1. Tutorial 11 Chemical Equilibrium

2. Chemical equilibrium -A state where the concentrations of all reactants and products remain constant with time. aA + bB cC + dD - Expressed Using The Equilibrium Constant [K]:  Species as a Pure liquid (l), a Pure Solid (s) or solvent Present in Excess; Don’t appear in the Equilibrium Constant Expression.

3. Question 1: N 2 (g) + 3Cl 2 (g) 2NCl 3 (g) An analysis of an equilibrium mixture is performed at a certain temperature. It is found that [NCl 3 ] 1.9 x 10-1 M, [N 2 ] =l.4 x 10-3 M and [Cl 2 ]= 4.3 x l0-4 M. Calculate K for the reaction at this temperature. Answer: [1.9x10 -1 ] 2 = [1.4x10 -3 ] [4.3x10 -4 ] 3 = 3.24x10 11

4. Question 2: The equilibrium constant, K, for the reaction has the value 2.1x103 at a particular temperature. When the system is at equilibrium at this temperature, the concentrations of H2 and F2 are both found to be 0.0021 M. What is the concentration of HF in the equilibrium system under these conditions? H 2 + F 2 2HF

5. Answer : [HF] 2 2.1x 10 3 = [0.0021] [0.0021] [HF] = 0.096 M

6. Reaction Quotient (Q) - To Determine the direction in which the system must shift to reach equilibrium. - Obtained by applying the law of mass action using initial concentrations instead of equilibrium concentrations. Ex: N 2(g) + 3H 2(g) 2NH 3(g) Where the subscript zeros indicate initial concentrations.

7. By Comparing The Values of Q and K: 1.Q = k, The system is at equilibrium, no shift will occur. 2. Q > K, The ratio of initial concentrations of products to initial concentrations of reactants is too large. The system shifts to the left to reach equilibrium. 3. Q < K, The ratio of initial concentrations of products to initial concentrations of reactants is too small. The system shifts to the Right to reach equilibrium.

8. Question 3: The following results were collected for an experiment occurred in a 3.00 Liter flask involving the reaction at 600 °C for the decomposition of gaseous sulfur trioxide into gaseous sulfur dioxide and oxygen:

9. Answer: Q = 0.22 K= 0.08 Q > K Rx Shift to the LEFT

10. Le Châtelier’s Principle -If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. 1- Effect of Change in Concentration: - Alter the equilibrium position but they do not alter the equilibrium constant. * Upon addition of CO, Reaction Shifts to the left. No Change in the K. As 4 O 6(s) + 6C (s) As 4(g) + 6CO (g)

11. 2- Effect of Change in Temperature: For Exothermic reactions: (Treat Energy as a Product) For Endothermic reactions: (Treat Energy as a Reactant) N 2(g) + 3H 2(g) 2NH 3(g) + 92kJ An increase in the temperature will cause the equilibrium to shift to the left and the value of k to decrease. 556kJ + CaCO 3(s) CaO (s) + CO 2(g) An increase in the temperature will cause the equilibrium to shift to the right and the value of k to increase.

12. Q.4: What will happen to the number of moles of SO 3 in equilibrium with SO 2 and O 2 in each of the Following Cases: a)Oxygen Gas is added. -The Equilibrium Position will shift to the Left. b) The Temperature is Decreased. -The Equilibrium Position will shift to the Left. c) Gaseous Sulfur Dioxide is Removed. - The Equilibrium Position will shift to the Right.

13. Suggest 4 ways to increase the yield of methanol commercially : 1- Add more CO. 2- Add more H 2. 3- Continuously remove the methanol from the reaction mixture. 4- Lower the temperature of the reaction.