1. Gas Equilibria Unit 10

2. Equilibrium Reactions are reversible Reactants are not consumed Equilibrium mixture containing both products and reactants is obtained. @ Equilibrium both (forward and reverse) reactions are taking place at the same time.

3. Equilibrium The rate of the forward reaction = rate of the reverse reaction. K = [products]^ coefficients [reactants]^ coefficients

4. Equilibrium K is the equilibrium constant; it shows the relationship between partial pressures of different GASES present at equilibrium We need the partial pressure of each species –PV=nRT –Pressure must be in atms –R value will 0.0821 –V in L –T in Kelvins

5. At equilibrium Partial pressures remain constant as long as volume of container and temperature remain the same. Kp is equilibrium using partial pressure Kc represents concentration in moles/L (This is when we are looking at aqueous solutions and ions) Kp= Kc(RT) ∆ng ng= moles of gases

6. At Equilibrium For the most part, we are only going to be concerned about Kp so when asked to solve for K it is asking about Kp When writing K expressions, you only include gases and aqueous solutions or ions.

7. Writing K expressions Let’s look at page 344 –#6 –#8 –#12 –HW#1

8. Solving for Kc N 2 (g) + O 2 (g) ↔ 2NO (g) Kp= 0.0255 T= 123 o C Solver for Kc On problems 37-40 you are just solving for Kc. Do not follow the directions. HW#1

9. Solving for K N 2 (g) + O 2 (g) ↔ 2NO (g) –P N 2 = 0.345 atm –P O 2 = 0.765 atm –P NO = 1.34 atm –HW#2

10. Le Chatelier’s Principle When a change (that is, a change in concentration, temperature, pressure, or volume, pH, etc) is imposed on a system at equilibrium, the system responds by attaining a new equilibrium condition that minimizes the impact of the imposed change. I.e., the system will shift to undo the change

11. Concentration If you increase the concentration of any species, the reaction goes in the direction to get rid of it. (Solids, liquids=no effect) A + B  C + D Add A,  Add C,  Add E, since not in the equilibrium, no effect!

12. Pressure When the pressure of a system increases, to reduce the pressure the equilibrium will shift toward the side with the least moles of gases. Decrease pressure, add more moles

13. Volume When the volume of the system increases, we need to fill it back up. The equilibrium will shift toward the side with more moles of gas. Decrease volume, decrease moles.

14. Temperature The effect of temperature depends on the sign of  H. Treat temperature like a chemical: add heat, go to the cool side. If  H is negative, products are exothermic (Hot side). If  H is positive, products are endothermic (Cool side)

15. Reaction Quotient Q = [products]^ coefficients [reactants]^ coefficients This relationship can determine if a system is at equilibrium. It will also be used to predict direction to shift the system.

16. Q vs K If K < Q, the reaction R  P Because for Q to be too large, there is presently too much product. To undo product, the reaction arrow shifts to make more reactant. Vice versa…. If K > Q, the reaction R  P Q is too small because there isn’t enough product.

17. CO + 3H 2 CH 4 + H 2 O Kp = 102 at 773 K The following are introduced into a flask at 773K: 0.902 atm CO, 1.794 atm H 2, 18.38 atm CH 4 and 27.57 atm H 2 O. In what direction will the reaction occur to reach equilibrium?

18. CO + 3H 2 CH 4 + H 2 O Kp = 102 at 773K The following are added to a 15.5 L reaction vessel at 773K: 25.2 g CO, 15.1 g H 2, 130.2 g CH 4 and 125 g H 2 O. In what direction will a net reaction occur to reach equilibrium?

19. CO 2 + CF 4 2COF 2 Kc = 0.50 at 1000 o C If a 2.50 L reaction vessel at 1000 o C contains 0.525 mol CO 2, 1.25 mol CF 4 and 0.75 mol COF 2, in what direction will a net reaction occur to reach equilibrium? HW#3

20. I C E When an equilibrium shifts, it MUST do so STOICHIOMETRICALLY! The coefficients determine the amount and the arrow determines the direction. I -- I nitial concentration or pressure C -- C hange by coefficients E -- E quilibrium is established YouTube - Ice Ice Tables

21. CO + H 2 O CO 2 + H 2 Kp = 23.2 at 600K If 0.250 mol of CO and H 2 O are introduced into a 1 L flask and equilibrium is established, how many moles of each will be present at equilibrium?

22. C (s) + H 2 O CO + H 2 Kp = 0.111 at 1100 K If 0.100 atm H 2 O and 0.100 atm of H 2 are mixed with excess carbon in a 1 L flask initially, what will be the equilibrium concentration of each?

23. CO + Cl 2 COCl 2 The Kp = 21.88 at 395 o C. Initially, 20.0 g CO and 35.5 g Cl 2 are placed in an 8.05 L reaction vessel at 395 o C. What is the partial pressure of each at equilibrium?

24. H 2 + I 2 2HI At 430 o C, the Kc = 54.3. A 1.0 L vessel is charged at this temperature with 1.4 atm of each. What is the partial pressure of each gas?

25. CO 2 + CF 4 2COF 2 The Kp at 1000 o C is 0.50. The vessel is charged initially with 0.45 atm of carbon dioxide and carbon tetrafluoride and 1.2 atm of COF 2. What is the partial pressure of each gas?

26. NO + CO 2  NO 2 + CO Initially a vessel is charged with 3.9 moles of NO and 0.88 moles of CO 2. At equilibrium, there is 0.11 mol of CO 2. What is the Kp? What is Kc? In another experiment, all gases are initially at 1.5 atm. What is the equilibrium pressure?

27. Van’t Hoff Equation Ln K2 – ln K1 = -∆H/R (T2 -1 -T1 -1 ) lnK2/K1= -∆/R(T2 -1 - T1 -1 ) R= 8.31 ∆H= Standard Enthalpy ∆G = -RTlnK ***

28. Van’t Hoff Equation For the reaction, H 2 + Br 2 → 2HBr the Kp is 2.18 x 10 6 at 730 o C. What is the Kp at 1000 o C (you must calculate ∆H first)?

29. Van’t Hoff Equation The hydrogenation of octene to octane is an important process in refining gasoline. The K value for the reaction is 1.7 x 1015 at 298 K and is only 3.2 at 900K. (a) What is the ∆H for this reaction? (b) What is the∆G for this reaction at 298K?

30. Adding Chemical Equations Reaction 1= K1 Reaction 2 = K2 Reaction 3= K3 (overall reaction) K3 = K1 * K2

31. Adding Chemical Equations Calculate K for the reaction: SnO2 (s) + 2CO → Sn (s) + 2CO2 Given SnO2 (s) + 2H2 → Sn (s) + 2H2O K = 8.12 H2 + CO2 → H2O + CO K = 0.771