Unit 1 Chem Mod 2 Chemical Equilibrium PDF

Unit1 Chem Mod2– Principles of Chemical Equilibrium Equilibrium is a condition when the backward and forward of an equilibrium reaction occur at the same rate. There are two types of equilibrium:- static and dynamic Static vs Dynamic equilibrium An example of static equilibrium is two people of e...

Unit1 Chem Mod2– Principles of Chemical Equilibrium Equilibrium is a condition when the backward and forward of an equilibrium reaction occur at the same rate. There are two types of equilibrium:- static and dynamic Static vs Dynamic equilibrium An example of static equilibrium is two people of equal weight on a seesaw. An example of dynamic equilibrium is a restaurant fully occupied with 200 people, however as people enter the restaurant, the same number is entering and thus the restaurant still contains 200 people. Characteristics of dynamic equilibrium 1. A closed system, where matter cannot enter or exit. 2. Forward and backward reactions occur at the same rate 3. On a macroscopic scale, nothing seems to be happening 4. On the microscopic scale, particles are in motion Principles of chemical equilibrium All equilibrium reactions have an equilibrium constant, where the ratio of reactants to products is a constant at a given temperature For example the general reaction aA +bB 🡪 cC + dD Kc = [C]c[D]d [A]a[B]b Kc applies to solution and subscript c represents concentration. A similar process occurs for Kp (p represents partial pressure as this applies to gases only, in Kc and Kp expressions solids are excluded!!!) e.g. 2SO2 + O2 🡪 2SO3 Kp = (ρSO3)2 (ρSO2)2 x ρO2 ALL EQUILIBRIUM REACTIONS ARE GOVERNED BY LE CHATELIER’S PRINCIPLE Le Chatelier’s Principle It states that if a system in equilibrium is subjected to a change (stressor), the system will try to counteract that change to re-establish equilibrium. In other words, the system will try to do the opposite of the change imposed upon it to restore equilibrium. Example: Consider the Haber Process N2 + 3H2 ↔ 2NH3 ΔH is –ve (meaning forward reaction is exothermic i.e. temperature increases) By lowering the temperature, the system will want to increase the temperature and to do this, the equilibrium would shift to the right i.e. the forward reaction (exothermic) will be favoured, this would also cause an increase in the yield of ammonia. By raising the temperature, the system will want to lower the temperature and the equilibrium would shift to the left and the endothermic reaction i.e. the backward reaction would be favoured. If pressure is increased, the system will favour the side which has the lower number of molecules i.e. to lower the pressure and which is the right side i.e. the equilibrium position shifts to the right. The reverse would occur if pressure was lowered. A similar explanation can be used for the Contact Process NB A catalyst does NOT shift the equilibrium position or constant, it only speeds up the rate at which equilibrium is reached! In other words, it speeds up both the forward and backward reaction. Only temperature can affect the equilibrium constant. Catalysts cause equilibrium to be reached more quickly but they do NOT affect the equilibrium constant Suppose you are given the following equilibrium: CO(g) + H2 O(g) CO2 (g) + H2 (g) Keq = 23.2 at 600 K If the initial amounts of CO and H2O were both 0.100 M, what will be the amounts of each reactant and product at equilibrium? For this type of problem, it is convenient to set a table showing the initial conditions, the change that has to take place to establish equilibrium and the final equilibrium conditions. Let's begin by showing the initial conditions: Initially, 0.100 M CO and 0.100 M H2O are present. Equilibrium hasn't been established yet, so the amounts of CO2 and H2 are assumed to be zero. To establish equilibrium, some CO and H2O has to react, so we will call the amount of CO and H2O reacted x, and the same x amount of CO2 and H2 must form: The amounts of reactants and products present at equilibrium will be the combination of the initial amounts and the change. Just add the quantities together: Substitute the above algebraic quantities into the mass action expression: Since the algebraic expression is a perfect square, begin solving for x by taking the square root of both sides of the equation: Multiply both sides by the denominator, 0.100 - x: Simplifying gives: 5.85x = 0.485 Solve for x by dividing both sides by 5.85: Recall that x represents the equilibrium quantities of both H2 and CO2. The equilibrium quantities of CO and H2O is given by: 0.100 - x = 0.100 - 0.0829 = 0.017 M = [CO] = [H2O] Please note For the questions following, if the volume given is NOT 1 dm3, you must FIRST convert the moles to molar concentration by dividing the # of moles by the volume (in dm3) to get the molar concentrations of the substances. Then you can proceed as normal with the calculations Worksheet Note L = 1 dm3 or 1000 cm3 1. 2. CO2 + H2 ↔ CO + H2O 3. H2 + I2 ↔ 2HI …………………………………………………………………………….………………………………… …………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. Checkpoint A 1. 2) Given the equation 4NH3 + 5O2 ↔ 4NO + 6H2O , give the equilibrium expression Kc for the reaction 3) i.e. purple colour is seen if the equilibrium position is on the left and colourless if the equilibrium position is on the right i) Would the equilibrium reaction shift to the left or to the right when more iodine is added? ……………………………. ii) Based on your answer from part i) what colour would be seen? ………………………….. iii) The forward reaction of the system is EXOTHERMIC. If the temperature of the system was increased, what would the system try to do? ………………………………………………………………….. iv) A catalyst would cause the forward reaction only to occur faster. True or False? ……….. b) 2NO2 (brown gas) N2O4 (colourless gas) ΔH = -ve Study the equilibrium reaction above and answer the questions below i) If the pressure is increased, what colour would most likely be seen? ……………………….. ii) If the temperature is decreased, would the backward or the forward reaction be favoured? ……………………………. iii) Write the equilibrium constant (K) for the reaction iv) If the equilibrium concentrations of NO2 & N2O4 are 0.4 and 0.2 mol dm-3 respectively, calculate the equilibrium constant and state its units Worksheet Remember set up your table as shown in the example to solve the problem! 1. A 2.00-L container contains 1.00 mole each of H2 and I2 gases. When the system reached equilibrium, the molar concentration of I2 is 0.11. The equilibrium equation is H2(g) + I2(g) 🡪 2 HI(g) What is the equilibrium constant? 2. A 1.00-L container contains 1.00 M of phosgene, which decomposes according to the reaction, COCl2(g) 🡪 CO(g) + Cl2(g). At equilibrium, the concentration of Cl2 is 0.028 M. What is the concentration of CO? 3. A 1.00-L container contains 1.00 M of phosgene, which decomposes according to the reaction, COCl2(g) 🡪 CO(g) + Cl2(g). At equilibrium, the concentration of Cl2 is 0.028 M. What is the equilibrium constant? 4. For the gas phase reaction H2(g) + I2(g) 🡪 2 HI(g)Kc = 50.3 at 731 K. Equal amounts (0.100 M each) is introduced to a container, and then the temperature is raised to 731 K. Calculate the value of K based on these concentration, is K equal to 50.3? If not, according to le Chatelier’s principle, what would the system try to do? 5. What is the Kp expression for the reaction …………………………………………………………………………….………………………………… …………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ……………………………………………………………………………. ………………………………………………