How To Understand Chemical And Physical Equilibrium

Transcription

1 Chapter 13 Chemical Equilibrium Equilibrium Physical Equilibrium refers to the equilibrium between two or more states of matter (solid, liquid and gas) A great example of physical equilibrium is shown by the concept of vapor pressure Chemical Equilibrium is achieved when two conditions are met: The rates of the forward and reverse reactions are equal The concentrations of the reactants and products remain constant Chapter 13 1

2 Reversible Reactions Previously, we have assumed that reactants convert completely to products. However, this is not always true. For some reactions, the reaction can proceed towards products (the forward reaction) or towards reatants (the reverse reaction). The reactions are called reversible reactions. Strictly speaking, ALL reactions are reversible to some degree Chapter 13 3 Chemical Equilibrium Most reactions are ongoing, reversible processes; preceding in both the forward direction to give products and in the reverse direction to give the original reactants. We indicate an equilibrium reaction with a double arrow: Reactants forward reaction Products reverse reaction Chapter 13 4

3 Chemical Equilibrium Chapter 13 5 Chemical Equilibrium and Reaction Rates In an equilibrium reaction, initially the rate of the forward reaction is very fast. As more products are formed, the rate of the reverse reaction begins to speed up. When the rates of the forward and reverse reactions are the same, the system is at equilibrium. Chapter

4 Chemical Equilibrium and Reaction Rates At Equilibrium, there is no further observable change in the concentration of the reactants or products. Chapter 13 7 Equilibrium Constant Expression (K c ) Consider the following general reaction: a A + b B c C + d D The constant, K c, is the general equilibrium constant. The c indicates that the amounts of reactants and products are concentrations (M) The value of K c varies with temperature. So a given value of K c is valid only for a specific temperature. Chapter

5 Homogeneous Equilibria A homogeneous equilibrium is a reaction where all of the products and reactants are in the physical same state. The amounts are given as molarity (aqueous or gas) or partial pressures (gas) What is the equilibrium constant expression for the homogeneous equilibrium? SO (g) + O (g) SO 3 (g) K C = [SO ] 3 [SO ] [O Chapter 13 9 ] Heterogeneous Equilibria A heterogeneous equilibrium is a reaction where one of the substances is in a different physical state. C(s) + H O(g) CO(g) + H (g) K C = [CO][H ] [HO] The concentrations of liquids and solids do not change, and they are therefore omitted from equilibrium constant expressions: Chapter

6 Equilibrium Constant Expressions What are the equilibrium constant expressions for the following reactions? NH 4 NO 3 (s) N O(g) + H O(g) CO(g) + H (g) CH 3 OH (l) HCl(aq) + O (g) H O(l) + Cl (g) Chapter Using Equilibrium Expressions: Calculating K c We can calculate the numerical value of K c if we know the equilibrium concentrations of all of the species in the reaction. If the concentrations at equilibrium are [H ] = 0.10 M, [I ] = 0.0 M, and [HI] = 1.07 M, what is K c? H (g) + I (g) HI(g) K c = [HI] [H ][I ] (1.07) = = 57. (0.10)(0.0) Chapter

7 Equilibrium Constant Expressions The following pictures represent mixtures of A molecules (red) and B molecules (blue), which interconvert according to the equation A B. If Mixture (1) is at equilibrium, which of the other mixtures is also at equilibrium? Chapter Equilibrium Constant K p Because gas pressures are easily measured, equilibrium expressions for gas-phase reactions are often written with partial pressures rather than concentrations. a A + b B c C + d D We call this version of the equilibrium expression K p c d ( P ) ( P ) K = C D p a b ( P ) ( P ) A B We can relate K p to K c by using the Ideal gas law (PV = nrt): K p c ( P )( P ) C D a ( P ) ( P ) c ([C]RT) ([D]RT) a ([A]RT) ([B]RT) d d c d [C] [D] = b = b = x RT a b [A] [B] A B K p ( ) n = K RT ( ) ( c + d ) ( a + b ) Chapter C 7

8 Equilibrium Constant Expressions Write the K p and K c expressions for: N O 5 (g) 4 NO (g) + O (g) The equilibrium concentrations for the reaction between CO and Cl to form carbonyl chloride (phosgene gas) at 74 C are [CO] = 1. x 10 M, [Cl ] = M, and [COCl ] = 0.14 M. Calculate K c and K p. CO(g) + Cl (g) COCl (g) Chapter Equilibrium Constant Expressions Methane (CH 4 ) reacts with hydrogen sulfide to yield H and carbon disulfide, a solvent used in manufacturing. What is the value of K p at 1000 K if the partial pressures in an equilibrium mixture at 1000 K are 0.0 atm of CH 4, 0.5 atm of H S, 0.5 atm of CS, and 0.10 atm of H? Chapter

9 Using Equilibrium Constants The value of K c or K p can indicate whether products or reactants are favored in a reaction > K 10-3 < K < 10 3 K > 10 3 The equilibrium constant (K c ) for the formation of nitrosyl chloride, from nitric oxide and chlorine gas is 6.5 x 10 4 at 35 C. What is favored? Chapter Reaction Quotient (Q c ) The Reaction Quotient (Q c ) is determined by substituting non-equilibrium concentrations into the equilibrium expression c [C] [D] Q = C a [A] [B] The direction of the reaction can be predicted by comparing Q c with K c : d b Q c > K c Q c = K c Q c < K c System proceeds to form reactants. System is at equilibrium. System proceeds to form products Chapter

10 Predicting the Direction of the Reaction Chapter Predicting the Direction of the Reaction The equilibrium constant (K c ) for the formation of nitrosyl chloride, from nitric oxide and chlorine gas is 6.5 x 10 4 at 35 C. NO(g) + Cl (g) NOCl(g) In an experiment,.0 x 10 moles of NO, 8.3 x 10 3 moles of Cl, and 6.8 moles of NOCl are mixed in a.0-l flask. In which direction will the system proceed to reach equilibrium? Chapter

11 Using K c to Calculate Concentrations Guide to Using K eq Value STEP 1 Write the K eq expression for the equilibrium equation STEP Solve the K eq expression for the unknown concentration. STEP 3 Substitute the known values into the rearranged K eq expression STEP 4 Check answer by using the calculated concentration in the K eq expression Chapter 13 1 Calculating Concentration Using K c Ethanol can be produced by reacting ethylene (C H 4 ) with water vapor. At 37 C, the K c = 9.00 x C H 4 (g) + H O (g) C H 5 OH (g) If an equilibrium mixture has concentrations of [C H 4 ] = 0.00 M and [H O] = M, what is the equilibrium concentration of C H 5 OH? Chapter 13 11

12 Using Equilibrium Constants We can also use the equilibrium constant to calculate equilibrium concentrations from initial concentrations. We use the Initial Change Equilibrium method (the ICE table!). H (g) + I (g) HI (g) Initial (M) Change (M) Equilibrium (M) X X X + X X Use K c =57.0 to determine equilibrium concentrations. -X 0 Chapter 13 3 Using Equilibrium Constants We can also use the equilibrium constant to calculate equilibrium concentrations from initial concentrations. We use the Initial Change Equilibrium method (the ICE table!). Initial Change Equilibrium A + B C X X X X 0 +X X STEP 1 Write the balanced equation for the reaction STEP Under the balanced equation, Set up your ICE table STEP 3 Write the K c expression and substitute the equilibrium concentrations into the expression STEP 4 Solve the K c expression for X. If you have made an assumption, make sure it is valid. If it is not, you must use the quadratic equation! STEP 5 Go back to your ICE table and calculate the Equilibrium concentrations using X STEP 6 Check answer by substituting your calculated equilibrium concentrations into the K c expression Chapter

13 Using Equilibrium Constants: The Assumption Once you have prepared your ICE table, and plugged your equilibrium concentrations into your K c expression, you may not be able to solve for x using simple algebra. If you cannot solve for x with simple algebra, there are two ways to proceed: The Assumption (Sometimes works) The Quadratic Equation (Always works) The Assumption is that x <<< than [ ] so you can ignore subtracting or adding it. If you make this assumption, you must check that you are correct once you find x. For the assumption to be valid x < 3% of the number your are comparing it to. If it is more than 3% you must go back and use the Quadratic Equation! Chapter 13 5 Using Equilibrium Constants: The Assumption At a certain temperature, the decomposition of PCl 5 (g) has an equilibrium constant K c = 5.8 x Calculate the equilibrium concentration of all species if the initial concentration of PCl 5 is.60 M. PCl 5 (g) Cl (g) + PCl 3 (g) K Initial Change Equilibrium [Cl ][PCl ] PCl 5 (g).60 -X.60 X (X) Cl (g) + PCl 3 (g) If you make the 3 = = assumption, you must C [PCl ] (.60 - X) (.60) show that it is valid! 5 Chapter 13 6 Assume X<<< X X (X) 0 + X X 13

14 Using Equilibrium Constants: The Quadratic If you find that your assumption is NOT valid (i.e. X > 3%) then you must go back and solve the Quadratic Equation for X The expression must first be rearranged to: ax The values are substituted into the quadratic and solved for a positive solution to x and ph. x = + bx + c b ± b a 4ac Chapter 13 7 = 0 For additional info, see Appendix A.4 on page A-9 in the back of your text Using Equilibrium Constants: The Quadratic For the example started on slide 6, our assumption was NOT valid so we pull out the Quadratic Equation: X b ± = X X b a = ac ± = OR (0.0580) (1) x 10 0 = X (X) a b c Chapter (X) = (.60 - X) 4(1)( 0.151) Which answer makes sense? 14

15 Using Equilibrium Constants Calculate the equilibrium concentration of all species when 1.00 mol of sulfur dioxide and.00 mol of chlorine are sealed in a 100 L reactor. The temperature is raised to 400 K where the K c =89.3. SO (g) + Cl (g) SO Cl (g) Initial Change Equilibrium SO (g) + Cl (g) SO Cl (g) X + X + X X X X K [SO Cl ] = [SO ][Cl ] C = X (0.01- X)(0.0 - X) Chapter 13 9 Changing Equilibrium Conditions We know that at equilibrium, the concentrations of the reactants and the products are not visibly changing. However, changes that occur in the reaction conditions will disturb the equilibrium Le Châteleir s principle states that when a reversible reaction at equilibrium is stressed by a change in concentration, temperature, or volume (pressure), the equilibrium must shift to relieve the stress. Chapter

16 Le Châteleir s: Effect of Concentration Let s look at the following reaction: A B + C When the concentration of A is increased (A is added!), the equilibrium must shift towards the products This shift allows the system to remove the extra A by making more product The concentration stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance (opposite side). The concentration stress of a removed reactant or product is relieved by reaction in the direction that replenishes the removed substance (same side). Chapter Le Châteleir s: Effect of Concentration Consider the effect of each of the following changes on the equilibrium: CO (g) + H O (g) CO (g) + H (g) Increasing [CO]? Increasing [H ]? Decreasing [H O]? Decreasing [CO ]? Chapter

17 Le Châteleir s: Common Ion Effect When two dissolved solutes that contain the same ion (cation or anion) that ion is termed a Common Ion The Common-Ion Effect is the shift in equilibrium caused by increasing the concentration of one of the ions in solution. Chapter Le Châteleir s: Effect of Pressure & Volume Changing the volume of a gas-state reaction has an effect on the pressure of the reaction (Boyle s Law!) PV = nrt tells us that increasing pressure or decreasing volume increases concentration. These changes will place stress on the reaction which must be relieved according to Le Châtelier. If the Pressure increases, the reaction will shift to the side with the fewest moles of gas If the Pressure decreases, the reaction will shift to the side with the most moles of gas CO(g) + H (g) CH 3 OH(g) 1 mole moles 1 mole 3 moles Low volume (high pressure) products High volume (low pressure) reactants Chapter

18 Le Châteleir s: Effect of Pressure & Volume Consider the reaction: N O 4 (g) NO (g), taking place in a cylinder with a volume = 1 L. [NO ] = mol/l = [N O 4 ] = 1 mol/l = 1 K = [NO ] = 4 [N O 4 ] Chapter Le Châteleir s: Effect of Pressure & Volume The Volume is then halved, which is equivalent to doubling the pressure. [NO ] = mol/0.5 = 4 [N O 4 ] = 1 mol/0.5 = Q = [NO ] = 8 [N O 4 ] Since Q > K, the [product] is too high and the reaction progresses in the reverse direction. Shifts towards the side with less moles of gas! Chapter

19 Le Châteleir s: Effect of Pressure & Volume Does the number of moles of products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume. 1. PCl 5 (g) PCl 3 (g) + Cl (g). CaO(s) + CO (g) CaCO 3 (s) 3. 3 Fe(s) + 4 H O(g) Fe 3 O 4 (s) + 4 H (g) Chapter Le Châteleir s: Effect of Temperature Of the stresses we have talked about, only changing the temperature results in a new K eq. The effect of a temperature change is dependent on whether the reaction is endothermic (requires heat) or exothermic (gives off heat). Endothermic processes are favored when temperature increases. Exothermic processes are favored when temperature decreases. Chapter

20 Le Châteleir s: Effect of Temperature Consider the reaction: N (g) + 3 H (g) NH 3 (g) H rxn = -9. kj What is favored as the temperature increases, Reactants or Products? Chapter Le Châteleir s: Effect of Temperature In the first step of the Ostwald process for synthesis of nitric acid, ammonia is oxidized to nitric oxide by the reaction: 4NH 3 (g) + 5O (g) 4NO(g) + 6H O(g) H rxn = kj How do the equilibrium amounts of the following vary with an increase in temperature: NO NH 3 H O O Chapter

21 Effect of a Catalyst If we add a catalyst to a reaction at equilibrium, what do you think will happen? A catalyst decreases the activation energy of the reaction This decrease effects results in the increase of the rate of both the forward and reverse reactions equally However, the same ratios of products and reactants are attained. Adding a catalyst has no effect on a system at equilibrium. Chapter Summary of Le Châtelier s Chapter

22 Link between Equilibrium and Kinetics k f a A + b B c C + d D When the system reaches equilibrium then the rate of the forward reaction is equal to the rate of the reverse reaction: k = a b [ A] [B] k f r [C] Rearrangement leads to: And we know: k k f r c [C] [D] = a [A] [B] d b So.. k k f k r = K C Chapter 13 r 43 c [D] c [C] [D] K = C a [A] [B] d d b