15.5 Calculating Equilibrium Constants

1. 2 HI (g) H2 (g) + I2 (g) • We often do not know the equilibrium concentrations of all chemical species in an equilibrium mixture. • Knowing the equilibrium concentration of at least one species allows us to use the stoichiometry of the reaction to deduce the equilibrium concentrations of the others. • Ex: A closed system initially containing 1.000 x 10−3M H2 and 2.000 x 10−3M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3M. • Calculate Kc at 448C for the reaction taking place, which is: 15.5 Calculating Equilibrium Constants

2. What Do We Know?

3. The change in [HI] is1.87 x 10-3M

4. Stoichiometry tells us [H2] and [I2] decrease by half as much

5. We can now calculate the equilibrium concentrations of the reactants.

6. Kc= [HI]2 [H2] [I2] (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) = = 51 Therefore, the equilibrium constant is. . .

7. 15.6 Applications of Equilibrium Constants • The magnitude of K indicates the extent to which a reaction will proceed. (Sec 15.3) • K also allows us to predict the direction in which a reaction mixture will proceed to reach equilibrium, and • K allows us to calculate the concentration of reactants and products when equilibrium has been reached.

8. 1. Predicting the Direction of Reaction • Given the following equation, Kc = 0.105 at 472oC: N2(g) + 3 H2(g) 2 NH3(g) • Calculate the reaction quotient, Q, a number obtained by substituting starting reaction and product concentrations (or partial pressures) into an equilibrium-constant expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

9. If Q = K, the system is at equilibrium.

10. If Q > K, there is too much product and the equilibrium shifts to the left.

11. If Q < K, there is too much reactant, and the equilibrium shifts to the right.

12. Example for predicting the direction of approach to equilibrium

13. 2. Calculating Equilibrium Concentrations Example 1: Calculate concentration of product

14. Example 2: Calculate Equilibrium Concentrations from Initial Concentrations

15. Calculations:

16. Practice Exercise:

17. 15.7 Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

18. 1. Change in a Reactant or Product Concentration • If a chemical system is at equilibrium and we add a substance (either reactant of product), the reaction will shift so as to reestablish equilibrium by consuming part of the added substance. Conversely, removing a substance will cause the reaction to move in the direction that forms more of the substance

19. Example: The Haber Process • N2(g) 3 H2(g) 2 NH3(g) • The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.

20. N2(g) 3 H2(g) 2 NH3(g) If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3.

21. N2(g) 3 H2(g) 2 NH3(g) • Removing NH3 causes a shift towards producing more NH3. • This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid.

22. 2. Effects of Volume and Pressure Changes • A system can reduce its pressure by reducing the total number of gas molecules. • Thus, at constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas. • See Fig 15.13

23. 3. The Effect of Changes in Temperature • Treat heat as if it were a chemical reagent. Endothermic: Reactants + heat products Exothermic: Reactants products + heat Endothermic: Increasing T results increase in K Exothermic: Increasing T results in decrease in K (Cooling has the opposite effect.)

24. 4. The Effect of Catalysts • Catalysts increase the rate of both the forward and reverse reactions. • Equilibrium is achieved faster, but the equilibrium composition remains unchanged. • The value of K is unaffected by the presence of a catalyst.