2 Reaction Quotient, QQ is obtained by applying the law of mass action to INITIAL CONCENTRATIONS!Q is useful in determining which direction a reaction must shift to establish equilibrium.K vs. Q.K is calculated from equilibrium concentrations or pressures.Q is calculated from initial concentrations or pressures.
3 Reaction Quotient, QK = Q; The system is at equilibrium. No shift will occur.K < Q; The system shifts to the left.Consuming products and forming reactants, until equilibrium is achieved.K > Q; The system shifts to the right.Consuming reactants and forming products, to attain equilibrium.Write K then Q, the greater/less than sign points the way the reaction shifts!
4 Change Shift Right 2 H2O ⇌ H3O+ + OH- Shift -2y +y +y Shift Left
5 K vs Q ProblemFor the synthesis of ammonia at 500o C, the equilibrium constant is 6.0 x Predict the direction in which the system will shift to reach equilibrium in each of the following cases: N2(g) + 3 H2(g) ⇌ 2 NH3(g)Conc. (M) [NH3]o [N2]o [H2]oTrial x x x10-3Trial x x x10-1Trial x x x10-3
6 ICE tables To do equilibrium problems set up an ICE table. Write the balanced equation. Underneath it label three rows.Initial Concentrations (pressure)ChangeEquilibrium concentration (pressure)
7 Gas ProblemAt a certain temperature a 1.00 L flask initially contained mol PCl3, 8.70 x 10-3 mol PCl5, and no Cl2. After the system had reached equilibrium, 2.00 x 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the following reactionPCl5(g) ⇌ PCl3(g) + Cl2(g)Calculate the equilibrium concentrations of all species and the value of K.
8 Answer PCl5(g) PCl3(g) + Cl2(g) 8.7 x10-3 M .298 M 0 -x +x +x I It must have shifted to the right because chlorine increased.x must = 2.0x10-3ICE.0067 M.300 M
10 Concentration Problem Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask.
11 Concentration Problem Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 102 at a certain temperature. In a particular experiment, mol of each component was added to a L flask. Calculate the equilibrium concentrations of all species.
12 Quadratic EquationTo solve some problems you need to use the quadratic equationor solver function on a calculatorFor a x2 + bx + c = 0x = -b b2 – 4ac2a
13 Solver Function, Quadratic program Graphing calculators (required for this course) come with a solver function to solve equations.You can also have a quadratic equation program.All of this is legal for the AP or any of my tests.*except on multiple choice, but I have not seen that.
14 Simplified Assumptions If you are using a solver function this is unnecessary.You do have to understand the concept for a possible multiple choiceFor some reactions the change will be very small compared to the initial amount.You always have to check!
15 ExampleGaseous NOCl decomposes to from the gases NO and Cl2. At 35o C, the K = 1.6 x The initial conc. Of NOCl is .5 M. What are the equilibrium concentrations?2NOCl(g) ⇌ 2NO(g) + Cl2(g)The small K value means this will favor the reactant side so there wouldn’t be a large shift to the right.
16 Problem 2NOCl(g) ⇌ 2NO(g) + Cl2(g) I .50 0 0 C -2x +2x + x E x x x1.6 x10 -5 = (2x)2 x / (.5-2x)2The algebra looks a little sticky on this problemIf the shift is really small then.50 -2x .5
17 Solve now 1.6 x10 -5 = (2x)2 x / (.5)2 x = .01 Of course this is ONLY ACCEPTABLE IF x .5.50 – 2 (.01) = .48If the change is less than 5% it is considered small enough to ignore.48/.5 = 96% or a 4% change.
18 Solver Function This equation 1.6 x10 -5 = (2x)2 x / (.5-2x)2 is not difficult to solve using a solver function.You have to have a decent guess, you may have to change your guess.You canNOT have a negative value for your shift.
19 Simplified Assumptions In a study of halogen bond strengths, 0.50 mol I2 was heated in a 2.5 L vessel, and the following reaction occurred:I2(g) ⇌ 2I(g).Calculate [I2]eq and [I]eq at 600 K; Kc = 2.94 xCalculate [I2]eq and [I]eq at 2000 K; Kc =