1. Chemical Equilibrium L. Scheffler

2. Chemical Equilibrium Chemical equilibrium occurs in chemical reactions that are reversible. In a reaction such as: CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g) The reaction can proceed in both directions CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g)

3. An Equilibrium System CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g)  After some of the products are created the products begin to react to form the reactants  At equilibrium there is no net change in the concentrations of the reactants and products  The concentrations do not change but they are not necessarily equal

4. Chemical Equilibrium CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) At the beginning of the reaction, the rate that the forward direction is higher As the reactants decrease the rate in the forward direction slows As the products form, the rate in the reverse direction increases When the two rates are the same equilibrium is achieved

5. Dynamic Equilibrium An equilibrium is Dynamic CH 4 (g) + H 2 O(g)  CO(g) + 3H 2 (g) The amount of products and the reactants are constant. (Note: The concentrations are not necessarily equal but constant. Both reactions are still occurring, but at the same rate)

6. The Equilibrium Constant aA + bB  cC + dD The upper case letters are the molar concentrations of the reactants and products. The lower case letters are the coefficients that balance the equation.

7. The Equilibrium Constant aA + bB  cC + dD

8. Equilibrium Constant Calculations Example N 2 (g) + 3H 2 (g)  2NH 3 (g) At equilibrium, a one-liter container has 1.60 moles NH 3,.800 moles N 2, and 1.20 moles of H 2. What is the equilibrium constant?

9. Equilibrium Constant Calculations At equilibrium, a one-liter container has 1.60 moles NH 3,.800 moles N 2, and 1.20 moles of H 2. What is the equilibrium constant?

10. Reaction quotient The equilibrium constant is a constant ratio only when the system is in equilibrium. If the system it not at equilibrium the ratio is known as a Reaction Quotient If the reaction quotient is equal to the equilibrium constant then the system is at equilibrium

11. Using Equilibrium Constants for other calculations If a solution is not at equilibrium the ratio of the right side over the left is called a reaction quotient. aA + bB  cC + dD

12. Equilibrium Constants and calculations If Q > Keq, the product side is too high and the equilibrium will shift to the left to restore equilibrium If Q < Keq, the product side is too low and the equilibrium will shift to the right to restore equilibrium

13. Equilibrium Calculations – Using I. C. E. Models Equilibrium constants and concentrations can often be deduced by carefully examining data about initial and equilibrium concentrations I nitial C hange E quilbrium

14. Equilibrium Calculations ICE Model problem 1 Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2  2HI Suppose that 1.5 mole of H 2 and 1.2 mole of I 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H 2 ] and [I 2 ] and the equilibrium constant.

15. Equilibrium Calculations ICE Model Problem 1 Solution Hydrogen and iodine are in equilibrium with Hydrogen iodide to this reaction: H 2 + I 2  2HI Suppose that 1.5 mole of H 2 and 1.2 mole of I 2 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.4 mole of HI. Calculate the equilibrium concentrations of [H 2 ] and [I 2 ] and the equilibrium constant. ICE Since 2x = 0.4, x = 0.2 [H 2 ] 1.5 -x 1.5- x [H 2 ] = 1.5 – 0.2 = 1.3 [ I 2 ] 1.2 -x 1.2 –x [I 2 ] = 1.2 – 0.2 = 1.0 [HI ] 0 +2x 0.4 Keq = [HI] 2. = (0.4) 2 = 0.123 [H 2 ] [I 2 ] (1.3) (1.0)

16. Equilibrium Calculations ICE Model Problem 2 Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO 2 + O 2  2SO 3 Suppose that 1.4 mole of SO 2 and 0.8 mole of SO 3 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.6 dm 3 of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant.

17. Equilibrium Calculations ICE Model Problem 2 Solution Sulfur dioxide reacts with oxygen to produce sulfur trioxide according to this reaction: 2 SO 2 + O 2  2SO 3 Suppose that 1.4 mole of SO 2 and 0.8 mole of SO 3 are placed in a 1.0 dm 3 container. At equilibrium it was found that there were 0.6 dm 3 of SO 3. Calculate the equilibrium concentrations of [SO 2 ] and [O 2 ] and the equilibrium constant. ICE Since 2x = 0.6, x = 0.3 [SO 2 ] 1.4 -2x 1.4-2x [SO 2 ] =1.4 – 2( 0.3) = 0.8 [O 2 ] 0.8 -x 0.8 –x [O 2 ] = 0.8 – 0.3 = 0.5 [SO 3 ] 0 +2x 0.6 Keq = [SO 3 ] 2. = (0.6) 2 = 0.281 [SO 2 ] 2 [O 2 ] (0.8) 2 (0.5)

18. Le Chatelier’s Principle Le Chatelier's Principle states: When a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. A change imposed on an equilibrium system is called a stress The equilibrium always responds in such a way so as to counteract the stress

19. Le Chatelier’s Principle Types of stresses Change in concentration of one or more reactants or products Change in temperature Change in pressure Addition of a catalyst

20. Effect of a Change in Temperature An increase in the temperature causes the equilibrium to shift in the direction of the endothermic reaction N 2 (g) + 3 H 2 (g)  2NH 3 (g)  H =-92 kJ mol -1 Since  H is negative the endothermic reaction is the reverse direction. An increase in temperature causes the reaction to shift to the left, resulting in an increase in N 2 and H 2 and a decrease in NH 3

21. Effect of a Change in Pressure Pressure affects only gases in an equilibrium PV = nRT An increase in pressure causes the equilibrium to shift in the direction that has the fewer number of moles N 2 (g) + 3 H 2 (g)  2NH 3 (g)  H =-92 kJ mol -1 An increase in pressure results in a an decrease in N 2 and H 2 and an increase in NH 3

22. Effect of a Change in one of the reactants or products The equilibrium responds in such a way so as to diminish the increase Substances on the same side of the arrow respond in opposite directions. Substances on the opposite side of the arrow move in the same direction N 2 (g) + 3 H 2 (g)  2NH 3 (g) An increase in [N 2 ] results in a decrease in N 2 and H 2 and an increase in NH 3

23. Effect of a Catalyst Catalysts affect both the forward and reverse directions equally A catalyst does not change the concentrations but reduces the time required for the system to come to equilibrium