Presentation on theme: "Equilibrium Follow-up. Equilibrium Problems The Haber process proceeds as follows: 2NH 3 (g) + 92KJ ↔ N 2 (g) + 3H 2 (g) If the equilibrium concentrations."— Presentation transcript:
Equilibrium Problems The Haber process proceeds as follows: 2NH 3 (g) + 92KJ ↔ N 2 (g) + 3H 2 (g) If the equilibrium concentrations are: [NH 3 ] = 3.1x10 -2 M [N 2 ] = 8.5x10 -1 M [H 2 ] = 3.1x10 -3 M Then what is the value for K? Is the forward reaction favored? How can we increase the amount of product?
Answer K = [H 2 ] 3 [N 2 ] [NH 3 ] 2 K =(3.1x10 -3 ) 3 (8.5x10 -1 ) (3.1x10 -2 ) 2 K = 2.6x 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 NH 3, or removing H 2 or N 2.
K p And K c In the reaction at 25 o C : 2NO (g) + Cl 2(g) ↔ 2NOCl (g) The equilibrium pressures were found to be: P NOCl = 1.2atm P NO = 5.0x10 -2 atm P Cl2 = 3.0x10 -1 atm What is the K p and K c for this reaction?
Answer K p = P NOCl 2 (P NO ) 2 (P Cl2 ) K p = (5.0x10 -2 ) 2 (3.0x10 -1 ) K p = 1.9x10 3 K p = K c (RT) ∆n ∆n = 2-3 = -1 So K P = K c / RT or K c = K p (RT) K c = 1.9x10 3 (0.0821)(298) = 4.6x10 4
The Reaction Quotient It 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.
You try! Haber again! N 2 (g) + 3H 2 (g) ↔2NH 3 (g) Give the concentrations: [NH 3 ] = 1.0x10 -3 M [N 2 ] = 1.0x10 -5 M [H 2 ] = 2.0x10 -3 M Calculate the value for Q and predict which direction the equilibrium will shift if the K is 6.0x10 -2.
Answer Q = (1.0x10 -3 ) 2 (1.0x10 -5 )(2.0x10 -3 ) 3 Q = 1.3x10 7 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. N 2 (g) + 3H 2 (g) ←2NH 3 (g)
Calculating Equilibrium Pressures Lets start easy: In the reaction N 2 O 4 ↔ 2NO 2 has a K p value of If the pressure of N 2 O 4 at equilibrium is 2.71atm, then what is the Equilibrium pressure of NO 2 ? Tougher: A 1.00L flask initially contained 0.298mol of PCl 3 and 8.70x10 -3 mol of PCl 5. After the system reached equilibrium, 2.00x10 -3 mol of Cl 2 was found. Calculate K (c) and the Equilibrium concentrations. PCl 5 ↔ PCl 3 + Cl 2
Answers K p = P NO2 2 /P N2O = P NO2 2 / 2.71 P NO2 2 = P NO2 = 0.600
Answer Part 2 K = [Cl 2 ][PCl 3 ]/[PCl 5 ] Concentrations must be found first. [Cl 2 ] o = 0 [PCl 3 ] o = 0.298mol/1L = 0.298M [PCl 5 ] o = 8.70x10 -3 M Next we find the change to reach equilibrium. ∆[Cl 2 ] = +2.00x10 -3 M Since they all have coefficients of 1, then ∆[PCl 3 ] = +2.00x10 -3 M ∆[PCl 5 ] = -2.00x10 -3 M
Part 2 finish Final concentrations of gases: [Cl 2 ] = 2x10 -3 M [PCl 3 ] = 0.300M [PCl 5 ] = 6.70x10 -3 M K = (2x10 -3 )(0.300)/(6.70x10 -3 ) K = 8.96x10 -2
ICE By 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
ICE problem For the reaction: CO + H 2 O ↔ CO 2 + H 2 O The K value is Calculate the equilibrium concentrations if the initial concentrations of all components is 1M.
You try! H 2 + F 2 ↔ 2HF If the equilibrium Constant is 1.15x10 2 then if 3.00 moles of each substance were added to a 1.5L flask, what will the equilibrium concentrations be?
answer Must find Q first to know which way reaction will shift. Q = [HF] 2 /[H 2 ][F 2 ] All are 2M gas solutions so Q = 2 2 /(2)(2) = 1 Since Q is less then K reaction shifts right. [H 2 ][F 2 ][HF] I2M C-x +2x E2M-x 2M + 2x
Answer finish X = so: [H 2 ] = [F 2 ] = 0.472M [HF] = 5.056M
Small K values If 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.
You Try 2NOCl (g) ↔ 2NO (g) + Cl 2(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?
Answer Initial concentration of NOCl is 1mol/2L or 0.5M. 2NOCl ↔ 2NO + Cl 2 K = [NO] 2 [Cl 2 ]/[NOCl] 2 1.6x10 -5 = (2x) 2 (x)/(0.5-2x) 2 Since 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][Cl 2 ] I0.500 C-2x+2x+x E0.5-2x+2x+x
Answer Cont. So the equations becomes: 1.6x10 -5 = 4x 3 /(0.5) 2 X = 1x10 -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 -2 M and [Cl 2 ] = 1.0x10 -2 M
One More time with Haber! N 2 (g) + 3H 2 (g) ↔2NH 3 (g) If the equilibrium concentrations of N 2, H 2, and NH 3 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?
Answer First use the given equilibrium concentrations to find the value for K. K = 5.96x10 -2 Clearly since N 2 was added, then the reaction will shift right. 5.96x10 -2 = ( x) 2 /((1.399-x)( x) 3 ) [N 2 ][H 2 ][NH 3 ] I C-x-3x+2x E1.399-x x x