Chapter 14 Chemical Equilibrium Hill, Petrucci, McCreary & Perry 4 th Ed. Chemical Equilibrium Many Reactions seem to STOP before all the reactants are used up. The Concentrations of Reactants and Products reach constant values. A State of Chemical Equilibrium is said to have been established, but nothing stopped! Chemical Equilibrium is a Dynamic State Rate Forward Reaction = Rate Reverse Reaction The Hydrogen Iodide Decomposition The equilibrium constant is actually the ratio of the rate constants for the forward and reverse reactions! [H 2][I 2 ] The exponents here are the same as the coefficients in the chemical equation. Only this form of the equilibrium expression gives a constant ratio. 1.84 x 10-2 at 698 K This value of the equilibrium constant is valid only at 698 K!
The Kinetic View of Equilibrium Rate of the forward reaction: Rate of the reverse reaction: At Equilibrium the rates are equal: Rate = k f Rate = k r k f = k r k f k r = [H 2][I 2 ] The Value of an Equilibrium Constant [H 2][I 2 ] = 1.84 x 10-2 at 698 K A value for any one concentration may be calculated if the others are known: = [H 2][I 2 ] 1.84 x 10-2 [HI] = 1.84 x 10-2 1 2 General Equation for Equilibria For the General Equation: a A + b B +... g G + h H +... The Equilibrium Expression will have the form: [G] g [H] h... [A] a [B] b... The Equilibrium Expression will always have the form: [products] n [reactants] m
Calculating an Equilibrium Constant Need: Balanced Chemical Equation Initial Concentrations of all species One Equilibrium Concentration Alternatively the initial moles of all species and one equilibrium mole plus the volume of the container Concentrations Decomposition of Hydrogen Iodide at 731 K 4.00 moles of HI (g) in a 5.00 L container was decomposed at 731 K. The amount of I 2(g) at equilibrium was found experimentally to be 0.442 mole. Calculate value of K c. balanced chemical equation: from the stoichiometry: -2x +x +x Initial concentrations from data: [HI] o = 4.00 mol 5.00 L = 0.800 M [H 2] o = [I 2 ] o = 0 [I 2 ] eq = 0.442 mol 5.00 L = 0.0884 M Can You Solve This Problem? balanced chemical equation: [initial] 0.800 M 0 M 0 M from the stoichiometry: -2x +x +x [equilibrium] 0.0884 M
Variations of Equilibrium Expressions balanced chemical equation: 2 HI (g) H 2(g) + I 2(g) If the equation is reversed; = 0.0201 H 2(g) + I 2(g) 2 HI (g) K'c = 1 = = 49.7 0.0201 K'c is the reciprocal of Kc Equilibrium Expressions other than K c balanced chemical equation: 2 HI (g) H 2(g) + I 2(g) = 0.0201 Quantities proportional to moles/liter can be substituted for Molarity: From PV = nrt the ideal gas equation n M = V = P RT = PH 2 PI 2 RT RT PHI RT 2 = PH 2 PI 2 Kp = Kc only when # moles reactants = # moles products. Kp = Kc(RT) n P HI 2 = K p Decomposition of PCl 5 PCl 5(g) PCl 3(g) + Cl 2(g) 3.26 x 10-2 at 191 o ; Calculate Kp Kp = Kc (RT) n n = # moles product - # moles reactant n = 2-1 = 1 Kp = Kc (RT) n T K = 191 + 273 = 464 K Kp = (3.26 x 10-2 M) [ (0.0821 L atm )( 464 K)] 1 K mol Kp = 1.24 atm
Value of Kc & The Extent of Reaction Kc >> 1 Reaction goes to completion! Kc << 1 Very little Product Forms! [products] n [reactants] m Value of Kc for Reactions Added Together Kc total = Kc 1 x Kc 2 x Homogeneous versus Heterogeneous Equilibria Homogeneous Equilibria All species have the same phase, all gas, liquid etc. Heterogeneous Equilibria Some species have different phases - Only the Principal Phase, usually the most mobile is in the Kc expression. 3 Fe (s) + 4 H 2 O (g) Fe 3 O 4(s) + 4 H 2(g) [H 2 ] 4 [H 2 O]4 Can You Write Equilibrium Expressions for the Following Reactions? CaCO 3(s) CaO (s) + CO 2(g) H 2 O (l) H 2 O (g) C (s) + H 2 O (g) CO (g) + H 2(g)
The Reaction Quotient, Q Predicting the Direction of Net Change For any real system at equilibrium: [H 2][I 2 ] = 1.84 x 10-2 at 698 K For any real system NOT at equilibrium: Qc = [H 2][I 2 ] = Any value If Qc > Kc the reaction will go toward more reactants If Qc < Kc the reaction will go toward more products Le Chatelier s Principle When a change (concentration, temperature, pressure or volume) is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. Adding reactants will cause more product to form, and conversely. Removing Products will cause more product to form, and conversely. Increasing temperature will cause a shift in the endothermic direction, and conversely. Le Chatelier s Principle At equilibrium: New equilibrium: = Kc [N 2 ] [H 2 ] 3 [N 2 ] Change: Add hydrogen! = Qc [H 2 ] 3 < Kc The reaction will react to reduce the excess H 2 by combining it with N 2 to produce more NH 3. From the new concentrations N 2 & H 2 will decrease and NH 3 will increase until Q c = Kc.
Le Chatelier s Principle At equilibrium: The reaction will react to produce more N 2 by decomposing NH 3, the concentrations of N 2 & H 2 will increase until Q c = Kc. Change: Remove Nitrogen! [N 2 ] [H 2 ] 3 = Qc > Kc New equilibrium: = Kc Le Chatelier s Principle At equilibrium: [NH 3 ] 2 Change: Remove ammonia! The reaction will react to to produce more NH 3 by combining H 2 with N 2. The N 2 & H 2 concentrations will decrease and NH 3 will increase until Q c = Kc. New equilibrium: = Q c < Kc = K c At equilibrium: [NH 3 ] 2 The forward reaction will speed up less than the endothermic reverse reaction. This change actually changes Kc. The new K'c will be smaller than the old. Le Chatelier s Principle k f k r K'c = k f k r H < 0 (exothermic) Change: Increase Temperature! = Q c > K'c New equilibrium: = K' c < Kc
Changing the Pressure by Compression of Volume N 2 O 4(g) 2 NO 2(g) n = 1 Increasing pressure by compression causes the [NO 2 ] 2 = [x]2 [N 2 O 4 ] [y] < [2x]2 numerator to become [2y] larger than denominator. The position of equilibrium will shift to the side having fewer molecules. n = 0 = [x][x] [y] 2 = [2x][2x] [2y] 2 Increasing pressure by compressing the volume has no effect on the ratio, numerator and denominator increase to the same extent. There will be no shift in the position of equilibrium when n = 0. Optimum Conditions H < 0 (exothermic) 4 moles reactants 2 moles products n = -2 Increasing pressure shifts to right, products. Decreasing pressure shifts to left, reactants. Increasing temperature shifts to left, reactants. Decreasing temperature shifts to right, products. Optimum Conditions: highest feasible pressure lowest feasible temperature The converse of the equilibrium shifts also apply. See the last two problems for the pressure effect. Effect of a Catalyst on Equilibria A catalyst changes the mechanism and lowers activation energies. Both forward and reverse reactions are affected precisely the same amount. A catalyst will not change an equilibrium composition A catalyst will change the rate that any equilibrium is achieved.
Using Kc to Calculate Equilibrium Concentrations Need: Balanced Chemical Equation -- Or thekcexpression. Value of Kc at a Specified Temperature. All Initial Concentrations -- Or All but one Equilibrium Concentrations Calculating Equilibrium Concentrations Calculate the equilibrium concentrations for the following reaction given the initial concentrations. Kc' = H 2(g) + I 2(g) = 49.7 at 731 C [H 2 ] [I 2 ] 2 HI (g) [HI] [initial] 0.500 M 0.500 M 0.0 M Calculating Equilibrium Concentrations Calculate the equilibrium concentrations for the reaction below given all but one [equil]. PCl 5(g) PCl 3(g) + Cl 2(g) [PCl 3 ] [Cl 2 ] = 0.415 at 250 o [PCl 5 ] Calculate [PCl 5 ] eq Given that [PCl 3 ] = [Cl 2 ] = 0.020 M
Which way will this Reaction Go? A 50.0 L reaction vessel contains 1.00 mol N 2 ; 3.00 mol H 2 and 0.500 mol NH 3. If 0.500 at 400EC, determine the direction of reaction. Note: P total = ~5 atm. First we need to calculate the concentrations of each reactant and product. Then we need to write the equilibrium expression. Then we will set the equilibrium expression equal to Q c since the system is not at equilibrium and calculate Q c from the values. From the value of Q c we shall determine the direction the reaction will go. Which way will this Reaction Go? A 5.0 L reaction vessel contains 1.00 mol N 2 ; 3.00 mol H 2 and 0.500 mol NH 3. If 0.500 at 400EC, determine the direction of reaction. Note: P total = ~50 atm.