Presentation on theme: "(a) Advanced Calculations using the Equilibrium Constant; (b) Le Chatelier’s Principle; (c) Equilibria of Real Gases Chemistry 142 B Autumn Quarter 2004."— Presentation transcript:
(a) Advanced Calculations using the Equilibrium Constant; (b) Le Chatelier’s Principle; (c) Equilibria of Real Gases Chemistry 142 B Autumn Quarter 2004 J. B. Callis, Instructor Lecture #19
Solving Equilibrium Problems Write the balanced equation for the reaction. Write the equilibrium expression. List the initial concentrations. Calculate Q and determine the direction of shift to equilibrium. Define the change needed to reach equilibrium and define the equilibrium concentrations. Substitute the equilibrium concentrations into the equilibrium expression and solve for the unknown. Check the solution by calculating K and making sure it is identical to the original K.
Problem 19-1 – Graphical Solution to the Quadratic Equation Let us begin by considering the equilibrium between NO 2, a red-brown gaseous pollutant formed by automobiles, and its dimer, N 2 O 4. This equilibrium can be expressed by the chemical equation: 2 NO 2 (g) = N 2 O 4 (g) The equilibrium constant for this reaction is where P N2O4 and P NO2 are the equilibrium partial pressures of the two gases. K P has the numerical value of 8.8 at T = 25 o C when the partial pressures are expressed in atmospheres.
Problem 19-1 – Graphical Solution cont. Let us now consider the problem of finding the equilibrium partial pressures of NO 2 and N 2 O 4, given the value of K P, and the initial pressures of NO 2 and N 2 O 4 ( (P NO2 ) 0 and (P N2O4 ) 0 ). Then if x atm is the additional amount of N 2 O 4 formed by the equilibrium, the partial pressure of NO 2 must decrease by 2x atm. Inserting the equilibrium partial pressures into the equilibrium expression results in the equation:
Finding the Solution to an Equilibrium Problem Using a Graphical Method From Graph: x = and x = The later is chosen because it gives all positive concentrations.
Problem 19-2: Using Exact Solution of Quadratic Equation
Problem 19-2: Solution(a) Pressure (atm) Isopropyl alcohol, IPAAcetone, AcHydrogen H 2 (g) Init. Change Equil.
Problem 19-2: Continued (b)
Problem 19-2: Continued (c)
Problem 19-3: Problems Involving Higher Order Polynomials
Problem 19-3: Solution(a) Pressure (atm) CH 4 (g)H 2 O(g)CO(g)3 H 2 (g) Init. Change Equil.
Problem 19-3: Continued (b)
Problem 3- Continued (c)
Le Chatelier’s Principle If a change in conditions (a ‘stress’) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions. Henri Le Chatelier, 1884
The Effect of a Change in Concentration If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to to reduce the concentration of the added component. If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to to increase the concentration of the removed component.
Problem 19-4: The Effect of a Change in Concentration Consider the following reaction: 2 H 2 S(g) + O 2 (g) = 2 S(s) + 2 H 2 O(g) What happens to: (a) [H 2 O] if O 2 is added? Ans: (b) [H 2 S] if O 2 is added? Ans: (c) [O 2 ] if H 2 S is removed? Ans: (d) [H 2 S] if S is added? Ans:
Effect of a Change in Pressure 1.Add or remove a gaseous product at constant volume. 2.Add an inert gas (one not involved in the reaction) at constant volume. 3.Change the volume of the container. There are three ways to change the pressure of a reaction system involving gaseous components at a given temperature:
Problem 19-5: The Effect of a Change in Pressure How would you change the total pressure of each of the following reactions to increase the yield of the products: (a) CaCO 3 (s) = CaO(s) + CO 2 (g) Ans: (b) S(s) + 3 F 2 (g) = SF 6 (g) Ans: (c) Cl 2 (g) + I 2 (g) = 2 ICl (g) Ans:
The Effect of a Change in Temperature Changes in concentration or pressure alter the equilibrium position. In contrast, changes in temperature alter the value of the equilibrium constant.
Exothermic Reactions Releases heat upon reaction. H is negative. Addition of heat to an exothermic reaction shifts the equilibrium to the left. The value of K decreases in consequence.
Endothermic Reactions Absorbs heat upon reaction. H is positive. Addition of heat to an endothermic reaction shifts the equilibrium to the right. The value of K increases in consequence.
Using Le Chatelier’s Principle to Describe the Effect of a Temperature Change on a System in Equilibrium Treat the energy as a reactant (in an exothermic process) or as a product (in an endothermic process). Predict the direction of the shift as if an actual reactant or product has been added or removed.
Problem 19-6: The Effect of a Change in Temperature on the Position of Equilibrium How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions: (a) CaO(s) + H 2 O (l) = Ca(OH) 2 (aq) H 0 =-82 kJ Ans: (b) (a) CaCO 3 (s) = + CaO(s) + CO 2 (g) H 0 = 178 kJ Ans: (c) SO 2 (g) = S(s) + O 2 (g) H 0 = 297 kJ Ans:
Equilibria Involving Real Gases We have thus far assumed that gas phase equilibria involve gases that behave as ideal gases. The effect of non-ideal behavior is to cause real equilibrium constants to become non- constant under differing conditions of total pressure.
The Activity Coefficient The activity of the ith gaseous component of the equilibrium system is represented as: Where i is called the activity coefficient for correcting P i obs to the ideal value. For equilibrium pressures of 1 atm or less, the value of K p calculated from the observed pressures is expected to be within about 1% of the true value
Answers to Problems in Lecture 19 1.From graph: x = and x = The later is chosen because it gives all positive concentrations (a) The reaction proceeds to the right so H 2 O increases. (b) Some H 2 S reacts with the added O 2 to move the reaction to the right, so [H 2 S] decreases. (c) The reaction proceeds to the left to re-form H 2 S, more O 2 is formed as well, O 2 increases. (d) S is a solid, so its concentration does not change. Thus, [H 2 S] is unchanged. 5.(a) The only gas is the product CO 2. To move the reaction to the right decrease the pressure. (b)With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF 6. (c) The number of moles of gas is the same on both sides of the equation, so a change in pressure will have no effect. 6. (a) Add heat to the right side. Adding heat shifts the system to the left. [Ca(OH) 2 ] and K will decrease. (b) Add heat to the left side. Adding heat shifts the system to the right. [CO 2 ] and K will increase. (c) Add heat to the left side. Adding heat shifts the system to the right. [SO 2 ] will decrease and K will increase.