I. The Reaction Quotient 1. 2. We can write K only for a reaction at equilibrium How can we describe a reaction at other times? N2 + 3H2 2NH3 2 Q 3. 4. 5. [NH3 ]0 3 [N 2 ]0 [H 2 ]0 Describes the reaction when not at equilibrium Q = reaction quotient has the same form as K, but is at non-equilibrium times. Q can tell us how a reaction will change to get to equilibrium 1) If Q = K, we are at equilibrium 2) If Q > K, then products > reactants and the reaction will shift left 3) If Q < K, then reactants > products and the reaction will shift right Q K Predicted Direction of Reaction 0.55 1.45 To the right (towards products) 2.55 1.45 To the left (towards reactants) 1.45 1.45 No change (at equilibrium) Example: For the above reaction, K = 6 x 10-2. Predict change: 1) [N2]0 = 1 x 10-5 M, [H2]0 = 0.002 M, [NH3]0 = 0.001M 2) [N2]0 = 1.5 x 10-5 M, [H2]0 = 0.354 M, [NH3]0 = 0.0002M 3) [N2]0 = 5 M, [H2]0 = 0.01 M, [NH3]0 = 0.0001 M II. Solving Equilibrium Problems A. Problems where equilibrium concentrations (or pressures) are given 1. Example: N2O4(g) 2NO2(g) a) KP = 0.133 b) P(N2O4) at equilibrium = 2.71 atm c) Find P(NO2) at equilibrium 2. B. Example: PCl5(g) PCl3(g) + Cl2(g) a) [PCl3]0 = 0.298 M, [PCl5]0 = 0.0087 M b) At equilibrium, [Cl2]0 = 0.002 M c) Calculate all concentrations and K Procedure when not given equilibrium concentrations 1. Write the balanced equation 2. Write the equilibrium expression 3. List the initial conditions 4. Calculate Q to decide which way the reaction will shift 5. Write equilibrium conditions using x as unknown changes in concentration 6. Substitute the equilibrium conditions into the equilibrium expression and solve for x 7. Check your answer to see if it gives the correct value of K A. Example: 3 mol of hydrogen and 6 mol of fluorine are mixed in a 3 L flask to give hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. 1. Write the balanced equation: H2 + F2 2HF [HF] 2 K 115 [H 2 ][F2 ] 2. Write the equilibrium expression: 3. List the initial conditions: [H2]0 = 1 M, [F2]0 = 2 M, [HF]0 = 0 4. Calculate Q to decide which way the reaction will shift 2 [HF]0 02 Q 0 [H 2 ]0 [F2 ]0 1 x 2 5. Write equilibrium conditions using x as unknown changes in concentration Initial Change Equilibrium [H2] = 1 -x 1-x [F2] = 2 -x 2-x [HF] = 0 +2x 2x Shorthand: H2 Initial: 1 Change -x Equilibrium 1-x • + F2 2 -x 2-x 2HF 0 +2x 2x Substitute and solve for x [HF] 2 [2x] 2 4x 2 K 2 115 [H 2 ][F2 ] [1 - x][2 - x] x - 3x 2 4x 2 115x 2 345x 230 111x 2 345x 230 0 - b b 2 - 4ac Quadratic Equation : x 2a a = 111, b = -345, c = 230 x = 2.14 M or x = 0.968 M Reject x = 2.14 M since [H2] = 1 – x = 1 – 2.14 = -1.14 M • x = 0.968M Check your answer to see if it gives the correct value of K [HF]2 [2(0.968)] 2 K 113 [H 2 ][F2 ] [1 - 0.968][2 - 0.968] B. Example: CO(g) + H2O(g) CO2 + H2 K = 5.1 Find equilibrium conditions if 1 mol of each is mixed in a 1L flask. C. Example: 3 mol of hydrogen and 3 mol of fluorine are mixed in a with 3 mol hydrogen fluoride gas. K = 115. Find the equilibrium concentrations. H2 + F2 2HF D. Example: H2 + I2 2HI KP = 100. If we mix HI (P = 0.5 atm), H2 (P = 0.01 atm), and I2 (P = 0.005) in a 5 L flask, what are eq. conc.? E. Approximations when x is small 2NOCl(g) 2NO + Cl2(g) Initial Equil. 0.5 M 0.5 – 2x 0 2x 0 x 1.5 L flask K = 1.6 x 10-5 Since K is small, x will be very small. [NO]2 [Cl2 ] (2x) 2 (x) (2x) 2 (x) 4x 3 5 K 1.6x10 [NOCl]2 (0.5 x) 2 (0.5) 2 0.25 x = 1 x 10-2 [NO]2 [Cl 2 ] (2 x10 -2 ) 2 (1 x10 -2 ) 5 K 1.6x10 [NOCl]2 (0.5) 2 II. Le Chatelier’s Principle A. We can change the position of an equilibrium by changing conditions 1. N2 + 3H2 2NH3 2. Low temperature and high pressure favor the forward reaction Temp. (oC) 300 atm 400 atm 500 atm 400 48% NH3 55% NH3 61% NH3 500 26% NH3 32% NH3 38% NH3 600 13% NH3 17% NH3 21% NH3 Le Chatelier’s Principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance. Effect of change in concentrations N2 + 3H2 2NH3 K = 0.0596 1) At equilibrium, [N2] = 0.399 M, [H2] = 1.197 M, [NH3] = 0.202 M 2) Let’s add 1 M N2. How does the equilibrium shift? 3. B. [NH3 ]2 (0.202) 2 Q 0.017 K 3 2 [N 2 ][H 2 ] (1.399)(1.197) a) b) c) Too much N2 for equilibrium Reaction must adjust to get back to equilibrium Reaction must shift to the right Eq. shifts to the right 3. We could do the x calculation and find the new concentrations if we wanted, but sometimes you just want to know which way the reaction will shift. Le Chatelier’s principle lets us do that simply. 4. Another Example: What would happen if we add N2O4? 5. C. Example: As4O6(s) + 6C(s) As4(g) + 6CO(g) a) Add CO b) Add or remove As4O6(s) c) Remove As4(g) Effect of Change in Pressure/Volume 1. If we add/remove reactant/product, we change concentration (see above) 2. Adding an inert gas, doesn’t change concentrations, so no effect on equilibrium 3. Changing the size of the flask: V↓ = P↑ V↑ = P↓ 4. a) b) The volume changes, so the concentrations must change To re-establish equilibrium, the concentrations will adjust i. Fewer gas particles are favored at higher pressures ii. This reduces the pressure (Le Chat’s principle) c) N2(g) + 3H2(g) 2NH3(g) Increase the pressure i. There are 4 particles on the left and 2 on the right ii. To reduce the pressure, the equilibrium will shift right d) If we increased the volume and decreased the pressure, the reaction will shift to increase the number of particles (shift left) Example: Reduce the volume on the following reactions: a) P4(s) + 6Cl2(g) 4PCl3(l) b) PCl3(g) + Cl2(g) c) PCl3(g) + 3NH3(g) PCl5(g) P(NH2)3(g) + 3HCl(g) D. Effect of change in Temperature 1. The above changes (P, Concentration) effect the equilibrium position, but not the equilibrium constant K 2. Changing Temperature changes the equilibrium constant K 3. We can predict the changes by treating heat as a product or reactant of every reaction 4. 5. 6. Exothermic reactions produce heat as a product a) Adding heat (increasing temp.) shifts away from heat, to left, K decreases b) Removing heat (decreasing temp.) shifts towards heat, to right, K increases c) N2 + 3H2 2NH3 + 93 kJ/mol (DH = - 93 kJ/mol) d) Temperature (oC): 500 600 700 800 K: 90 3 0.3 0.04 Endothermic reactions require heat as a reactant a) Add heat = shift to right, K increases b) Remove heat = shift to left, K decreases c) 556 kJ/mol + CaCO3(s) CaO(s) + CO2(g) (DH = + 556 kJ/mol) Examples: N2(g) + O2(g) 2SO2(g) + O2(g) 2NO(g) DH = 181 kJ/mol 2SO3 DH = - 198 kJ/mol)