Chapter 14 Chemical Equilibrium Forward reaction H 2 (g) + I 2 (g) 2HI(g) Reverse reaction 2HI(g) H 2 (g) + I 2 (g) At equilibrium H 2 (g) + I 2 (g) 2HI(g) Chemical equilibrium is reached when reactants combine to form products at the same rate as the products react to form reactants A forward reverse B rate f =k f [A] A B rate r =k r [B] at equilibrium, rate f = rater, k f [A] = k r [B] Ch 14 Equilibrium 1
Reaction Quotient, Q, and Equilibrium Constant, K In general, aa + bb cc + dd [ Q = C ] c [ D] d Q = [ A] a [ B] b Pr oducts Re actants Generally use Q when a reaction has not reached equilibrium. At equilibrium, all [ ] do not change over time, and Q = K 3H 2 (g) + N 2 (g) 2NH 3 (g) Whatever effects rates, effects equilibrium -Temp, concentration aa + bb cc + dd At equilibrium: ratef = rater if ratef = kf [A][B] and rater = kr[c][d] and kf[a][b] = kr[c][d] and [ C] c [ D] d [ A] a [ B] b = k f = K k r The first quotient is known at the equilibrium constant expression. Ch 14 Equilibrium 2
[ K = C ] c [ D] d [ A] a [ B] b Law of Mass Action Products in the numerator and reactants in the denominator, "Products over Reactants" Raise each concentration to the power of the stoichiometric coefficient in the balanced equation K «1 favors reactants (10 3 ) K» 1 favors products (10 3 ) If 10 3 < K < 10 3, products and reactants present Note: For the reaction PCl 3 (g) + Cl 2 (g) PCl 5 (g) at 400K K = 2.89 For PCl 5 (g) K = 1 2.89 = 0.346 PCl 3 (g) + Cl 2 (g) The equilibrium constant is dependent on the direction of the reaction Ch 14 Equilibrium 3
If a chemical equation can be expressed as the sum of two or more chemical equations, the equilibrium constant for the overall reaction is the product of the equilibrium constants for the component reactions. 2 P( g) + 3 Cl 2 ( g) 2 PCl 3 ( g) K 1 = PCl 3 ( g) + Cl 2 ( g) PCl 5 ( g) K 2 = 2 P( g) + 5 Cl 2 ( g) 2 PCl 5 ( g) K 3 = [ PCl 3 ] 2 P [ ] 3 [ ] 2 Cl 2 [ PCl 5 ] [ PCl 3 ] Cl 2 [ ] [ PCl 5 ] 2 P [ ] 5 [ ] 2 Cl 2 2 P g ( ) + 3 Cl 2 ( g) 2 PCl 3 ( g) ( ) + 2 Cl 2 ( g) 2 PCl 5 ( g) 2 PCl 3 g 2 P( g) + 5 Cl 2 ( g) 2 PCl 5 ( g) [ PCl K 3 = 5 ] 2 [ P] 2 Cl 2 [ ] 5 = [ PCl 3 ] 2 [ [ P] 2 [ Cl 2 ] 3 PCl 5 ] [ PCl 3 ][ Cl 2 ] = K 1 K 2 K 2 = K 1 K 2 2 [ PCl 5 ] [ PCl 3 ] Cl 2 [ ] Relations Between Equilibrium Constants Chemical Equation Equilibrium Constant aa + bb cc + dd K 1 cc + dd aa + bb naa + nbb ncc + ndd K 2 = 1 K 1 = K 1 1 K 3 = K 1 n H 2 S, K a1 = 9.5x10-8, K a2 = 1x10-19 Ch 14 Equilibrium 4
Be aware of Kc and Kp K c is used with concentration expressed in molarity or moles per liter. Kp is used with pressures, if you are given partial pressures in atmospheres. P = n RT V 3H 2 (g) + N 2 (g) n is a measure of concentration V 2NH 3 (g) K P = P 2 NH 3 P 3 = H2 P N2 ([ NH 3 ]RT) 2 ([ H 2 ]RT) 3 ([ N 2 ]RT) = K c ( RT) 2 RT ( ) 3 ( RT) = K c( RT) 2 K p = K c ( RT) n n = moles products moles reac tan ts For above reaction, n = 2 - (3 + 1) = -2 K p = K c ( RT) 2 Ex: In the synthesis of ammonia from nitrogen and hydrogen, K c = 9.60 at 300 C. Calculate K p for this reaction at this temperature. Ch 14 Equilibrium 5
Heterogeneous Equilibrium CaCO 3 (s) CaO(s) + CO 2 (g) The concentration of a solid is proportional to its density and remains constant as long as the temperature does not change. Density M = g cm3 g mol = mol cm 3 Since concentrations of solids are constant, they are included in the equilibrium constant and don't appear in the reaction quotient. [ ] K = [ CaO] CO 2 [ CaCO 3 ] [ [ ] = Kx CaCO 3] CO 2 [ CaO] [ CO 2 ] = K P CO2 = K p Write the equilibrium-constant expression for Kc and Kp for each of the following reactions. CO 2 (g) + H 2 (g) SnO 2 (s) + 2CO (g) CO(g) + H 2 O(l) Sn(s) + 2CO 2 (g) Ch 14 Equilibrium 6
Calculating Equilibrium Constants Substitute the equilibrium concentrations into the equilibrium concentrations. Ex: Nitryl chloride, NO 2 Cl, is in equilibrium with NO 2 and Cl 2, 2 NO 2 Cl(g) 2 NO 2 (g) + Cl 2 (g) At equilibrium the concentrations of the substances are [NO 2 Cl] = 0.00106 M, [NO 2 ] = 0.0108 M, and [Cl 2 ] = 0.00538 M. From these data calculate the equilibrium constant, K c. Predicting the Direction of a Reactions. Q < K, not equilibrium, reactants will be converted to products Q = K, equilibrium Q > K, no equilibrium, products will be converted to reactants Ex: At 448 C the equilibrium constant, Kc for the reaction, H 2 (g) + I 2 (g) 2 HI(g), is 51. Predict how the reaction will proceed to reach equilibrium at 448 C if we start with 2.0x10-2 mol of HI, 1.0x10 2 mol of H 2, and 3.0x10 2 mol of I 2 in a 2.0 L container. Ch 14 Equilibrium 7
Calculating Equilibrium Concentrations Follow flow chart on page 580 Balanced Equation Difference Table: include initial conc., change in conc. [ ], and equilibrium conc. Substitute equilibrium conc. into equilibrium equation and solve for x. If you use quadratic equation, use the root that makes sense. Calculate equilibrium concentrations by inserting value of x. Check your results by substituting your answer into equilibrium equation. EX: The equilibrium constant, K c, for the reaction of H 2 with I 2 is 57.0 at 700 K: H 2 (g) + I 2 (g) 2 HI(g) K c = 57.0 at 700 K. If 1.00 mol of H 2 is allowed to react with 1.00 mol of I 2 in a 10.0 L reaction vessel at 700 K, what are the concentrations of H 2, I 2 and HI at equilibrium? What is the equilibrium composition of the equilibrium mixture in moles? Ch 14 Equilibrium 8
EX: For the reaction I 2 (g) + Br 2 (g) 2 IBr(g), K c = 280 at 150 C. Suppose that 0.500 mol of IBr in a 1.00 L flask is allowed to reach equilibrium at 150 C. What are the equilibrium concentrations of IBr, I 2 and Br 2. Ch 14 Equilibrium 9
Factors that Effect Equilibrium Concentrations change Pressures change Temperature changes Le Châtelier's Principle If a stress (such as a change in concentration, pressure, or temperature) is applied to a system in equilibrium, the equilibrium shifts in a way that tends to undo the effect of the stress. Example A change in concentration will effect the value of Q. The concentrations will shift until Q again equals K. 3H 2 (g) + N 2 (g) 2NH 3 (g) Ch 14 Equilibrium 10
K = [ NH 3 ] 2 [ H 2 ] 3 [ N 2 ] = 1.2 at 375 C H 2 (g) + I 2 (g) 2HI(g) K = [ HI] 2 [ H 2 ] I 2 [ ] = 50.0 If H 2 is added, Q will decrease. The equilibrium will shift so that more HI is produced by consuming H 2 and I 2 until Q = K Endothermic: Reactants + heat = products Increasing T results in increase in K. Exothermic: Reactants = products + heat Increasing T results in decrease in K. Equilibrium shifts in the direction that absorbs heat. Ch 14 Equilibrium 11
Effects of Disturbances on Equilibrium and K Disturbances Addition of reactant Addition of Product Decrease in volume, increase in pressure Increase in volume, decrease in pressure Rise in temperature Drop in temperature Change as Mixture Returns to Equilibrium Some of Added reactant is consumed Some of added product is consumed Pressure decreases Effect on Equilibrium Shift to right Shift to left Shift toward fewer gas molecules Pressure increases Shift toward more gas molecules Heat energy is consumed Heat energy is generated Shift in the endothermic direction Shift in exothermic direction Effect on K No Change No Change No change No change Change Change The Effect of Catalyst A catalyst increases the rate at which equilibrium is achieved, but it does not change the composition of the equilibrium mixture. Ch 14 Equilibrium 12
Le Châtelier's Examples N 2 (g) + O 2 (g) 2NO(g) H = +180.5 kj Equilibrium Constant, K Temperature -31 298 K 4.5 x 10-10 900 K 6.7 x 10-3 2300 K 1.7 x 10 2 NO 2 N 2 O 4 (g) H = -57.2 kj Equilibrium Constant, K Temperature 1300 273 K 70 298 K 3H 2 (g) + N 2 (g) 2NH 3 (g) Which way does equilibrium shift when: Add H 2? Add NH 3? Volume increased? Volume decreased? Ch 14 Equilibrium 13
The Link Between Chemical Equilibrium and Chemical Kinetics K = k f k r k = Ae E a RT ln k = E a RT ln k 1 T = E a R lnk slope = E a /R lnk slope = E a /R 1/T 1/T (1/K) K = k 1 k 1 Potential Energy Potential Energy Reaction Coordinate Reaction Coordinate Ch 14 Equilibrium 14