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There are essentially two types of problems which can be analyzed by ICE tables.

There are essentially two types of problems which can be analyzed by ICE tables. Type 1 A. The initial or equilibrium concentration of some substances must be determined.

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There are essentially two types of problems which can be analyzed by ICE tables.

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  1. There are essentially two types of problems which can be analyzed by ICE tables. Type 1 A. The initial or equilibrium concentration of some substances must be determined. B. Initial or equilibrium concentrations of some substances are given, but not both. Change is therefore treated as an unknown (x) C. The equilibrium constant is given. Type 2 A. The equilibrium constant or concentration must be determined. B. Initial and equilibrium concentrations of at least one substance are given so that change can be calculated directly. C. All other initial and equilibrium concentrations of substances are determined directly from the table.

  2. AType 1 problem was reviewed in the previous Powerpoint (Equilibrium 5 – ICE tables A). Here is a second example. Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF The equilibrium constant, Kc , for the reaction at 200 C is 1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium? ICE Table H2 + F2 → 2 HF Initial 2.00 2.00 0 Change -x -x 2x Equil (M) 2.00 -x 2.00 - x 2x

  3. ICE Table H2 + F2 → 2 HF Initial 2.00 2.00 0 Change -x -x 2x Equil (M) 2.00 -x 2.00 - x 2x Note certain characteristics of the ICE table: 1. Only initial concentrations of some substances are provided. The change and equilibrium values are therefore based on an unknown value (x). 2. Since the equilibrium constant is large (1.00 x 10 2), the value of x is significant relative to the initial concentration (2). The equilibrium concentration of the reactants cannot be simplified to approximately 2.

  4. Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF The equilibrium constant, Kc , for the reaction at 200 C is 1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium? H2 + F2 → 2 HF Initial 2.00 2.00 0 Change -x -x 2x Equil (M) 2.00 -x 2.00 - x 2x Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2 [ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2

  5. Hydrogen gas reacts with fluorine gas to produce hydrogen fluoride gas as shown: H2 + F2 → 2 HF The equilibrium constant, Kc , for the reaction at 200 C is 1.00 x 10 2. If 4.00 mol of each reactant are added to a 2.00 L flask, what would be the concentration of all three substancesat equilibrium? Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2 [ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2 The equation is a perfect square and can be easily simplified. 10 = 2x 20 – 10x = 2x 20 = 12x x = 1.67 2-x

  6. H2 + F2 → 2 HF Initial 2.00 2.00 0 Change -x -x 2x Equil (M) 2.00 -x 2.00 - x 2x Kc = 1.00 x 10 2 = [[ HF ] 2= [ 2x ] 2 = [ 2x ] 2 [ H2 ] [ F2 ] [2.0 – x] [2.0 – x] [2.00-x] 2 10 = 2x 20 – 10x = 2x 20 = 12x x = 1.67 2-x HF= 2(1.67) = 2.79 M H2 = 2-1.67 = 0.33 M F2 = 2-1.67 = 0.33 M

  7. Type 2 Problems A. The equilibrium constant or concentrations must be determined. B. Initial and equilibrium concentrations of some substances are both given. C. All other initial and equilibrium concentrations of substances are determined directly from the table. For the reaction: H2 + I2 → 2HI Initial concentrations of hydrogen and iodine were both 0.4 M. At equilibrium and 100 C, the concentration of hydrogen iodine was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium constant for the reaction?

  8. For the reaction: H2 + I2 → 2HI Initial concentrations of hydrogen and iodine were both 0.40 M At equilibrium and 100 C, the concentration of hydrogen iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium constant for the reaction? H2 + I2 → 2HI Initial 0.4 0.4 0 Change Equil 0.13

  9. H2 + I2 → 2HI Initial 0.4 0.4 0 Change Equil 0.13 Note certain characteristics of the ICE table. Although not complete, the table includes some values for initial concentrations and some values for equilibrium conditions. As a result, the table can be completed with no unknown values.

  10. Since HI went from 0 initially to 0.13 at equilibrium, the change for HI must be 0.13. H2 + I2 → 2HI Initial 0.4 0.4 0 Change 0.13 Equil 0.13

  11. The ratios of change for all substances in a reaction must match the coefficients in the balanced chemical equation. Since HI has a change of 0.13, the change of H2 and I2 is one-half of 0.13 or 0.065. H2 + I2 → 2HI Initial 0.4 0.4 0 Change - 0.065 - 0.0650.13 Equil 0.13

  12. The equilibrium concentrations equal the sum of the initial concentrations and the change. Since both H2 and I2 decreased by 0.065, the equilibrium concentrations must be 0.4-0.065 = 0.335 H2 + I2 → 2HI Initial 0.4 0.4 0 Change - 0.065 - 0.0650.13 Equil 0.335 0.335 0.13 Note that the table is completed without any unknown values. This is because partial information was available about both initial and equilibrium concentration.

  13. For the reaction: H2 + I2 → 2HI Initial concentrations of hydrogen and iodine were both 0.40 M At equilibrium and 100 C, the concentration of hydrogen iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium constant for the reaction? H2 + I2 → 2HI Initial 0.4 0.4 0 Change - 0.065 - 0.0650.13 Equil 0.335 0.3350.13 H2 = I2 = 0.335 M

  14. For the reaction: H2 + I2 → 2HI Initial concentrations of hydrogen and iodine were both 0.40 M At equilibrium and 100 C, the concentration of hydrogen iodide was 0.13 M. Determine the concentrations of hydrogen and iodine at equilibrium and determine the equilibrium constant for the reaction? H2 + I2 → 2HI Initial 0.4 0.4 0 Change- 0.065 - 0.0650.13 Equil 0.335 0.3350.13 H2 = I2 = 0.335 M Kc = [HI] 2 = [0.13] 2 = 0.15 [H2] [I2] [0.335] 0.335]

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