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Unformatted text preview: Chapter 15: Chemical Equilibrium
A system is in chemical equilibrium any time two opposite reactions occur at the same rate. Describing Chemical Equilibrium 1. Chemical Equilibrium: A Dynamic Equilibrium Chemical equilibrium is a state in which the forward and reverse rates are equal. 2. Equilibrium Constant The equilibrium constant expression for a general reaction aA + bB <===> cC + dD c d K = [C] [D]
C [A] [B] a b Example Write the equilibrium constant expression for the following reaction: CO(g) + 3 H2 (g) <==> CH4(g) + H2O(g) KC = [CH4][H2O] [CO][H2]
3 The value of the equilibrium constant is obtained by substituting equlibrium concentrations into the equilibrium constant expression. The value will be constant at a given temperature. 3. Heterogeneous Equilibria Heterogeneous equilibrium is one in which reactants and products are in more than one physical state (phase). Since the density (mass/volume) of pure solids and pure liquids is constant at a given temperature, the molarity of pure solids and pure liquids will also be constant at a given temperature. These constants are incorporated into the value of the equilibrium constant, KC, to give KC'. The "place" of the concentration of the solid and/or liquid is "held" by using a "1." Example Write the equilibrium constant expression for the following reaction: 2+ PbI2(s) <==> Pb (aq) + 2 I (aq) KC = [Pb ][I ] 1
2+ 2 = [Pb ][I ] 2+ 2 CHM 1220/1225 Chapter 15 1 of 9 Using the Equilibrium Constant 4. Interpreting the Value of the Equilibrium Constant The value of the equilibrium constant can give general information about what is present in the equilibrium mixture: When KC > 10, the mixture is predominantly products When KC < 0.10 the mixture is predominantly reactants When KC is between 0.1 and 10, there are significant amount of both reactants and products present in the equilibrium mixture. 5. Predicting the Direction of Reaction To predict the direction a reaction that is not at equilibrium will proceed, we compare the value of the equilibrium constant, KC to the value of the reaction quotient, QC. QC has the same expression as KC, but uses nonequilibrium concentrations. When KC > QC, the reaction will proceed forward to the right toward products When KC = QC, the mixture is at equilibrium When KC < QC the reaction will proceed reverse to the left toward reactants Using the Equilibrium Constant 6. Interpreting the Value of the Equilibrium Constant The value of the equilibrium constant can give general information about what is present in the equilibrium mixture: When KC > 10, the mixture is predominantly products When KC >> 10, the mixture is essentially all products When KC < 0.10 the mixture is predominantly reactants When KC << 0.10, the mixture is essentially all reactants When KC is between 0.1 and 10, there are significant amount of both reactants and products present in the equilibrium mixture. 7. Finding the Value of the Equilibrium Constant, KC To find the value of the equilibrium constant, substitute equilibrium concentrations (Molarity) into the equilibrium constant expression. CHM 1220/1225 Chapter 15 2 of 9 Example: Carbon dioxide decomposes at high temperatures to give carbon monoxide and oxygen. At 3000 K, 2.00 mol CO2 is placed in a 1.00L container. At equilibrium 0.90 mol CO2 remain. What is the value of KC? Begin by writing the balanced chemical equation for the reaction. Then establish a table using the reactants and products for columns. 2 CO2(g) <===> 2 CO(g) + O2(g) Initial Change Equilibrium To fill in the chart, calculate the initial concentration of the carbon dioxide: 2.00 mol/1.00 L = 2.00 M. 2 CO2(g) <===> 2 CO(g) + O2(g) Initial 2.00 0 0 Change Equilibrium Use the stoichiometry to put in an expression that relates the change in concentration of reactants and products. Reactants will be used, while products are produced. 2 CO2(g) <===> 2 CO(g) + 1 O2(g) Initial 2.00 0 0 Change 2 x +2 x +1 x Equilibrium To find the expression for the equilibrium concentration, add the initial and change. 2 CO2(g) <===> 2 CO(g) + O2(g) Initial 2.00 0 0 Change 2 x +2 x +x Equilibrium 2.00 2x 2x x The problem gives us the equilibrium amount of carbon dioxide: 0.90 mol/1.00 L = 0.90 M 2 CO2(g) <===> 2 CO(g) + O2(g) Initial 2.00 0 0 Change 2 x +2 x +x Equilibrium 2.00 2x 2x x 0.90 We can solve for x by using the two different expressions for the equilibrium concentration of carbon dioxide: 2.00 2x = 0.90 so, x = 0.55 M Substitute this value into the other equilibrium concentration expressions: [CO2]EQ = 0.90 M [CO]EQ = 2x = 1.10 M [O2]EQ = 0.55 M Substitute these values into the equilibrium constant expression: 2 2 KC = [CO] [O2] = (1.10) (0.55) = 0.82 by convention units are not reported 2 2 [CO2] (0.90) CHM 1220/1225 Chapter 15 3 of 9 8. Predicting the Direction of Reaction To predict the direction a reaction that is not at equilibrium will proceed, compare the value of the equilibrium constant, KC to the value of the reaction quotient, QC. QC has the same expression as KC, but uses nonequilibrium concentrations. When KC > QC, the reaction will proceed forward to the right toward products When KC = QC, the mixture is at equilibrium When KC < QC the reaction will proceed reverse to the left toward reactants Example CH4(g) + 2 H2S(g) <===> CS2(g) + 4 H2(g) At 900C KC = 3.59. What will be the direction of the reaction if we start with [CH4] = 1.26 M, [H2S] = 1.32 M, [CS2] = 1.43, and [H2] = 1.12 M? 4 4 KC = [CS2] [H2] QC = (1.43) (1.12) = 1.02 2 2 [CH4][H2S] (1.26)(1.32) KC QC 3.59 > 1.02 So, the reaction will go forward. 9. Calculating Equilibrium Concentrations a. Equilibrium amount given for one substance (stoichiometry!) Example A 1.250mol sample of phosphorus pentachloride, PCl5, dissociates at 160C and 1.00 atm to give 0.169 mol phosphorus trichloride, PCl3, at equilibrium. What is the composition of the final equilibrium mixture? PCl5(g) <===> PCl3(g) + Cl2(g) Procedure: Set up a table, using the reaction for columns. In the first row put the initial number of moles. In the second row, put the change due to reaction: this is related to the stoichiometry, so if x moles of PCl5 react, the change is x; the corresponding change for PCl3 and Cl2 is +x. In the third row put the expression for the equilibrium amount present: the sum of the initial amount and the change. PCl5(g) 1.250 mol x 1.250 mol x <===> PCl3(g) + 0 mol +x x mol Cl2(g) 0 mol +x x mol Initial Change Equilibrium Now, compare these expressions of the equilibrium amount to other information given about equilibrium.
CHM 1220/1225 Chapter 15 4 of 9 In this case, we're told that there is 0.169 mol of PCl3 at equilibrium. We now have two ways to express that amount: x and 0.169 mol, so we know that x = 0.169 mol Having the value of x allows us to find the other moles present at equilibrium PCl5: 1.250 0.169 = 1.081 mol PCl3: 0.169 mol Cl2: 0.169 mol Example You place 0.750 mol nitrogen, N2, and 2.250 mol of hydrogen, H2, into a reaction vessel at 450C and 10.0 atm. What is the composition of the final equilibrium mixture if you obtain 0.060 mol NH3 from it? N2(g) + 3 H2(g) <===> 2 NH3(g) Procedure: Set up a table, using the reaction for columns. In the first row put the initial number of moles. In the second row, put the change due to reaction: this is related to the stoichiometry, so if x moles of PCl5 react, the change is x; the corresponding change for PCl3 and Cl2 is +x. In the third row put the expression for the equilibrium amount present: the sum of the initial amount and the change. N2(g) + 0.750 mol x 0.750 mol x 3 H2(g) 2.250mol 3x 2.250 mol 3x <===> 2 NH3(g) 0 mol +2x 2x mol Initial Change Equilibrium Now, compare these expressions of the equilibrium amount to other information given about equilibrium. In this case, we're told that there is 0.060 mol of NH3 at equilibrium. We now have two ways to express that amount: 2x and 0.060 mol, so we know that x = 0.030 mol Having the value of x allows us to find the other moles present at equilibrium N2: 0.750 0.030 = 0.720 mol H2: 2.250 3(0.030) = 2.160 mol NH3: 2(0.030) = 0.060 mol b. No equilibrium concentrations; perfect square, linear solution Set up table. Substitute expressions for equilibrium concentration into equilibrium constant expression. Note that the right side is a perfect square. Take the square root of both sides. Solve the resulting linear equation. Substitute the value of x into the expressions for equilibrium concentration. CHM 1220/1225 Chapter 15 5 of 9 Example Hydrogen iodided decomposes to give hydrogen and iodine gases. At 800 K, KC = 0.016. What are the equilibrium concentrations if we begin with 0.500 mol HI in a 5.00L flask? 2 HI(g) <===> H2(g) + I2(g) Procedure: Set up a table as in the previous example. Be sure to convert moles to Molarity. 2 HI(g) ) <===> 0.100 M 2x 0.100 2x H2(g) + 0 +x x I2(g) 0 +x x Initial Change Equilibrium Now, substitute these expressions into the expression for the equilibrium constant. 2 KC = [H2][I2] 0.016 = x [HI] (0.100 2x) Taking the square root of both sides gives: Solving this, we first cross multiply:
2 2 0.126 = x/(0.100 2x) 0.0126 0.252x = x 0.0126 = 1.252x and x = 0.0100M Substitute this value into the expressions for equilibrium concentration from the table: [HI]EQ = (0.100 2x) = 0.080 M and [H2]EQ = [I2]EQ = x = 0.0100 M c. No equilibrium concentrations; quadratic solution We will omit this application. Changing the Reaction Conditions: Le Chtelier's Principle LeChtelier's Principle: When a stress is applied to a system in equilibrium, the system will respond to reduce the impact of the stress. In other words: If we add something, the reaction will use it by shifting ... If we remove something, the reaction will produce it by shifting... A reaction can only shift in two ways: Forward or to the right or toward products Reverse or to the left or toward reactants 10. Adding or Removing Reactants or Products An image that may assist you in creating a mental model for thinking about LeChtelier's Principle is a Utube containing a liquid. If liquid is added to the reactant side, some of the reactants "reactants" "products" will shift toward the product side. If liquid is added to the produce side, some of the products will shift toward the reactant side. If liquid is removed from the reactant side, some of the products will shift toward the reactant side. If liquid is removed from the product side, some of the reactants will shift toward the product side. CHM 1220/1225 Chapter 15 6 of 9 11. Changing the Pressure Changes in pressure only affect reactions in which the number of moles of gas on the reactant side is different from the number of moles of gas on the product side. If pressure is increased, the system will decrease pressure by shifting toward the side with fewer moles of gas. If pressure is decreased, the system will increase pressure by shifting toward the side with more moles of gas. Note: decreasing container volume increases pressure; increasing container volume decreases pressure. Example Determine the impact of decreasing the size of the reaction vessel on the following equilibrium reactions: (a) 2 HI(g) <==> H2(g) + I2 (g)
There are 2 mole of gas on the reactant side; there are 2 mol of gas on the product side. Decreasing the volume increases pressure, but this has no effect on the equilibrium. (b) CO(g) There are 3 mole of gas on the reactant side; there is 1 mol of gas on the product side. Decreasing the volume increases pressure, the reaction shifts toward fewer moles of gas, or forward (toward the products, to the right). + 2 H2 (g) <===> CH3OH(g) (c) CH4 (g) 2 H2 (g) ) <===> There are 3 mole of gas on the reactant side; there are 5 mol of gas on the product side. Decreasing the volume increases pressure, the reaction shifts toward fewer moles of gas, or reverse (toward the reactants, to the left). CS2 (g) + 4 H2 (g) 12. ChangingTemperature To consider the impact of temperature, we need to look at the enthalpy change for a reaction. When DH is positive, the reaction is endothermic. We can consider heat a reactant. When DH is negative, the reaction is exothermic. We can consider heat a product. Now, we can think of increasing temperature as adding heat and When we add heat, the system uses heat by shifting away from the side with heat. decreasing temperature as removing heat. When we remove heat, the system produces heat by shifting toward the side with heat. CHM 1220/1225 Chapter 15 7 of 9 Example Given: 2 H2O(g) <==> 2 H2(g) + O2 (g); DH = 484 kJ Write the reaction showing heat as a reactant or product. heat + 2 H2O(g) <==> 2 H2(g) + O2 (g) What is the impact of (a) removing O2? (b) (c) (d) (e) Adding H2O?
If we add water, the system will use water by shifting forward. If we remove oxygen, the system will produce oxygen by shifting forward. Decreasing pressure?
If we decrease pressure, the system will increase pressure by shifting forward (toward more moles of gas). Decreasing volume?
If we decrease the volume, we increases the pressure; the system will decrease the pressure by shifting reverse (toward fewer moles of gas). Raising the temperature?
If we raise the temperature, we are adding heat; the system will use heat by shifting forward. 13. Effect of a Catalyst A catalyst has no impact on the equilibrium composition. It only affects the rate at which equilibrium is reached. Example Given: 2 IBr(g) M<===> I2(g) + Br2(g) We begin with 2.56 x 105 mol of IBr in a 4.00L vessel. At equilibrium, there are 3.00 x 106 mol I2. What are the equilibrium concentrations of all species? First: Calculate molarity. [IBr]0 = 2.56 x 105 mol/4.00 L = 6.40 x 106 M; [I2]EQ = 3.00 x 106 mol/4.00 L = 7.50 x 107 M Second: Construct the table. 2 IBr(g) M<===> I2(g) + Br2(g) Initial 6.40 x 106 M 0 0 Change 2x +x +x 6 Equilibrium 6.40 x 10 M 2x x x Experimental data 7.50 x 107 M So, x = 7.50 x 107 M Third: Substitute the value of x into the expressions for equilibrium concentration. [I2]EQ = [Br2]EQ = x = 7.50 x 107 M [IBr]EQ = 6.40 x 106 M 2x = 6.40 x 106 M 2(7.50 x 107 M) = 4.90 x 106 M CHM 1220/1225 Chapter 15 8 of 9 Example Given: 2 IBr(g) M<===> I2(g) + Br2(g); KC = 0.026 at 100C If we begin with 0.25 M of IBr, what will be the equilibrium, concentrations of all species? First: We are given Molarity. Second: Construct the table. 2 IBr(g) M<===> I2(g) + Br2(g) Initial 0.25 M 0 0 Change 2x +x +x Equilibrium 0.25 2x x x Third: Write the expression for the equilibrium constant; substitute the value of KC and the equilibrium values from the table into it; solve for x. KC = [I2][Br2] 0.026 = x2 so 0.161 = x [IBr]2 (0.25 2x)2 (0.25 2x) Solving this: 0.(0.25 2x) = x 0.0403 0.161x = x 0.0403 = 1.161 x x = 0.0347 M Fourth: Substitute the value of x into the expressions for the equilibrium concentrations: [IBr]EQ = 0.25 2x = 0.25 0.0694 = 0.18 M [I2]EQ = [Br2]EQ = x = 0.347 M CHM 1220/1225 Chapter 15 9 of 9 ...
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 Fall '07
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 Equilibrium, Dynamic Equilibrium

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