1. Reaction Equilibrium Do any reactions truly go to completion??

2. Introduction Any reaction in a closed system never truly stops reacting due to the fact that the molecules involved are still interacting, still colliding. Reactions in open systems can go to completion, however. Reactions in closed systems will eventually appear to reach completion. At this point, the reaction has reached a point called equilibrium.

3. Equilibrium  A point in a reaction in which the rate of the forward reaction is the same as the rate of the reverse reaction.  Another way to express this concept is that equilibrium is the point at which reactants becomes products at the same rate as products become reactants.

4. Refer to the graph below – NOTE: This shows reaction rate vs. time

5. Refer to the graph below – NOTE: This plots Concentration of substances vs. Time

7. Extra Practice  Look at and answer question #10 on the practice test.

8. There are two main types of equilibria  Homogeneous equilibrium – has everything present in the same phase. The usual examples include reactions where everything is a gas, or everything is in the same solution.  Heterogeneous equilibrium – has substances present in more than one phase. The usual examples include reactions involving solids & liquids, or solids & gases.

9. How do we represent a reaction at equilibrium?  Reactions at equilibrium are represented using the double arrow.  Examples: 2CO(g)  CO 2 (g) + C(s) 2CO(g)  CO 2 (g) + C(s) H 2 (g) + I 2 (g)  2HI(g) H 2 (g) + I 2 (g)  2HI(g) N 2 (g) + O 2 (g)  2NO(g) N 2 (g) + O 2 (g)  2NO(g)

10. How do we write an equilibrium expression?  If you recall from the Kinetics unit, the proposed rate of a reaction is dependent on the concentration of the reactants raised to the power of the coefficients of each reactant in the equation. The equilibrium expression is treated the same way, except the rate of the reaction in both directions is considered.  See the derivation of the expression in class, but first…

11. Writing an equilibrium expression – for a homogeneous equilbrium expression  We are going to look at a general case with the equation: aA + bB  cC + dD  No state symbols have been given, but they will be all (g), or all (l), or all (aq) if the reaction was between substances in solution in water.  If you allow this reaction to reach equilibrium and then measure the equilibrium concentrations of everything, you can combine these concentrations into an expression known as an equilibrium constant.  The equilibrium constant always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with. It is also unaffected by a change in pressure or whether or not you are using a catalyst.

13. Writing an equilibrium expression – for a heterogeneous equilibrium expression  The equilibrium expression is written the same as that on the previous slide, except, pure solids are excluded from the expression.  Example: CaCO 3 (s)  CaO(s) + CO 2 (g) Kc = [CO 2 (g)]

14. Homework!!  On the practice test given in class, answer questions 1 – 4.

15. What does Kc or Kp tell us?  First, you can use K c or K p compared to Q, the reaction quotient, to determine the direction a reaction is going.  Second, you can use K c or K p to determine the extent to which a reaction will proceed.  Third, you can use Le’Chatlier’s Principle to predict the direction of a reaction at equilibrium after changing one of the reaction conditions.

16. What is a reaction quotient?  A reaction quotient is found using the same equation as is used to find the equilibrium constant. The difference is that the reaction quotient uses actual concentrations and pressures, whereas equilibrium constants are calculated using concentrations and pressures of substances at equilibrium.

17. How to we compare equilibrium constants to reactions quotients?  If Q > Kc (or Kp), then the amount of products is too high and the reaction needs to go to the left to use up product and make more reactant.  If Q = Kc (or Kp), the reaction is at equilibrium.  If Q < Kc (or Kp), the amount of product is too small and the reaction is favoring the formation of product, and therefore going to the right.

18. Let’s Practice comparing Q to Kc  Let’s look at the synthesis of Hydrogen Iodide from it’s elements: H 2 (g) + I 2 (g)  2HI(g) For each of the following sets of concentrations, determine whether the reaction is at equilibrium. If it isn’t, decide what direction it must go to reach equilibrium. a) [H 2 ] = [I 2 ] = [HI] = 0.010 M b) [HI] = 0.30 M; [H 2 ] = 0.01 M; [I 2 ] = 0.15 M c) [H 2 ] = [HI] = 0.10 M; [I 2 ] = 0.0010 M

19. Let’s try one more involving Q and Kc  Nitrosyl Chloride is an orange gas that dissociates at high temperatures into chlorine and nitric oxide: 2NOCl(g)  2NO(g) + Cl 2 (g) In a certain experiment, 2.0 moles No, 3.0 moles Cl2, and 1.4 moles NOCl were introduced into a 10 liter container. A)What is the value of the reaction quotient under these conditions? B)If Kc for this reaction under the same temperature conditions is 4.25, determine the direction of this reaction.

20. You Try!!!  Answer Question #5 on the practice test.

21. What about the extent a reaction is proceeding?  If Kc or Kp is very large (>> 1), the reaction favors the formation of product. What this means is that there is more product at equilibrium than reactant.  If Kc or Kp is small (<< 1), the reaction does not favor the formation of product. An example of a reaction of this type may be the following: N 2 (g) + 2O 2 (g)  2NO 2 (g) Kc @ room temperature = 1 x 10 -30

22. Do you understand??  Answer question 10 on the practice test

23. What isLe’Chatlier’s Principle and how is it used to determine shifts in reactions at equilibrium?  There are three factors affecting the position of equilibrium and hence Kc: Temperature, pressure and concentration. If a reaction at equilibrium is subjected to a change in any of these factors the position of equilibrium will shift to counteract the change. Think of it as a naughty child - ask a naughty child to sit down and it stands up! This is known as Le Chateliers Principle

24. Temperature  To predict what effect this will have on a reaction we will need to know the enthalpy change of the reaction. In the example below the enthalpy change for the reaction is a negative number, indicating that the forward reaction is exothermic. The backward reaction is therefore endothermic.  Example:N 2 (g) + 3H 2 (g)  2NH 3 (g) + energy  An increase in temperature will result in the backward reaction speeding up (to use up the excess heat energy) so the equilibrium will shift to the left hand side, Kc will decrease.

25. Pressure  As with temperature, it helps if you add some simple labels to the equilibrium equation.  Example: N 2 (g) + 3H 2 (g)  2NH 3 (g) HIGH PRESSURE LOW PRESSURE HIGH PRESSURE LOW PRESSURE  This side has a total of This side has a total of  This side has a total of This side has a total of  4 moles of reactants 2 moles of products  An increase in pressure would cause the equilibrium to shift to the low pressure side, Kc would increase. Note that pressure only affects a reaction involving gases.

26. Concentration  If a substance becomes more concentrated in a reaction, the equilibrium will shift to reduce its concentration.  Example: N 2 (g) + 3H 2 (g)  2NH 3 (g)  If the ammonia made in the above reaction is removed, then the equilibrium position will shift to the right hand side to make additional ammonia to replace what was taken away.  If extra ammonia is added to the reaction, the reaction shifts to the left to use up the excess ammonia gas.

27. Catalyst  A catalyst speeds up both the forward and backward reactions, it doesn't change the position of equilibrium but does allow the reaction to reach equilibrium quicker.

28. What is a practical application of Le’Chatlier’s Principle??  What are the optimal conditions under which to make ammonia gas from it’s elements?  In the manufacture of ammonia in the Haber process, the conditions that would give the best yield of ammonia would be a low temperature and a high pressure.  In reality a temperature of 500 degrees centigrade and a pressure of 200 atmospheres is used. The high temperature is used because it speeds up the formation of ammonia and the relatively low pressure is used because running a plant at a high pressures requires specialized and expensive machinery.

29. You Try!!!  On the practice test, complete question #6

30. More sample questions involving equilibrium constants!!  Given the equilibrium constant and the initial concentrations of the reactants, can you find the equilibrium concentrations of the reactions?  Can you look at solubility-product constants (K sp ) and qualitatively determine the extent of solubility?

31. How do I find Kc or the equilibrium concentrations?  Calculating Kc from a known set of equilibrium concentrations seems pretty clear. You just plug into the equilibrium expression and solve for Kc.  Calculating equilibrium concentrations from a set of initial conditions takes more calculation steps. In this type of problem, the Kc value will be given as well as the initial concentrations of reactants, you simply need to find the EQUILIBRIUM CONCENTRATIONS!!

32. The best way to explain how to find equilibrium concentrations from initial values and Kc is by example.  Given this equation: H 2 (g) + I 2 (g)  2 HI(g)  Calculate all three equilibrium concentrations when [H 2 ] o = [I 2 ] o = 0.200 M and Kc = 64.0.

33. Here’s the example:  The solution technique involves the use of an ICE box. Here is an empty one: [H 2 (g)] [I 2 (g)]  [HI] Initial Change Equilibrium

34. Let’s keep working!!  Fill in the first row with given data: Note – the initial concentration of HI = 0.0 M as the reaction has not started yet! Note – the initial concentration of HI = 0.0 M as the reaction has not started yet! [H 2 (g)] [I 2 (g)]  [HI] Initial 0.200 M 0.0 M Change Equilibrium

35. Keep it up!!! Now for the change row. This is the one that causes the most difficulty in understanding: [H 2 ][I 2 ][HI] Initial0.200 0 Change- x + 2x Equilibrium The minus sign comes from the fact that the H 2 and I 2 amounts are going to go down as the reaction proceeds. X signifies that we know some H 2 and I 2 get used up, but we don't know how much. What we do know is that an EQUAL amount of each will be used up. We know this from the coefficients of the equation. For every one H 2 used up, one I 2 is used up also. The positive signifies that more HI is being made as the reaction proceeds on its way to equilibrium. The 2 is important. HI is being made twice as fast as either H 2 or I 2 are being used up. In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms. In problems such as this one, never use more than one unknown. Since we have only one equation (the equilibrium expression) we cannot have two unknowns.

36. Keep going!!!  The equilibrium row should be easy. It is simply the initial conditions with the change applied to it: [H 2 (g)] [I 2 (g)]  [HI] Initial 0.200 M 0.0 M Change - X - X -X-X-X-X + 2X Equilibrium 0.200 – X -.200 – X 2X

37. Keep going!!! Now we are ready to put values into the equilibrium expression. For convenience, here is the equation again: H 2 (g) + I 2 (g)  2 HI(g) The equilibrium expression is: Kc = [HI] 2 [H 2 ][I 2 ] Plugging values into the expression gives 64.0 = (2x) 2 ((0.200 - x) (0.200 - x))

38. Keep going!!  Two points need to be made before going on:  1) Where did the 64.0 value come from? It was given in the problem. 2) Make sure to write (2x) 2 and not 2x2. As you well know, they are different. This mistake happens a LOT!!

39. Almost Done!!  Both sides are perfect squares (done so on purpose), so we square root both sides to get:  8.00 = (2x) / (0.200 - x)  From there, the solution should be easy and results in x = 0.160 M

40. Not we are finished!!!  This is not the end of the solution since the question asked for the equilibrium concentrations, so:  [H 2 ] = 0.200 - 0.160 = 0.040 M [I 2 ] = 0.200 - 0.160 = 0.040 M [HI] = 2 (0.160) = 0.320 M

41. What if both sides are NOT perfect squares???  This example will involve the use of the quadratic formula. (Pause for whining and moaning. Go ahead, get it out of your system! Now, back to chemistry.)  Given this equation: PCl 3 + Cl 2 PCl 5  Calculate all three equilibrium concentrations when Kc = 16.0 & [PCl 5 ] o = 1.00 M.

42. Here is the completed ICE box: [PCl 3 ] [Cl 2 ] [PCl 5 ]  Initial 0 M 0 M 1.00 M Change + x + x - x Equilibrium x x 1.00 - x

43. The equilibrium expression is:  Kc = [PCl 5 ] ([PCl 3 ] [Cl 2 ]) ([PCl 3 ] [Cl 2 ])  Substituting gives:  16.0 = 1.00 - x (x)(x) (x)(x)

44. After suitable manipulation (which you can perform yourself), we arrive at this quadratic equation in standard form:  16x 2 + x - 1 = 0  Using the quadratic formula, which is: x = (- b ± square root[b 2 - 4ac]) / 2a, x = (- b ± square root[b 2 - 4ac]) / 2a, we obtain:  x = (- 1 + square root[12 - (4) (16) (-1)]) / 32  After suitable calculations, we find x = 0.221.

45. Why was the one answer not used???  Please notice that the negative root was dropped, because negative b turned out to be negative one. The answer obtained in this type of problem CANNOT be negative.  Why?  Because we are dealing with the amount of a physical substance in mol / L. Amounts of substances are always represented with positive numbers. An amount of a substance with physical reality cannot be represented with negative numbers.

46. What if we get two positive answers??  The question then becomes how to determine which root is the correct one to use?  Let’s look at one final example:  Given this equation: COCl 2 CO + Cl 2  Calculate all three equilibrium concentrations when Kc = 0.680 with [CO] o = 0.500 and [Cl 2 ] o = 1.00 M.

47. Here is the completed ICE box:  [COCl 2 ] [CO] [Cl 2 ] Initial 0 M 0.500M 1.00 M Change + x - x - x Equilibrium x 0.500 - x 1.00 - x

48. Use your Kc expression:  The equilibrium expression is: Kc = ([CO] [Cl 2 ]) Kc = ([CO] [Cl 2 ]) [COCl 2 ] [COCl 2 ]  Substituting into the expression gives: 0.680 = ((0.5 - x) (1 - x)) 0.680 = ((0.5 - x) (1 - x)) x

49. After some manipulation, we arrive at this quadratic equation, in standard form: x 2 - 2.18x + 0.5 = 0  Using the quadratic formula, we have this to start:  x = (2.18 ± square root[(2.18) 2 - (4) (1) (0.5)] ) / 2  After some manipulation we arrive at: (2.18 ± 1.66) 2  Both roots yield positive values, so how do we pick the correct one?

50. What is the final answer???  The answer lies in the fact that x is not the final answer, whereas (0.5 - x) is. It is the term (0.5 - x) which must be positive.  So the root of 1.92 is rejected in favor of the 0.26 value and the day is saved!!!

51. You Try!!!  Answer questions 8 & 9 on the practice test!!!