Chemical Equilibrium

1. Chemical Equilibrium Learning Goals: I will understand what dynamic equilibrium means, and relate it to a) phase, b) solubility and c) reaction equilibrium I will also be able to calculate equilibrium concentrations using ICE tables

2. Equilibrium is reached in a system when the reactants combine to form products at the same rate at which the products combine to form reactants. There becomes no apparent change in amounts of products or reactants

3. 2NO2 N2O4 • A reaction that is in equilibrium is indicated by the use of a double arrow. • All reactions are considered reversible, but the extent to which they proceed in both directions varies • When reactants are strongly favored, the percent reaction is much less than 1% • When products are strongly favored, %reaction >99%

4. A) Phase Equilibrium • Phase Equilibrium – a dynamic equilibrium between different physical states of a pure substance in a closed system

6. The system is initially at equilibrium. The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again. B) Solubility Equilibrium • The equilibrium between a solute and a solvent in a saturated solution in a closed system

7. Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium. The reaction continues as time passes and more reactants are changed to products. The reaction begins to occur, and some products (H2 and CO2) are formed. Equal numbers of moles of H2O and CO are mixed in a closed container. C) Chemical Equilibrium • The equilibrium between reactants and products of a chemical reaction in a closed system

8. H2O + CO H2 + CO2

9. Solving Equilibrium Problems

10. The I stands for initial, the C for change and the E for equilibrium. • An ICE box looks like this:

11. Example 1 • Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2+ F2 → 2 HF • If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the equilibrium concentration of hydrogen and hydrogen fluorine if the equilibrium concentration of fluorine gas is 0.48mol/L? [F2] 2 – x = 0.48 - x = - 1.52 x = 1.52 mol/L [H2] = 2 – x = 2 – 1.52 = 0.48 [HF] = 2x = 2(1.52) = 3.04 mol/L C = mol/L CH2 = 4/2 = 2 mol/L ICE Table H2 + F2 → 2 HF Initial 2.00 2.00 0 Change -x -x 2x Equil (M) 2.00 -x 2.00 - x 2x

12. Homework • Pg 428 Q 4, 5

13. Equilibrium Constant (K) and Equilibrium Law Equation Learning Goal: I will understand what the equilibrium constant (K) and the equilibrium law equation are, and be able to use them to complete calculations solving for either one.

14. Equilibrium does not mean that concentrations are all equal!! • However, we can quantify concentrations at equilibrium. • Every equilibrium has its own equilibrium constant.

15. The Equilibrium Constant • equilibrium constant (K): the ratio at equilibrium of the [ ]’s of products raised to their stoichiometric coefficients divided by the [ ]’s of reactants raised to their stoichiometric coefficients.

16. For a general equilibrium aA + bB cC + dD, the equilibrium expression is: • Note that units are not included when calculating K’s. • Thus, equilibrium constants are unitless.

17. Example 1 • Write the equilibrium constant expression for the reaction: 2H2(g) + O2(g) 2H2O(g)

18. Example 2 • Nitrogen and hydrogen combine to form ammonia. Calculate the value of the equilibrium constant for this reaction if the concentrations were measured at equilibrium, at 500oC to be: N2= 1.50 x 10-5 mol/L H2= 3.45 x 10-1 mol/L NH3= 2.00 x 10-4 mol/L

19. Physical Meaning of K • K>>1 means that the equilibrium favors products, i.e. there are high [ ]’s of products and low [ ]’s of reactants at equilibrium • K=1 means [products] is approximately equal to [reactants] at equilibrium • K<<1 means that the equilibrium favors reactants, i.e. there are low [ ]’s of products and high [ ]’s of reactants at equilibrium

20. Rules for Manipulating K • If the equation is reversed, the equilibrium constant is inverted.

21. Rules for Manipulating K • If the equation is multiplied by a factor, the equilibrium constant is raised to the same factor.

22. Heterogeneous Equilibria • When reactants and products are not all in the same phase (opposite of homogeneous equilibria) • If an equilibrium contains pure solids or pure liquids, they are not included in the equilibrium constant expression.

23. Example 1 • A 10.0L container is filled with 4.0 mol of NO(g) and 2.0 mol of O2(g). 2NO(g) + O2(g) ↔ 2NO2(g) • At equilibrium, the container only had 2.8 mol of NO(g). Calculate the Keq for this reaction. • The concentrations change according to the mol ratio in the balanced equation, so O2 will change half as much as NO. • Then we calculate the Keq. Keq = (0.12)2/(0.28)2(0.14) = 1.3 • Since the Keq is greater than one, the products are favoured. If it was less than one the reactants would be favoured.

24. Example 2 • For the following reaction, find all equilibrium concentrations if you started off with 0.5M of H2 and 0.5M of I2 0.5M and Keq = 32. H2(g) + I2(g) ↔ 2HI(g) 32 = (2x)2/(0.5-x)2 • Solve for x and you get x = 0.37. • Then you go back to the equilibrium concentrations and plug in x to get 0.13M for H2 and I2 and 0.74M for HI.

25. Example 3 • Sometimes to find x, you must use the quadratic formula: x = -b ± √(b2-4ac) 2a • Find all equilibrium concentrations for the following reaction if you begin with 0.2M of PCl5 and the equilibrium constant is 1.3. PCl5(g) ↔ PCl3(g) + Cl2(g) 1.3 = (x)2/(0.2-x) • This equilibrium expression is a quadratic so you need the quadratic formula to solve for x. Once you solve, you get x = -1.48 and x = 0.176. x = -1.48 gives you negative concentrations so you need to discard it which leaves x = 0.176 as your answer. Plug it back into the equilibrium concentrations and you get 0.024M for PCl5 and 0.176M for PCl3 and Cl2.

26. Homework • Pg 436 Q2, 5