Dynamic Equilibrium Illustrated

Transcription

1 שו וי מ שק ל Equilibrium Reactants Products In an equilibrium, the forward and reverse processes continue to occur but at equal rates! The reactant and product concentrations remain constant We are usually concerned with the situation after equilibrium is reached. 1 Dynamic Equilibrium Illustrated NaCl containing radioactive Na + is added to a saturated NaCl solution. After a time, the solution contains radioactive Na + NaCl dissolves and recrystallizes continuously. and the salt now contains some radioactive Na

2 Concentration vs. Time If we begin with only 1 M HI, the [HI] decreases and both [H 2 ] and [I 2 ] increase. Beginning with 1 M H 2 and 1 M I 2, the [HI] increases and both [H 2 ] and [I 2 ] decrease. 2 HI(g) H 2 (g) + I 2 (g) Beginning with 1 M each of H 2, I 2, and HI, the [HI] increases and both 3[H 2 ] and [I 2 ] decrease. Regardless of the starting concentrations; once equilibrium is reached the expression with products in numerator, reactants in denominator, where each concentration is raised to the power of its coefficient, appears to give a constant. 4 2

3 קבוע ש.מ. The Equilibrium Constant For the general reaction: aa+ bb yy+ zz The equilibrium expression is: Each concentration is raised to the power of its stoichiometric coefficient K eq = [Y] y [Z] z [A] a [B] b K eq : is independent on the initial concentrations depends on temperature depends on the way the equation is written! has units of M x Products in numerator. Reactants in denominator. 5 If the equilibrium concentrations of COCl 2 and Cl 2 are the same at 395 C, find the equilibrium concentration of CO in the reaction: CO(g) + Cl 2 (g) COCl 2 (g) K eq = 1.2 x 10 3 M -1 at 395 C 6 3

4 K eq of the Reverse Reaction Consider the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K eq = = 4.67 x M -1 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: 2 NO 2 (g) 2 NO(g) + O 2 (g) What will be the equilibrium constant K' eq for the new reaction? [NO] 2 [O 2 ] K' eq = [NO 2 ] = = = = x M [NO K eq 4.67 x ] 2 [NO] 2 [O 2 ] 7 K eq of a Multiplied Equation Consider the reaction: 2 NO(g) + O 2 (g) 2 NO 2 (g) [NO 2 ] 2 K eq = = 4.67 x M -1 (at 298 K) [NO] 2 [O 2 ] Now consider the reaction: NO(g) + ½ O 2 (g) NO 2 (g) What will be the equilibrium constant K" eq for the new reaction? [NO 2 ] K" eq = [NO][O 2 ]½ = K eq ½ = 4.67 x = 6.83 x 10 6 M -½ 8 4

5 K eq and the Modifying Equation: Summary For the reverse reaction, K is the reciprocal of K for the הופכי reaction. forward When an equation is multiplied by a common factor n to produce a new equation, K eq is raised to the power of n to obtain the new equilibrium constant. It should be clear that K eq always refers to the way the balanced chemical equation is written. The equilibrium concentrations do not depend on the way the equation is written! 9 The equilibrium constant for the reaction: ½H 2 (g) + ½ I 2 (g) at 718 K is HI(g) (a) What is the value of K eq at 718 K for the reaction HI(g) ½ H 2 (g) + ½ I 2 (g) (b) What is the value of K eq at 718 K for the reaction H 2 (g) + I 2 (g) 2 HI(g) 10 5

6 The Equilibrium Constant for an Overall Reaction Consider the following reactions: N 2 O(g) + ½ O 2 (g) 2 NO(g) K 1 = 1.7 x M ½ 2 NO(g) + O 2 (g) 2 NO 2 (g) K 2 = 4.67 x M -1 N 2 O(g) + 3/2 O 2 (g) 2 NO 2 (g) K 3 =?? What would be the equilibrium constants for the overall reaction? K 3 = K 1 K 2 = 7.9 M -½ Adding the given equations gives the desired equation. Multiplying the given values of K gives the equilibrium constant for the overall reaction. 11 Equilibria Involving Pure Solids and Liquids The concentration of pure solid and liquid phases remains constant during the reaction (although their amount changes). Why? Concentrations of pure solid and liquid phases get incorporated into the value of K eq and do not appear in the expression Example: CaCO 3 (s) CaO(s) + CO 2 (g) [CaO] [CO 2 ] K eq = [CaCO 3 ] K eq = [CO 2 ] 12 6

7 Equilibrium Constants: When Do We Need Them and When Do We Not? A very large numerical value of K eq signifies that a reaction goes (essentially) to completion. A very small numerical value of K eq signifies that the forward reaction, as written, occurs only to a slight extent. An equilibrium constant expression applies only to a reversible reaction at equilibrium. 13 מנת הריכוזים The Reaction Quotient, Q For non-equilibrium conditions, the expression having the same form as K eq is called the reaction quotient, Q. The reaction quotient is useful for predicting the direction in which a net change must occur to establish equilibrium Q < K eq reaction moves in the forward direction Q > K eq reaction moves in the reverse direction 14 7

8 The Reaction Quotient, Q 15 Le Châtelier s Principle When any change in 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. Henri Le Châtelier

9 Changing the Amounts of Reactants At equilibrium, Q = K eq. If [reactant] Q Q is now less than K eq. This condition is only temporary, however, because the concentrations of all species must change in such a way so as to make Q = K eq again. In order to do this, the concentrations of the products increase; the equilibrium is shifted to the right. 17 C 8 H 17 OH + CH 3 COOH the acetic acid concentration first increases CH 3 COO-C 8 H 17 + H 2 O then the concentrations of both reactants decrease When acetic acid (a reactant) is added to the equilibrium mixture and the concentrations of both products increase, until a new equilibrium is established. 18 9

10 Equilibrium Calculations When we have: 1. initial amounts of the reactants 2. equilibrium amount of the product(s) We can calculate: a) amounts of the species in equilibrium b) K eq How? Use the "ICE" table! (Initial/Change/Equilibrium) 19 Equilibrium Quantities from K eq Values When we have: We can calculate: 1. initial amounts of the reactants Amounts of the species 2. K eq in equilibrium Consider the reaction: H 2 (g) + I 2 (g) 2 HI(g) K eq = 54.3 at 698 K If we start with mol I 2 (g) and mol H 2 (g) in a 5.25-L vessel at 698 K, how many moles of each gas will be present at equilibrium? 20 10

11 Carbon monoxide and chlorine react to form phosgene, COCl 2, which is used in the manufacture of pesticides, herbicides, and plastics: CO(g) + Cl 2 (g) COCl 2 (g) K eq = 1.2 x 10 3 M -1 at 668 K How much of each substance, in moles, will there be at equilibrium in a reaction mixture that initially has mol CO, mol Cl 2, and mol COCl 2 in a 10.0-L flask? 21 Summary of Concepts At equilibrium, the concentrations of all reactants and products remain constant with time Equilibrium constant is expressed as the ratio of concentration of products to reactants For the general reaction: aa+ bb yy+ zz The expression for K eq is: K eq = [Y] y [Z] z [A] a [B] b 22 11

12 Summary of Concepts If the chemical equation for a reversible reaction is modified, the equilibrium constant expression must also be modified If the equation is reversed, the K eq expression is inverted If the coefficients of the equation are multiplied by a common factor, the K eq expression is raised to the corresponding power The expression for K eq does not include pure liquids and solids 23 Summary of Concepts In general, if K eq is very large, the forward reaction goes to completion; if K eq is very small, the forward reaction occurs to a very limited extent Usually, calculations based on the equilibrium constant expression are necessary only when K eq values lie between these extremes The reaction quotient is a ratio of concentrations (Q) having the same form as the equilibrium constant expression but using non-equilibrium concentrations If Q < K eq reaction will proceed in the forward direction If Q > K eq reaction will proceed in the reverse direction 24 12

13 Summary of Concepts Qualitative predictions about the effect of various changes (amounts of reactants or products, volume, external pressure, or temperature) can be based on Le Châtelier s principle When a change is imposed on a system at equilibrium, the system responds by attaining a new equilibrium in which the impact of the change is minimized The most common types of equilibrium calculations are determining the value of an equilibrium constant from initial and equilibrium conditions using initial conditions and the equilibrium constant to determine equilibrium conditions 25 13