1. 7.5 Equilibrium Law Calculations (with RICE charts)

2. Recall the calculation of K the larger the value of K, the more the reaction, as written, favours products Thus if we compare the two reaction the first favours the products more heavily.

3. Is a system at Equilibrium? We may want to determine whether the system is at equilibrium—and, if not, in which direction the system will shift to reach equilibrium We can use a trial value that is called a reaction quotient, Q to examine if a system is at equilibrium Think of Q as being similar to K,with the difference being that K is calculated using concentrations at equilibrium, whereas Q may or may not be at equilibrium Q = K, system is at equilibrium Q is > K, system must shift left to reach equilibrium Q is < K, system must shift right to reach equilibrium

4. In a container at 450°C, nitrogen and hydrogen react to produce ammonia. The equilibrium constant, K, is 0.064. When the system is analyzed, the concentrations are found to be as follows: [N2(g)] is 4.0 mol/L, [H2(g)] is 2.0 ×10 -2 mol/L, and [NH3(g)] is 2.2 × 10 –4 mol/L. Determine whether the system is at equilibrium, and if it is not, predict the direction in which the reaction will proceed to reach equilibrium. Since the value of Q (0.0015) is smaller than the value of K (0.064), the reaction is not at equilibrium. In order to attain equilibrium, the reaction will shift to the right.

5. Try Questions 1& 2 pg. 452

6. Problem type: Calculations at Equilibrium Hydrogen iodide, HI(g), a compound used in the production of hydroiodic acid, HI(aq), is produced by reacting hydrogen gas and iodine vapour according to the following equation: H 2 (g) + I 2 (g) → 2HI(g) The equilibrium constant, K, for this reaction is 49.70 at 458°C. Calculate [HI(g)] at equilibrium if the equilibrium concentration of the other two entities is 1.07 mol/L. K = [HI] 2 [H 2 ] [I 2 ] 49.70 = [HI] 2 [1.07] [1.07] [HI] 2 = 49.70×1.1449 [HI] = √56.901 = 7.54 At equilibrium the concentration of hydrogen iodide is 7.54 mol/L

7. In a container at 250°C, iodine and hydrogen react to produce hydrogen iodide. The equilibrium constant, K, is 49.5. Calculate the concentrations of all entities at equilibrium if 0.200 mol of each are initially placed in a 1.000-L closed container. Problem type: Calculating Equilibrium Concentrations from Initial Concentrations Balanced Equation: H 2 + I 2  2HI Calculate Q

8. I C E H2H2 I2I2 2HI Since Q is not equal to K, we know that the reaction is not at equilibrium and we can set out to solve what the concentrations will be at Equilibrium.

10. To check your answer sub you calculated values in to the equilibrium expression. [H 2 ][I 2 ] Q = [HI] 2

11. Try 1-3 on pg 454

12. Problem type: Equilibrium calculations when K is very small Thus far, problems have been designed so that the solution for x is straightforward If the problems were not so carefully designed we might have to use quadratic equation (or calculus) to solve the problem. If K is very large or very small we can use a simplification to make calculating x simple Setting up the ICE chart is the same, but the calculation of K is now slightly different

13. Equilibrium calculations when K is small Looking at the equilibrium law: [0.100 - 2x] 2 4x 3 = small Kc For Kc to be small, top must be small, bottom must be large (relative to top) For top to be small, x must be small If x is small, then 0.100 - 2x  0.100 Notice that we can only ignore x when it is in a term that is added or subtracted. Can we ignore x in: 4x, 3+x, 0.1-3x, 3x-x, x 2 +1?

14. The Hundred’s rule The hundred’s rule is an assumption that we can make in chemistry which will allow us to ignore x. To carry out the 100’s rule we take the initial concentration of the chemical species and divide it by K. If the number is greater than 100 it is generally safe to assume that x is negligible. If the answer is less than 100 we cannot assume that x is small and we have to proceed with using the quadratic or another method of solving.

15. When nitrogen and oxygen gas are inserted into a reaction vessel nitrogen monoxide is formed. If 0.33 mol of nitrogen and 0.0810 mol of oxygen are initially placed in a 10.0L reaction vessel, what is the equilibrium concentration of nitrogen monoxide? The K for the reaction is 4.8 x 10 -31. Balanced Eq’n: Solve for Q and compare to K: Example of Equilibrium calculations when K is small

16. N 2 + O 2  2NO I C E N2N2 O2O2 NO Make an Ice table:

18. Test the validity of the earlier assumption = 0.033-x [N 2 ] = 0.00810-x [O 2 ] Percent error: 5% is generally the acceptable margin of error - See pg. 717 bottom

19. HClH2H2 Cl 2 I C E 2HCl  H 2 + Cl 2 Kc= 3.2 x 10 –34 determine [equil], if [initial] are 2.0 M, 1.0 M, 0 M Can we use the 100’s rule?

21. Plug back into [equil] from table and we get that 2, 1 and 1.3 x 10 –33

22. Try questions 1-3 on pg.

23. Example Solving with the Quadratic If 0.50 mol of N 2 O 4 (g) is placed in a 1.0- L closed container at 150°C, what will be the concentrations of N 2 O 4 (g) and NO 2 (g) at equilibrium? (K = 4.50)

24. Since 0.11 << 100, the assumption that 0.50 x 0.50 is not warranted. We must solve the equation using the quadratic formula.

25. The negative root would result in a negative concentration for NO 2(g). Since negative concentrations are impossible, x=0.375 is the only acceptable solution to the quadratic equation.

26. Check to see if the answer works. Since Q equals K, the solution is at equilibrium when [NO 2(g) ] 0.12 mol/L, and [N 2 O 4(g) ] 0.75 mol/L.

27. Try questions 1-3 on pg. 458 and 3-5 on 459