Equilibrium Practice Problems

Transcription

1 Equilibrium Practice Problems 1. Write the equilibrium expression for each of the following reactions: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 I 2 (s) + Cl 2 (g) 2 ICl (g) K = [ICl] 2 [Cl 2 ] NO 2 (g) NO (g) + ½ O 2 (g) K = [NO] [O 2 ] 1/2 [NO 2 ] 2. The dissociation of acetic acid, CH 3 COOH, has an equilibrium constant at 25 C of 1.8 x The reaction is CH 3 COOH (aq) CH 3 COO - (aq) + H + (aq) If the equilibrium concentration of CH 3 COOH is 0.46 moles in L of water and that of CH 3 COO - is 8.1 x 10-3 moles in the same L, calculate [H + ] for the reaction. K = [CH 3 COO - ] [H + ] Plug in known values and solve. [CH 3 COOH] 1.8 x 10-5 = [8.1 x 10-3 moles/0.500 L] [H + ] [0.46 moles/0.500 L] [H + ] = 1.0 x 10-3 M 3. Indicate the effect of a catalyst, pressure, temperature and concentration on each of the following on: catalyst pressure temperature concentration a. speed presence direct direct direct increase relationship relationship relationship b. pos n nothing based on moles based on inverse enthalpy of rxn relationship

2 4. Given the initial partial pressures of (P PCl5 ) = atm, (P PCl3 ) = atm, and (P Cl2) = atm at 250 C for the following reaction, what must each equilibrium partial pressure be? PCl 5 (g) PCl 3 (g) + Cl 2 (g) K p = 2.15 Determine reaction quotient, Q = x = Shift to products PCl 5 (g) PCl 3 (g) + Cl 2 (g) Initial Change -x +x +x Equil x x x 2.15 = (.150 +x)( x) so x = ( x) PCl 5 (g) PCl 3 (g) + Cl 2 (g) After x x x Substitute Confirm value of K with these values to check your answer. 5. The reaction has a value of K= 2400 at 2000 K. If 0.61 g of NO are put in a previously empty 3.00 L vessel, calculate the equilibrium concentrations of NO, N 2, and O 2. Initial 6.8 x Change -2x +x +x Equil 6.8 x x x x 2400 = [x] [x] so x = [6.8 x x] 2 Equil 6.8 x X X X Substitute

3 Does this make sense considering the value of K. 6. Using the same reaction as in #5, calculate the equilibrium concentrations of NO, N 2, and O 2 if the initial concentrations of each species are: [NO] = 0 M, [N 2 ] = M, [O 2 ] = M. Initial M M Change +2x -x -x Equil 2x M - x M - x 2400 = [0.850 M - x] [0.560 M - x] so x = [2x] 2 Equil 2x M - x M - x Substitute Does this make sense considering the value of K. 7. If the equilibrium constant for the following reaction is 0.10, determine the final concentration of ICl, if 4.0 moles of I 2 and Cl 2 were entered initially into an empty 1.0 liter flask. I 2 (g) + Cl 2 (g) 2 ICl (g) Initial 4.0 M 4.0 M 0 Change -x -x +2x Equil 4.0 x x 2x 0.10 = [2x] 2 so x = [4.0 x] 2 Equil (0.546) Substitute The following reaction has an equilibrium constant of 620 at a certain temperature. Calculate the equilibrium concentrations of all species if 4.5 mol of each component were added to a 3.0 L flask. H 2 (g) + F 2 (g) 2 HF (g) Determine molarity of solutions [4.5 mol / 3.0L ] = 1.5 M of all 3 solutions H 2 (g) + F 2 (g) 2 HF (g) Initial 1.5 M 1.5 M 1.5 M Change -x -x +2x

4 Equil 1.5 x 1.5 x x 620 = [1.5 +2x] 2 so x = 1.3 [1.5 x] 2 After 1.5 x 1.5 x x Substitute 0.2 M 0.2 M 4.1 M 9. Ammonia undergoes hydrolysis according to the following reaction: NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) K = 1.8 x 10-5 Calculate [NH 3 ], [NH 4 + ] and [OH - ] in a solution originally M NH 3. + NH 3 (aq) + H 2 O NH 4 (aq) + OH - (aq Initial M 0 0 Change -x +x +x Equil x x x 1.8 x 10-5 = [x] 2 so x =.0018 [0.200 x] Equil x x x Substitute M.002 M.002 M 10. The equilibrium constant for the following reaction is 600 C is 4.0. Initially, two moles of CO and one mole of H 2 O were mixed in a 1.0 liter container. Determine the concentration of all species at equilibrium. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) Initial 2.0 M 1.0 M 0 0 Change -x -x +x +x Equil 2.0 x x x x 4.0 = [x] 2 so x = 0.85 [1.0 x][2.0 x] Equil Substitute If 5.0 moles of O 2 and 4.0 moles of NO were entered into an empty 1.0 liter flask, calculate the equilibrium constant if the amount of NO 2 found at equilibrium was 1.5 moles. 2 NO (g) + O 2 (g) 2 NO 2 (g) Initial 4.0 M 5.0 M 0

5 Change -2x -x +2x Equil 4.0 2x x 2x Using 1.5 = 2x and therefore x = 0.75 M Equil Substitute K = [1.5] 2 = [2.5] 2 [4.25] 12. Two moles of NH 3 were entered in a 1.0 liter container at 650 C. At equilibrium only 71% of the original NH 3 was found. Determine the equilibrium constant of the following reaction. 2 NH 3 (g) N 2 (g) + 3 H 2 (g) Initial 2.0 M 0 0 Change -2x +x +3x Equil 2.0 2x +x +3x Using 2.0 2x = 2.0(71%) so x = 0.29 Equil 2.0 2(.29) (0.29) Substitute K = (0.29)(0.87) 3 = (1.42) The reaction of carbon disulfide with chlorine is as follows: CS 2 (g) + 3 Cl 2 (g) CCl 4 (g) + S 2 Cl 2 (g) + Heat H = -238 kj Predict the effect of the following changes to the system on the direction of equilibrium. a. The pressure on the system is doubled by halving the volume. b. CCl 4 is removed as it is generated. c. Heat is added to the system. 14. The reaction of nitrogen gas with hydloric acid is as follows: Heat + N 2 (g) + 6 HCl (g) 2 NH 3 (g) + 3 Cl 2 (g) H = 461 kj Predict the effect of the following changes to the system on the direction of equilibrium a. Triple the volume of the system. b. The amount of nitrogen is doubled. c. Heat is added to the system. Sometimes it s our fault that things are out of balance. (See below.)

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