2Equilibrium Problems The Haber process proceeds as follows: 2NH3 (g) + 92KJ ↔ N2 (g) + 3H2 (g)If the equilibrium concentrations are:[NH3] = 3.1x10-2M[N2] = 8.5x10-1M[H2] = 3.1x10-3MThen what is the value for K?Is the forward reaction favored?How can we increase the amount of product?
3Answer K = [H2]3[N2] [NH3]2 K = (3.1x10-3)3 (8.5x10-1) (3.1x10-2)2 -No it is not favored, the K value demonstrates that the reactant is more likely to be present at equilibrium than the products.-Product can be increased by increasing the temperature, decreasing the pressure, increasing the amount of NH3, or removing H2 or N2.
4Kp And Kc In the reaction at 25oC : 2NO(g) + Cl2(g) ↔ 2NOCl (g)The equilibrium pressures were found to be:PNOCl = 1.2atmPNO = 5.0x10-2atmPCl2 = 3.0x10-1atmWhat is the Kp and Kc for this reaction?
6The Reaction QuotientIt is possible to measure the concentrations of the reactants and products in a reaction vessel at any time.They may not necessarily be at equilibrium yet.When plugging those values into the Law of Mass Action, you will still get a value, but it might not be K.We call this value the reaction quotient, and comparing this value to K, we can see which direction the reaction is favored to move.
7You try! Haber again! N2 (g) + 3H2 (g) ↔ 2NH3 (g) Give the concentrations:[NH3] = 1.0x10-3M[N2] = 1.0x10-5M[H2] = 2.0x10-3MCalculate the value for Q and predict which direction the equilibrium will shift if the K is 6.0x10-2.
8Answer Q = (1.0x10-3)2 (1.0x10-5)(2.0x10-3)3 Q = 1.3x107 Since K = 6.0x10-2 and Q is significantly higher, then the concentrations of the products needs to go down, and reactants up meaning the reaction will shift left.N2 (g) + 3H2 (g) ← 2NH3 (g)
9Calculating Equilibrium Pressures Lets start easy:In the reaction N2O4 ↔ 2NO2 has a Kp value of If the pressure of N2O4 at equilibrium is 2.71atm, then what is the Equilibrium pressure of NO2?Tougher:A 1.00L flask initially contained 0.298mol of PCl3 and 8.70x10-3 mol of PCl5. After the system reached equilibrium, 2.00x10-3 mol of Cl2 was found. Calculate K(c) and the Equilibrium concentrations.PCl5 ↔ PCl3 + Cl2
11Answer Part 2 K = [Cl2][PCl3]/[PCl5] Concentrations must be found first.[Cl2]o = 0[PCl3]o = 0.298mol/1L = 0.298M[PCl5]o = 8.70x10-3MNext we find the change to reach equilibrium.∆[Cl2] = +2.00x10-3MSince they all have coefficients of 1, then∆[PCl3] = +2.00x10-3M∆[PCl5] = -2.00x10-3M
12Part 2 finish Final concentrations of gases: [Cl2] = 2x10-3M [PCl3] = 0.300M[PCl5] = 6.70x10-3MK = (2x10-3)(0.300)/(6.70x10-3)K = 8.96x10-2
13ICEBy knowing starting concentrations and K values, it is possible to determine the concentrations once equilibrium is reached.It is based on the stoichiometry of the problem.“I” stands for initial concentration“C” stands for change in concentration“E” stands for equilibrium concentration
14ICE problem For the reaction: CO + H2O ↔ CO2 + H2O The K value is Calculate the equilibrium concentrations if the initial concentrations of all components is 1M.
16You try!H2 + F2 ↔ 2HFIf the equilibrium Constant is 1.15x102 then if 3.00 moles of each substance were added to a 1.5L flask, what will the equilibrium concentrations be?
17answer Must find Q first to know which way reaction will shift. Q = [HF]2/[H2][F2]All are 2M gas solutions soQ = 22/(2)(2) = 1Since Q is less then K reaction shifts right.[H2][F2][HF]I2MC-x+2xE2M-x2M + 2x
19Small K valuesIf K is very small, and concentrations of substances are large, then the change is insignificant.X can be dropped at that point from the change of the reactants which makes the problem easier to solve.
20You Try 2NOCl(g) ↔ 2NO(g) + Cl2(g) If the K is 1.6x10-5, what will the equilibrium concentrations be if 1.0mol of NOCl is added to an empty 2.0L flask?
21Answer Initial concentration of NOCl is 1mol/2L or 0.5M. 2NOCl ↔ 2NO + Cl2K = [NO]2[Cl2]/[NOCl]21.6x10-5= (2x)2(x)/(0.5-2x)2Since K is to the -5 and the starting concentration is 4 powers of ten greater, then we can ignore the change in the reactant concentration.[NOCl][NO][Cl2]I0.5C-2x+2x+xE0.5-2x
22Answer Cont. So the equations becomes: 1.6x10-5 = 4x3/(0.5)2 As long as the change in concentration is smaller than 5%, it is valid to ignore the change.Plugging the change into the law of mass action and demonstrating that the K hasn’t changed is also a valid way to check.So the final concentrations are [NOCl] = 0.5M, [NO] = 2.0x10-2M and [Cl2] = 1.0x10-2M
23One More time with Haber! N2 (g) + 3H2 (g) ↔ 2NH3 (g)If the equilibrium concentrations of N2, H2, and NH3 were .399M, 1.197M, and 0.202M respectively, then what would the new equilibrium expression be under the same conditions if enough nitrogen gas was added to the system in order to raise the concentration of nitrogen to 1.399M?
24AnswerFirst use the given equilibrium concentrations to find the value for K.K = 5.96x10-2Clearly since N2 was added, then the reaction will shift right.5.96x10-2 = ( x)2/((1.399-x)( x)3)[N2][H2][NH3]I1.3991.1970.202C-x-3x+2xE1.399-xxx