - Solution Calculate the value of Q Since K = 6.0×10–2, Q is much greater than K

Interactive Example 13.7 - Solution (Continued) To attain equilibrium: The concentrations of the products must be decreased The concentrations of the reactants must be increased Therefore, the system will shift to the left

Consider the reaction represented by the equation: EXERCISE! Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Trial #1: 6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN2+(aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

Set up ICE Table Fe3+(aq) + SCN–(aq) FeSCN2+(aq) Initial 6.00 10.00 0.00 Change – 4.00 – 4.00 +4.00 Equilibrium 2.00 6.00 4.00 K = 0.333 The value for K is 0.333. Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Trial #2: Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq) (same temperature as Trial #1) Equilibrium: ? M FeSCN2+(aq)  5.00 M FeSCN2+ The answer for Trial #2 is 5.00 M. The students can solve this with the quadratic but it is not necessary. The numbers have been chosen to be relatively easy to solve (trial and error works well and will only take a short period of time). Tell the students the problems will not always have numbers like these, so take the time to understand what is going on (so the math doesn’t “get in the way”). Have the students write ICE tables for these. Copyright © Cengage Learning. All rights reserved

EXERCISE! Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Trial #3: Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq) Equilibrium: ? M FeSCN2+(aq) 3.00 M FeSCN2+ The answer for Trial #3 is 3.00 M. The students can solve this with the quadratic but it is not necessary. The numbers have been chosen to by relatively easy to solve (trial and error works well and will only take a short period of time). Tell the students the problems will not always have numbers like these, so take the time to understand what is going on (so the math doesn’t “get in the way”). Have the students write ICE tables for these. Copyright © Cengage Learning. All rights reserved

Interactive Example 13.8 - Calculating Equilibrium Pressures I Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions In the gas phase, it decomposes to gaseous nitrogen dioxide:

Interactive Example 13.8 - Calculating Equilibrium Pressures I (Continued) Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature where Kp = 0.133 At equilibrium, the pressure of N2O4 was found to be 2.71 atm Calculate the equilibrium pressure of NO2(g)

Interactive Example 13.8 - Solution The equilibrium pressures of the gases NO2 and N2O4 must satisfy the following relationship: Solve for the equilibrium pressure of NO2

Interactive Example 13.8 - Solution (Continued) Therefore,

Interactive Example 13.9 - Calculating Equilibrium Pressures II At a certain temperature, a 1.00 L flask initially contained 0.298 mole of PCl3(g) and 8.70×10–3 mole of PCl5 (g) After the system had reached equilibrium 2.00×10–3 mole of Cl2 (g) was found in the flask Gaseous PCl5 decomposes according to the reaction Calculate the equilibrium concentrations of all species and the value of K

Interactive Example 13.9 - Solution The equilibrium expression for this reaction is To find the value of K: Calculate the equilibrium concentrations of all species Substitute the derived quantities into the equilibrium expression

Interactive Example 13.9 - Solution (Continued 1) Determine the initial concentrations

Interactive Example 13.9 - Solution (Continued 2) Determine the change required to reach equilibrium Apply these values to the initial concentrations

Interactive Example 13.9 - Solution (Continued 3) Determine the equilibrium concentrations [Cl2]0 [PCl3]0 [PCl5]0

Interactive Example 13.9 - Solution (Continued 4) Determine the value of K Substitute the equilibrium concentrations into the equilibrium expression

Interactive Example 13.11 - Calculating Equilibrium Concentrations II Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15×102 at a certain temperature In a particular experiment, 3.000 moles of each component were added to a 1.500 L flask Calculate the equilibrium concentrations of all species

Interactive Example 13.11 - Solution The balanced equation for this reaction is The equilibrium expression is The initial concentrations are

Interactive Example 13.11 - Solution (Continued 1) The value of Q is Since Q is much less than K, the system must shift to the right to reach equilibrium To determine what change in concentration is necessary, define the change needed in terms of x

Interactive Example 13.11 - Solution (Continued 2) Let x be the number of moles per liter of H2 consumed to reach equilibrium The stoichiometry of the reaction shows that x mol/L F2 also will be consumed and 2x mol/L HF will be formed

Interactive Example 13.11 - Solution (Continued 3) Determine the equilibrium concentrations in terms of x

Interactive Example 13.11 - Solution (Continued 4) The concentrations can be expressed in a shorthand table as follows: To solve for the value of x, substitute the equilibrium concentrations into the equilibrium expression

Interactive Example 13.11 - Solution (Continued 5) The right side of this equation is a perfect square, so taking the square root of both sides gives

Interactive Example 13.11 - Solution (Continued 6) The equilibrium concentrations are Reality check Checking the values by substituting them into the equilibrium expression gives the same value of K

Solving Equilibrium Problems Write the balanced equation for the reaction. Write the equilibrium expression using the law of mass action. List the initial concentrations. Calculate Q, and determine the direction of the shift to equilibrium. Copyright © Cengage Learning. All rights reserved

Solving Equilibrium Problems 5) Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. Check your calculated equilibrium concentrations by making sure they give the correct value of K. Copyright © Cengage Learning. All rights reserved

Interactive Example 13.12 - Calculating Equilibrium Pressures Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00×102 Suppose HI at 5.000×10–1 atm, H2 at 1.000×10–2 atm, and I2 at 5.000×10–3 atm are mixed in a 5.000 L flask Calculate the equilibrium pressures of all species

Interactive Example 13.12 - Solution The balanced equation for this process is The equilibrium expression in terms of pressure is

Interactive Example 13.12 - Solution (Continued 1) The initial pressures provided are PHI0 = 5.000×10–1 atm PH20 = 1.000×10–2 atm PI20 = 5.000×10–3 atm The value of Q for this system is

Interactive Example 13.12 - Solution (Continued 2) Since Q is greater than K, the system will shift to the left to reach equilibrium Use pressures for a gas-phase system at constant temperature and volume Pressure is directly proportional to the number of moles Constant if constant T and V

Interactive Example 13.12 - Solution (Continued 3) Let x be the change in pressure (in atm) of H2 as the system shifts left toward equilibrium This leads to the following equilibrium pressures:

Interactive Example 13.12 - Solution (Continued 4) Determine the value of Kp Multiply and collect terms that yield the quadratic equation where a = 9.60×101, b = 3.5, and c = –2.45×10–1

Interactive Example 13.12 - Solution (Continued 5) From the quadratic formula, the correct value for x is 3.55×10–2 atm The equilibrium pressures can now be calculated from the expressions involving x

Interactive Example 13.12 - Solution (Continued 6) Reality check This agrees with the given value of K (1.00×102) Thus, the calculated equilibrium concentrations are correct

Treating Systems That Have Small Equilibrium Constants Consider the decomposition of gaseous NOCl at 35°C with an equilibrium constant of 1.6×10–5 The following steps determine the equilibrium concentrations of NOCl, NO, and Cl2 when one mole of NOCl is placed in a 2.0 L flask: The balanced equation is

Treating Systems that have Small Equilibrium Constants (Continued 1) The equilibrium expression is The initial concentrations are

Treating Systems that have Small Equilibrium Constants (Continued 2) Since there are no products initially, the system will move to the right to reach equilibrium Let x be the change in concentration of Cl2 needed to reach equilibrium The changes in the concentrations can be obtained from the following balanced equitation:

Treating Systems that have Small Equilibrium Constants (Continued 3) The concentrations can be summarized as follows: The equilibrium concentrations must satisfy the following equilibrium expression

Treating Systems that have Small Equilibrium Constants (Continued 4) In this situation, K is so small that the system will not proceed far to the right to reach equilibrium x represents a relatively small number, so when x is small Simplify the equilibrium expression using this approximation

Treating Systems that have Small Equilibrium Constants (Continued 5) Solving for x3 gives Test the validity of the approximation If x = 1.0×10–2, then

Treating Systems that have Small Equilibrium Constants (Continued 6) The difference between 0.50 and 0.48 is 0.02 This discrepancy will have little effect on the outcome Since 2x is very small compared with 0.50, the value of x obtained should be very close to the exact value Use the approximate value of x to calculate equilibrium concentrations

Treating Systems that have Small Equilibrium Constants (Continued 7) Reality check Since the given value of K is the same, these calculations are correct The small value of K and the resulting small shift to the right to reach equilibrium allowed simplification

Critical Thinking You have learned how to treat systems that have small equilibrium constants by making approximations to simplify the math What if the system has a very large equilibrium constant? What can you do to simplify the math for this case? Use the example from the text, change the value of the equilibrium constant to 1.6×105, and rework the problem Why can you not use approximations for the case in which K = 1.6?

EXERCISE! Consider the reaction represented by the equation: Fe3+(aq) + SCN-(aq) FeSCN2+(aq) Fe3+ SCN- FeSCN2+ Trial #1 9.00 M 5.00 M 1.00 M Trial #2 3.00 M 2.00 M 5.00 M Trial #3 2.00 M 9.00 M 6.00 M Find the equilibrium concentrations for all species. Copyright © Cengage Learning. All rights reserved

EXERCISE! Answer Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M Trial #2: [Fe3+] = 4.00 M; [SCN-] = 3.00 M; [FeSCN2+] = 4.00 M Trial #3: [Fe3+] = 2.00 M; [SCN-] = 9.00 M; [FeSCN2+] = 6.00 M Trial #1: [Fe3+] = 6.00 M; [SCN-] = 2.00 M; [FeSCN2+] = 4.00 M Trial #2: [Fe3+] = 4.00 M; [SCN-] = 3.00 M; [FeSCN2+] = 4.00 M Trial #3: [Fe3+] = 2.00 M; [SCN-] = 9.00 M; [FeSCN2+] = 6.00 M This problem will provide a good discussion of Q vs. K. Trial #1 proceeds to the right to reach equilibrium, Trial #2 proceeds to the left, and Trial #3 is at equilibrium. Watch for students setting up an ICE chart without thinking about which direction the reaction must proceed initially. Be prepared for some discussion about the fact that in the Change row we can have a “-x” on the right side and a “+x” on the left side and still use the same expression for K. Use this so that students can think about the direction the reaction must proceed initially so that they needed memorize a relationship between Q and K. The students can solve this with the quadratic but it is not necessary. The numbers have been chosen to be relatively easy to solve. Tell the students the problems will not always have numbers like these, so take the time to understand what is going on (so the math doesn’t “get in the way”). Copyright © Cengage Learning. All rights reserved

At equilibrium 1.00 mol of ammonia remains. CONCEPT CHECK! A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: NH3(g) N2(g) + H2(g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69 K = 1.69 This answer assumes the students have balanced the equation with relative coefficients of 2:1:3. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A 1.00 mol sample of N2O4(g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N2O4(g) 2NO2(g) K = 4.00 × 10-4 Calculate the equilibrium concentrations of: N2O4(g) and NO2(g). Concentration of N2O4 = 0.097 M Concentration of NO2 = 6.32 × 10-3 M Concentration of N2O4 = 0.097 M Concentration of NO2 = 6.32 x 10-3 M (without quadratic) or Concentration of N2O4 = 0.094 M Concentration of NO2 = 6.22 x 10-3 M (with quadratic) Use this problem to discuss the 5% allowable error (so we can assume x is negligible). Make sure the students understand we are NOT saying x is equal to zero but that x is negligible. This is a subtle but very important point. Note: Use the red box animation to assist in explaining how to solve the problem. Copyright © Cengage Learning. All rights reserved

If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. Helps in the qualitative prediction of the effects of changes in concentration, pressure, and temperature on a system at equilibrium Copyright © Cengage Learning. All rights reserved

Effects of Changes on the System Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2. Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product). Copyright © Cengage Learning. All rights reserved

Effects of Changes on the System Pressure: The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. Addition of inert gas does not affect the equilibrium position. Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas. Copyright © Cengage Learning. All rights reserved

Interactive Example 13.13 - Using Le Châtelier’s Principle I Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As4O6, which is then reduced using carbon

Interactive Example 13.13 - Using Le Châtelier’s Principle I (Continued) Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions: Addition of carbon monoxide Addition or removal of carbon or tetraarsenic hexoxide (As4O6) Removal of gaseous arsenic (As4)

Interactive Example 13.13 - Solution Le Châtelier’s principle predicts that the shift will be away from the substance whose concentration is increased Equilibrium position will shift to the left when carbon monoxide is added The amount of a pure solid has no effect on the equilibrium position

Interactive Example 13.13 - Solution (Continued) Changing the amount of carbon or tetraarsenic hexoxide will have no effect If gaseous arsenic is removed, the equilibrium position will shift to the right to form more products In industrial processes, the desired product is often continuously removed from the reaction system to increase the yield

The Effect of a Change in Pressure - Key Points Addition of an inert gas increases the total pressure Does not affect the concentrations or partial pressures of the reactants or products When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume Total number of gaseous molecules is reduced

The Effect of a Change in Pressure - Key Points (Continued) Rearranging the ideal gas law gives At constant temperature (T) and pressure (P), V ∝ n Equilibrium position shifts toward the side of the reaction that involves smaller number of gaseous molecules in the balanced equation

Interactive Example 13.14 - Using Le Châtelier’s Principle II Predict the shift in equilibrium position that will occur during the preparation of liquid phosphorus trichloride Assume that the volume is reduced

Interactive Example 13.14 - Solution Since P4 and PCl3 are a pure solid and a pure liquid, respectively, we need to consider only the effect of the change in volume on Cl2 Volume is decreased, so the position of the equilibrium will shift to the right Reactant side contains six gaseous molecules, and the product side has none

The Effect of a Change in Temperature Value of K changes with the temperature Consider the synthesis of ammonia, an exothermic reaction According to Le Châtelier’s principle, the shift will be in the direction that consumes energy Concentration of NH3 decreases and that of N2 and H2 increases, thus decreasing the value of K Copyright © Cengage Learning. All rights reserved

The Effect of a Change in Temperature (Continued) Consider the decomposition of calcium carbonate, an endothermic reaction Increase in temperature causes the equilibrium to shift to the right Value of K increases Copyright © Cengage Learning. All rights reserved

To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Equilibrium Decomposition of N2O4 To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved