AP Chem Chapter 12 Notes: Gaseous Equilibrium

Transcription

1 AP Chem Chapter 12 Notes: Gaseous Equilibrium Equilibrium I. Equilibrium is reached when both the and reactions are occurring at. A. Dynamic Equilibrium: reactions are still occurring but the of reactants are. Example: 2NO 2(g) N 2 O 4(g) B. When the reaction starts, the following happen simultaneously: 1. concentration 2. reaction rate 3. concentration 4. reaction rate C. As equilibrium is reached, the degree at which the (rate of the reaction). D. Once equilibrium is reached: 1. all 2. forward and reverse reactions are still occurring at a 3. all E. Equilibrium is shown using a :. Both reactions are being carried out. II. More about equilibrium A. Equilibrium can be reached from. B. If the reaction starts with all products,, or a, the ratio of products and reactants will remain. C. Many reactions are 1. a salt 2. Transfer of in 3. Transfer of in reactions. 4. Biology: binding of or molecules binding 5. Environment: transfer of between the atmosphere and biosphere or transfer of dissolved between atmosphere and hydrosphere Page 1 of 7

2 Equilibrium Constant ( ) I. Rate Laws (Example reaction: ) A. Rate law of forward rxn: B. Rate law of reverse rxn: C. at equilibrium,, therefore After rearranging: 3. The at a certain temperature is the at that temperature: II. For the reaction: A. the equilibrium constant expression is: B. Equilibrium expression notes: 1. Pure and are 2. Use or to write the expression (k eq ) a. for (all gases-use ) b. for [solutions-use ] c. and for d. for 3. In general, it s written as, but what is considered the products and reactants is. 4. Remember,!!! C. Examples: Write the equilibrium expressions for the following: Page 2 of 7

3 III. Calculating using the expression A. For every reaction, write the. B. What is? What are you? C. Plug in to find the unknown. D. Be careful with. E. Important notes: 1. K of the forward reaction is equal to the reciprocal of the reverse reaction. 2. If the coefficients are changed, the exponents change! 3. The K eq for a multistep reaction is the product of the constants of the individual steps!!! 4. Example: Given the reactions determine the value of K c for the reaction. IV. Relating k c to k p A. Recall: Concentration is or B. Use to relate them C. If of gaseous reactants and products (in the equation) and does not change, k c will be k p D. Plugging in pv= nrt into k c and k p, gives where Δn =(moles of gaseous ) - (moles of gaseous ) Page 3 of 7

4 V. Reaction quotient, q A. Obtained by plugging in into the expression. B. Used to predict the direction of the reaction. 1. if, the system is at 2. If, the reaction will shift from because the amount of products is too and the amount of reactants is too. 3. If, the reaction will shift from because the amount of reactants is too and the amount of products is too. C. At 448 C the equilibrium constant K c for the reaction is Predict in which direction the reaction will proceed to reach equilibrium at 448 C if we start with mol of HI, mol of H 2, and mol of I 2 in a 2.00-L container. D. Example: At 1000 K the value of K p for the reaction is Calculate the value for Q p, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures are VI. Finding equilibrium concentrations: A. Sometimes the given cannot be plugged into the equilibrium constant expression. B. The 1. i = ; c= ; e= 2. To find reactant is used or product is formed. 3. the change row maintains Page 4 of 7

5 C. Example: A closed system initially containing M H 2 and x 10 3 M I 2 at 448 C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is M. Calculate K c at 448 C for the reaction taking place, which is D. Sulfur trioxide decomposes at high temperature in a sealed container: 2 SO 3(g) 2 SO 2(g) + O 2(g) Initially, the vessel is charged at 1000 K with SO 3(g) at a partial pressure of atm. At equilibrium the SO 3 partial pressure is atm. Calculate the value of K p at 1000 K. E. A L flask is filled with mol of H 2 and mol of I 2 at 448 C. The value of the equilibrium constant Kc for the reaction belowat 448 C is What are the equilibrium concentrations of H 2, I 2, and HI in moles per liter? Page 5 of 7

6 Le Chatelier s Principle I. If a system is by a change in,, or, the system will position to the disturbance II. See-Saw model: A. Depending on what is done, the see-saw will be and the system will shift to a certain direction to equilibrium. B. Equilibrium shifts away from what is and towards what is. III. Example of change in concentration: 2 NO 2 (g) N 2 O 4 (g) H = kj Change to System Heavy Side of See-Saw Equil. Shift By... Add NO 2 Add N 2 O 4 Remove NO 2 IV. Example of change in temperature: 2 NO 2 (g) N 2 O 4 (g) H = kj Change to System Heavy Side of See-Saw Equil. Shift By... Decrease temp. Increase temp. A. Changing! This is because concentration and pressure are. B. Treat energy as a product or a reactant to determine equilibrium shift. V. Example of change in pressure: 2 NO 2 (g) N 2 O 4 (g) H = kj A. Only applies to!!! B. Increasing pressure favors the side with, so the equilibrium position shifts that direction (, producing N 2 O 4 and heat). Count of gas particles! C. Decreasing pressure favors side with (, producing NO 2 and absorbing heat). D. Volume has the since it is proportional to pressure. Page 6 of 7

7 V. Catalysts: lowers the of a reaction. A. This affects the. B. Therefore, adding a catalyst does not affect equilibrium position, thus will occur 2. But, the it takes to reach! VII. Inert gases: A. Inert means that it is, so they are in the reaction. B. They will not affect reaction so nothing will happen to the or. VIII. Addition of acids/bases: A. Strong acids and bases resulting in (or )and, respectively. B. These ions will if it is present in the equation. C. Example: 6m nitric acid is added to the following: NH 3 + H 2 O NH OH - 1. reacts with, causing the of it. 2. Reaction shifts to the, creating more. 3. Same goes for substances that result in. Page 7 of 7