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ICE Tables

ICE Tables. Reaction Quotient, Q. Q 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 .

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ICE Tables

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  1. ICE Tables

  2. Reaction Quotient, Q • Q 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, Q • K= 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 • 2 H2O⇌H3O+ + OH- • Shift +2y -y -y

  5. K vs Q Problem • For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 x 10-2. 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]o • Trial 1 1.0 x 10-31.0 x 10-52.0 x10-3 • Trial 2 2.0 x 10-41.5 x 10-3 3.54 x10-1 • Trial 3 1.0 x 10-45.0 x10-1 1.0 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) • Change • Equilibrium concentration (pressure)

  7. Gas Problem • At a certain temperature a 1.00 L flask initially contained 0.298 molPCl3, 8.70 x 10-3molPCl5, and no Cl2. After the system had reached equilibrium, 2.00 x 10-3mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to the following reaction • PCl5(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 • 2.0x10-3 M • It must have shifted to the right because chlorine increased. • x must = 2.0x10-3 I C E .0067 M .300 M

  9. K • K = .3 (.002) / .0067 • = .0896

  10. Concentration Problem • Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-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, 3.000 mol of each component was added to a 1.500-L flask. Calculate the equilibrium concentrations of all species.

  12. Quadratic Equation • To solve some problems you need to use the quadratic equation • or solver function on a calculator • For a x2 + bx + c = 0 • x = -b  b2 – 4ac • 2a

  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 choice • For some reactions the change will be very small compared to the initial amount. • You always have to check!

  15. Example • Gaseous NOCl decomposes to from the gases NO and Cl2. At 35o C, the K = 1.6 x10 -5. The initial conc. Of NOCl is .5 M. What is 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 .50-2x 2x x • 1.6 x10 -5= (2x)2 x / (.5-2x)2 • The algebra looks a little sticky on this problem • If 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 .50 -2x  .5 • .50 – 2 (.01) = .48 • If 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 x 10-10. • Calculate [I2]eq and [I]eq at 2000 K; Kc = 0.209.

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