Chapter 15 Chemical Equilibrium

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1 Equilibrium To be in equilibrium is to be in a state of balance: Chapter 15 Chemical Equilibrium - Static Equilibrium (nothing happens; e.g. a tug of war). - Dynamic Equilibrium (lots of things happen, but the net effect is zero; e.g. cars traveling). Chemical Equilibrium It occurs when a reaction and its reverse reaction proceed at the same rate The Concept of Equilibrium Many reactions go to completion, we predict the amount of product formed using stoichiometry. Many reactions (particularly organic reactions) DO NOT go to completion, but seem to stop before the reaction is complete. E.g. CO (g) + 3H 2 (g) CH 4 (g) + H 2 O (g) If we mix 1 mol CO + 3 mol H2, we would expect to get 1 mol CH4 and 1 mol of H2O. However, we get: 1.8 mol H2, 0.6 mol CO, 0.4 mol CH4, 0.4 mol H2O (at 1200 K in a 10.0L container), which is a mixture of reactants and products in a dynamic equilibrium. Once the two reactions (forward and reverse) get to the same rate, the concentrations do not change (again, this is a dynamic equilibrium). N 2 O 4 (g) 2NO 2 (g) 15.2 The Equilibrium Constant 15.3 Understanding and Working with Equilibrium Constants Equilibrium constants depend on the reaction (its stoichiometry, not its mechanism) and T. K C = [CH 4][H 2 O] ; [ ] Conc. in M (mol [CO][H 2 ] 3 L ) CO (g) + 3H 2 (g) CH 4 (g) + H 2 O (g) K = [Products] [Reactants] ; Coefficients become exponents. K P = P CH 4 P H2 O 3 ; P Pressure in atm. P CO P H2 K P K C in general ; K P = K C (RT) n n = (#moles of gases) ; products reactants 1

2 Taking: CO (g) + 3H 2 (g) CH 4 (g) + H 2 O (g) For the above reaction: n gas = 2 4 = 2 K P = K C (RT) 2 = K C (RT) 2 ; Latm R = molk ; T[Kelvin] If: n gas = 0 ; K P = K C. Equilibrium constants are reported without units since they are related not only to the kinetics, but also to the thermodynamics of the reaction. Equilibrium constants derived from thermodynamic measurements have no units, so, it is customary to write all types of equilibrium constants without units. Nevertheless, for gases we use P and for solutes we use M units (could be mixed units). Solids, liquids and solvents do not appear in K expressions (their concentrations are constant). Changing the amount of a solid or liquid has no effect on the equilibrium, as long as there is some present. + For: H 3 O (aq) + HCO 3 (aq) 2H 2 O (l) + CO 2 (g) K eq = P CO2 [H 3 O + ][HCO 3 ] Q. Write the Keq expression for: 2NO 2 (g) + 7H 2 (g) 2NH 3 (g) + 4H 2 O (l) Meaning of K If the system has reached equilibrium and the M s/p s are substituted into the K expression, the result is a constant; conversely, if the values are plugged in and the result equals K, then the system is at equilibrium. There are infinite possibilities for the equilibrium mixture. N 2 O 4 (g) 2NO 2 (g) What Does the Value of Keq Imply? When the value of Keq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules. When the value of Keq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules. 2

3 Q. Write the equilibrium constant expressions, KC, and predict the position of equilibrium for the following: 2 SO2(g) + O2(g) 2 SO3(g) K = 8 x N2(g) + 2 O2(g) 2 NO2(g) K = 3 x Manipulating Equilibrium Constants When the reaction is written backward, the equilibrium constant is inverted. For the reaction aa + bb cc + dd the equilibrium constant expression is: For the reaction cc + dd aa + bb the equilibrium constant expression is: K forward = [C]c [D] d [A] a [B] b K backward = 1 K forward K backward = [A]a [B] b [C] c [D] d When the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor. For the reaction aa + bb cc the equilibrium constant expression is: For the reaction 2aA + 2bB 2cC the equilibrium constant expression is: K original = [C]c [A] a [B] b K new = [C]2c [A] 2a [B] 2b K new = (K original ) n = ( [C]c [A] a [B] b) 2 When you add equations to get a new equation, the equilibrium constant of the new equation is the product of the equilibrium constants of the old equations For the reactions (1) aa bb and (2) bb cc the equilibrium constant expressions are: K 1 = [B]b [A] a ; K 2 = [C]c [B] b K new = K 1 K 2 For the reaction aa cc the equilibrium constant expression is: K new = [C]c [A] a = [B]b [C]c [A] a [B] b 3

4 Q. For N2(g) + 3 H2(g) 2 NH3(g), K = 3.7 x 10 8 at 25 C; find K for: NH3(g) 0.5N2(g) + 1.5H2(g), at 25 C. Q. Using: H 2 + Br 2 2HBr K p = H 2 2H K p = Br 2 2Br K p = Calculate KP for: H + Br HBr 15.4 Heterogeneous Equilibria Again: Solids, liquids and solvents do not appear in K expressions (their concentrations do not change); however, they must be present for equilibrium to occur. CaCO 3 (s) CaO(s) + CO 2 ; K C = [CO 2 ] and K P = P CO2 H 2 O (l) + CO 3 2 (aq) OH (aq) + HCO 3 (aq) ; K C = [OH ][HCO 3 ] [CO 3 2 ] 15.5 Calculating Equilibrium Constants 1. Easy: Given all equilibrium amounts, just substitute the values into the K expression and solve it. 2. Given initial amounts and one equilibrium amount, calculate K. You have to set up a chart. 4

5 Q. 2H 2 S (g) 2H 2 (g) + S 2 (g) Start with mol H2S in a 10.0 L vessel at 1132 C. At equilibrium, there s mol H2 present. What is the value of KC at this temperature? (K C = ) 15.6 Applications of Equilibrium Constants If a system is not in equilibrium, the reaction will go in whichever direction needed ( or ) to get to equilibrium. Reaction quotient: Q Q has the same form as K, but actual concentrations are used instead of equilibrium concentrations. Predicting the Direction of Reaction If Q > K, the reaction will proceed fastest in the reverse direction the [products] will decrease and [reactants] will increase If Q < K, the reaction will proceed fastest in the forward direction the [products] will increase and [reactants] will decrease If Q = K, the reaction is at equilibrium the [products] and [reactants] will not change If a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction If a reaction mixture contains just products, Q =, and the reaction will proceed in the reverse direction Note: if any product or reactant has a concentration of 0, the reaction must proceed to make some. 5

6 Q. CO 2 (g) + C (s) 2CO(g) ; K C = 1000 If mol CO2 and 0.10 mol CO are combined with C(s) in a 10.0 L 1000 C, will more CO form? Calculating Equilibrium Concentrations 1. Easy given K and some eq. concentrations (or P s), calculate the missing concentration (or P). 2. Given K and initial concentrations (or P s), find KC or KP: Set up a chart. Q. H 2 (g) + I 2 (g) 2HI(g) ; K C = 458 Start with moles of each (H2 and i2) in a 2.00 L flask. Calculate the equilibrium concentrations. For some problems, we will need to solve a quadratic equation: ax 2 + bx + c = 0 ; x = b ± b2 4ac 2a 2 solutions: find both, but keep the one that makes sense physically. 6

7 Q. PCl 5 (g) PCl 3 (g) + Cl 2 (g) ; K C = 160 If initial [PCl5] is 1.00M, what is the equilibrium composition at 160 C. Approximations to Simplify the Math When the equilibrium constant is very small, the position of equilibrium favors the reactants; therefore, if we have relatively large initial concentrations of reactants, the reactant concentration will not change significantly when it reaches equilibrium. So: [X]equilibrium = ([X]initial ax) [X]initial Example: 1.00 x ; if x = , then: = 1.00 to 2 or 3 sig. figs. we are approximating the equilibrium concentration of reactant to be the same as the initial concentration assuming the reaction is proceeding forward We can check our approximation by comparing the approximate value of x to the initial concentration: If the approximate value of x is less than 5% of the initial concentration, the approximation is valid. approximate x if 100 < 5% valid approximation initial concentration 7

8 Q. For the reaction I2(g) 2 I(g) the value of Kc = 3.76 x 10 5 at 1000 K. If 1.00 mole of I2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I2] and [I]? (use the simplifying assumption to solve for x) Q. For the reaction 2 H2S(g) 2 H2(g) + S2(g) at 800 C, Kc = 1.67 x If a L flask initially containing 1.25 x 10 4 mol H2S is heated to 800 C, find the equilibrium concentrations. 8

9 For reactions with large K values, two steps are required (two charts) 1. Assume the reaction goes to completion. 2. Consider the reverse reaction at equilibrium. Example: A (g) + 3B (g) C (g) + 2D (g) K P = Mix atm A and atm B. Determine all pressures at equilibrium. 1. The reaction goes to completion, so, the LR gets used up. Find the new P s. 2. Consider equilibrium in the reverse direction. A 3B C 2D Initially atm atm 0 0 Change -X -3X +X +2X End (not Eq!) X X X 2X L. R.? need: 3B A 0.2B ; have: 0.1A = 2B 1A ; B is L. R. ; x = 0 ; x = = atm A = = atm ; C = x = atm ; D = 2x = atm ; B = 0 We now consider the reverse reaction: C (g) + 2D (g) A (g) + 3B (g) ; 1 K P = = C 2D A 3B Initially atm atm 0 Change -Y -2Y +Y +3Y Equilibrium Y Y atm+Y 3Y K is small. The reactions does not proceed much. Y is small compared to other numbers Y ; Y We can neglect Y. Reduces to: K P = P 3 AP B 2 P C P = ( y)(3y) 3 = D ( y)( y) 2 ( )(3y) 3 ( )( ) 2 = ; (3y) 3 = 27y 3 3 y = ( )( )( ) 2 ( )(27) = atm Equilibrium Amounts (from table) P A = atm ; P C = atm P B = atm ; P D = atm 9

10 15.7 Le Châtelier s Principle Once at equilibrium (when the concentrations of all the reactants and products remain the same), what happens if the conditions are changed? The concentrations of all the chemicals will change until equilibrium is restored. The new concentrations will be different, but the equilibrium constant will be the same Unless the Temperature is changed. Le Châtelier's Principle guides us in predicting the effect various changes in conditions have on the position of equilibrium. If a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance. (A new equilibrium will ensue.) 1. Changing Concentrations This does not change the value of K. Equilibrium shifts away from side with added chemicals or toward side with removed chemicals If something is added to the mixture, the reaction shifts to use up some. If something is removed from the mixture, the reactions shifts to create more. Adding/removing solids or liquids has no effect (as long as there s still some present.) Q. H 2 (g) + I 2 (g) 2HI(g) Which way will the equilibrium shift if you : a. Add HI b. Add I 2 c. Remove H 2 d. Add NaOH 2. Changing Pressure/Volume If VContainer is decreased, P increases. The equilibrium shifts so as to decrease P: towards the side with fewer moles of gas. CO (g) + 3H 2 (g) CH 4 (g) + H 2 O (g) If V, P, Eq. will shift to right, toward fewer moles of gas (takes up less space ) Note: adding an inert gas won t change partial P s; therefore, there is no shift in equilibrium. 10

11 3. Changing Temperature When we change the temperature, the value of K also changes. If H(+)(endothermic), if T, K or T, K If H( )(exothermic), if T, K or T, K In other words, if T is increased, the reaction will go in the endothermic direction; if T is decreased, the reaction will go in the exothermic direction. Q. 2SO 2 (g) + O 2 (g) 2SO 3 (g) H = 198 kj If T, which way does this eq. shift? Q. 2CO 2 (g) 2CO(g) + O 2 (g) H = 566 kj What conditions of P and T would give the best yield of CO(g)? Be careful! Language What is the difference between saying that: the equilibrium shifts to the left and the equilibrium lies to the left? shifts to the left lies to the left It means the reaction It means at eq. we will go. have mostly reactants: Q > K K eq 1 The Effect of Catalysts Catalysts increase the rate of the forward and reverse reactions by the same factor; therefore, catalysts do not affect the position of equilibrium (it only is achieved faster). Kinetic vs Thermodynamic Control Changing conditions to affect the reaction rates. Adjusting the conditions to ensure that only the desired product or products are present in significant concentrations at equilibrium. 11