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. 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.

8. Are the following reactions Heterogeneous or Homogeneous?  2 HgO (s)  2 Hg ( ) + O 2(g)  Heterogeneous reaction  N 2(g) + 3 H 2(g)  2 NH 3 (g)  Homogeneous reaction

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

10. How do we write an equilibrium expression?  The proposed rate of a reaction is dependent on the concentration of the reactants and products raised to the power of the coefficients of each reactant and product in the equation.

11. Writing an equilibrium expression – for a homogeneous equilibrium reaction  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 (aq) if the reaction was between substances in solution.  When the reaction reaches equilibrium, the equilibrium concentrations can be measured and then used then used in the equilibrium expression to calculate the equilibrium constant.  The equilibrium constant always has the same value (provided you don't change the temperature) or amounts of the reactants or products. The equilibrium is unaffected by the addition of a catalyst.

13. Writing an equilibrium expression – for a heterogeneous equilibrium reactions  The equilibrium expression is written the same as the one the previous slide with the exception of the fact that pure solids and pure liquids are excluded from the expression.  Example: CaCO 3(s)  CaO (s) + CO 2(g) Kc = [CO 2(g) ]

14. What is Le’Chatelier’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. This is known as Le Chateliers Principle

15. 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 reverse reaction is therefore endothermic.  Example:N 2(g) + 3 H 2(g)  2 NH 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.

16. Pressure  As with temperature, it helps if you add some simple labels to the equilibrium equation.  Example: N 2(g) + 3 H 2(g)  2 NH 3 (g) HIGH PRESSURE LOW PRESSURE HIGH PRESSURE LOW PRESSURE This side has a total of This side has a 4 moles of gaseous total of 2 moles of gaseous reactantsgaseous 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.

17. Concentration  If a substance becomes more concentrated in a reaction, the equilibrium will shift to reduce its concentration.  Example: N 2(g) + 3 H 2(g)  2 NH 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.

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

19. What is a practical application of Le’Chatelier’s Principle??  What are the optimal conditions under which to make ammonia gas from its 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 pressure requires specialized and expensive machinery.

20. 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?

21. 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!!

22. 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.

23. 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

24. 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

25. 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.

26. 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 2x

27. 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))

28. 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 2x 2. As you well know, they are different. This mistake happens a LOT!!

29. 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

30. Now 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