Equilibrium The equilibrium constant K p for the reaction is 158 at 1000K. What is the equilibrium pressure of O 2 if the P NO = atm and P NO = atm? 2 2NO 2 (g) 2NO (g) + O 2 (g) 14.2 K p = 2 P NO P O 2 P NO 2 2 POPO 2 = K p P NO POPO 2 = 158 x (0.400) 2 /(0.270) 2 = 347 atm
Equilibrium The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = x = 220 K p = K c (RT) n n = 1 – 2 = -1 R = T = = 347 K K p = 220 x ( x 347) -1 =
Equilibrium A + B C + D C + D E + F A + B E + F K c = ‘ [C][D] [A][B] K c = ‘ ‘ [E][F] [C][D] [E][F] [A][B] K c = KcKc ‘ KcKc ‘‘ KcKc KcKc ‘‘ KcKc ‘ x If a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions. 14.2
Equilibrium Sample Exercise 15.5 Combining Equilibrium Expressions Given the reactions determine the value of K c for the reaction Solution Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium constant for a third equation, which is related to the first two. Plan We cannot simply add the first two equations to get the third. Instead, we need to determine how to manipulate the equations to come up with the steps that will add to give us the desired equation.
Equilibrium Sample Exercise 15.5 Combining Equilibrium Expressions Continued Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives Now we have two equations that sum to give the net equation, and we can multiply the individual K c values to get the desired equilibrium constant.
Equilibrium Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. CaCO 3 (s) CaO (s) + CO 2 (g) K c = ‘ [CaO][CO 2 ] [CaCO 3 ] [CaCO 3 ] = constant [CaO] = constant K c = [CO 2 ] = K c x ‘ [CaCO 3 ] [CaO] K p = P CO 2 The concentration of solids and pure liquids are not included in the expression for the equilibrium constant. 14.2
Equilibrium At C the equilibrium constant (K c ) for the reaction Is 1.1 x If the initial concentrations are [Br 2 ] = M and [Br] = M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) x-x+2x x x [Br] 2 [Br 2 ] K c = ( x) x = 1.1 x Solve for x 14.4
Equilibrium K c = ( x) x = 1.1 x x x = – x 4x x = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac 2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) x-x+2x x x x = x = At equilibrium, [Br] = x = Mor M At equilibrium, [Br 2 ] = – x = M 14.4
Equilibrium Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations Solution Analyze We are given the initial concentrations of H 2 and l 2 and the equilibrium concentration of HI. We are asked to calculate the equilibrium constant K c for Solve First, we tabulate the initial and equilibrium concentrations of as many species as we can. We also provide space in our table for listing the changes in concentrations. As shown, it is convenient to use the chemical equation as the heading for the table. Second, we calculate the change in HI concentration, which is the difference Between the equilibrium and initial values: A closed system initially containing 10 3 M H 2 and 10 3 M I 2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87 10 3 M. Calculate K c at 448 C for the reaction taking place, which is Plan We construct a table to find equilibrium concentrations of all species and then use the equilibrium concentrations to calculate the equilibrium constant. Change in [HI] = 1.87 10 3 M 0 = 1.87 10 3 M
Equilibrium Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H 2 ] and [I 2 ]: Fourth, we calculate the equilibrium concentrations of H 2 and I 2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed: Our table now looks like this (with equilibrium concentrations in blue for emphasis): Continued [H 2 ] = 10 3 M 10 3 M = 10 3 M [I 2 ] = 10 3 M 10 3 M = 10 3 M Notice that the entries for the changes are negative when a reactant is consumed and positive when a product is formed. Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Equilibrium Finally, we use the equilibrium-constant expression to calculate the equilibrium constant: Continued Sample Exercise 15.9 Calculating K from Initial and Equilibrium Concentrations
Equilibrium Q c > K c system proceeds from right to left to reach equilibrium Q c = K c the system is at equilibrium Q c < K c system proceeds from left to right to reach equilibrium 14.4
Equilibrium Sample Exercise Predicting the Direction of Approach to Equilibrium Solution Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium. Plan We can determine the starting concentration of each species in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the reaction quotient, Q c. Comparing the magnitudes of the equilibrium constant, which is given, and the reaction quotient will tell us in which direction the reaction will proceed. Solve The initial concentrations are The reaction quotient is therefore At 448 C the equilibrium constant K c for the reaction is Predict in which direction the reaction proceeds to reach equilibrium if we start with 2.0 10 2 mol of HI, 1.0 10 2 mol of H 2, and 3.0 10 2 of I 2 in a 2.00-L container. Because Q c < K c, the concentration of HI must increase and the concentrations of H 2 and I 2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.
Equilibrium Le Châtelier’s Principle Changes in Concentration continued ChangeShifts the Equilibrium Increase concentration of product(s)left Decrease concentration of product(s)right Decrease concentration of reactant(s) Increase concentration of reactant(s)right left 14.5 aA + bB cC + dD Add Remove
Equilibrium Le Châtelier’s Principle Changes in Volume and Pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas 14.5
Equilibrium uncatalyzedcatalyzed 14.5 Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium sooner Le Châtelier’s Principle