Chemistry 201. Working with K. NC State University. Lecture 11

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1 Chemistry 201 Lecture 11 Working with K NC State University

2 Working With K What is the relationship between pressure and concentration in K? How does one calculate K or components of K? How does one calculate equilibrium concentrations from initial concentrations? Text : Sections

3 Converting from pressure to molarity aa + bb gc

4 Converting from pressure to molarity aa + bb gc

5 Temperature-dependent equilibrium Example of Le Chatelier s principle Exothermic reaction: heat drives it to the left

6 Solving for a reactant/product pressure 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K = 245 at equilibrium : What is P(SO 3 )? P(SO 2 ) = 0.40 atm P(O 2 ) = 0.10 atm

7 Solving for a reactant/product 2 SO 2 (g) + O 2 (g) 2 SO 3 (g) K = 245 at equilibrium : What is P(SO 3 )? P(SO 2 ) = 0.40 atm P(O 2 ) = 0.10 atm

8 Solving for a reactant/product H 2 (g) + I 2 (g) 2 HI(g) A sealed container is filled with 0.60 atm of H 2 and 0.30 atm of I 2. When equilibrium is reached: P(HI) = 0.05 atm at 100 K. What is K?

9 Calculating an equilibrium constant H 2 (g) + I 2 (g) 2 HI(g) A sealed container is filled with 0.60 atm of H 2 and 0.30 atm of I 2. When equilibrium is reached: P(HI) = 0.05 atm at 100 K What is K? For this type of calculation we put in the fixed values of pressure using the stoichiometry.

10 Calculating an equilibrium constant H 2 (g) + I 2 (g) 2 HI(g) A sealed container is filled with 0.60 atm of H 2 and 0.30 atm of I 2. When equilibrium is reached: P(HI) = 0.05 atm at 100 K Construct a reaction table Species H 2 I 2 HI Initial Change -x -x 2x Equilibrium 0.6-x 0.3-x 2x

11 Calculating an equilibrium constant H 2 (g) + I 2 (g) 2 HI(g) A sealed container is filled with 0.60 atm of H 2 and 0.30 atm of I 2. When equilibrium is reached: P(HI) = 0.05 atm at 100 K Construct a reaction table Species H 2 I 2 HI Initial Change -x -x 2x Equilibrium 0.6-x 0.3-x 2x Note that 2x = 0.05 and therefore x = 0.025

12 Calculating an equilibrium constant H 2 (g) + I 2 (g) 2 HI(g) A sealed container is filled with 0.60 atm of H 2 and 0.30 atm of I 2. When equilibrium is reached: P(HI) = 0.05 atm at 100 K Construct a reaction table Species H 2 I 2 HI Initial Change Equilibrium Insert the appropriate values in the table.

13 Calculating an equilibrium constant H 2 (g) + I 2 (g) 2 HI(g) A sealed container contains atm of H 2, atm of I 2, and 0.05 atm of HI Calculate the equilibrium constant. Calculate K by inserting the values from the table into the equilibrium constant.

14 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K A sealed container is filled with 1.0 atm of I 2 and 2.0 atm of H 2. What are the equilibrium concentrations?

15 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K A sealed container is filled with 1.0 atm of I 2 and 2.0 atm of H 2. What are the equilibrium concentrations? Write the equilibrium constant. Make a reaction table. Species H 2 I 2 HI Initial Change -x -x 2x Equilibrium 2.0-x 1.0-x 2x

16 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Substitute the values

17 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Substitute the values Solve for x

18 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Substitute the values Solve for x

19 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Substitute the values Solve for x

20 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Substitute the values Solve for x

21 Determining equilibrium concentrations H 2 (g) + I 2 (g) 2 HI(g) K = 14 at 155 K Solve for x

22 Quantitative Le Chatelier s Principle N 2 O 4 (g) 2 NO 2 (g) At equilibrium : P(N 2 O 4 ) = 1.00 atm P(NO 2 ) = 0.33 atm in a fixed volume. The volume is doubled. Which way does the reaction proceed? What are the new equilibrium pressures?

23 Quantitative Le Chatelier s Principle At equilibrium : N 2 O 4 (g) 2 NO 2 (g) P(N 2 O 4 ) = 1.00 atm P(NO 2 ) = 0.33 atm The volume is doubled. Which way does the reaction proceed? What are the new equilibrium pressures? 1. Calculate equilibrium constant 2. Calculate total pressure at equilibrium 3. Calculate the new pressure after doubling 4. Use the mole fraction to solve for the new equilibrium pressure

24 Quantitative Le Chatelier s Principle N 2 O 4 (g) 2 NO 2 (g) At equilibrium : P(N 2 O 4 ) = 1.00 atm P(NO 2 ) = 0.33 atm

25 Quantitative Le Chatelier s Principle N 2 O 4 (g) 2 NO 2 (g) Double the volume. The total pressure decreases to 1.33/2 = 0.67 atm

26 Quantitative Le Chatelier s Principle N 2 O 4 (g) 2 NO 2 (g) Double the volume. The total pressure decreases to 1.33/2 = 0.67 atm The pressure of the reactant N 2 O 4 is

27 Phase transition equilibria

28 Phase transition equilibria H 2 O (l) H 2 O (g) The equilibrium of a liquid with its vapor is known as the vapor pressure. The vapor pressure is the equilibrium constant. We do not include the pure liquid in the equilibrium constant. The vapor pressure at the normal boiling point is 1 atm. Therefore, the K = 1 at the normal boiling point. Since This tells us that:

29 Understanding phase transitions Since the free energy of a phase transition is zero, we can conclude that there is the following relationship between the entropy and the enthalpy. This equation defines the entropy of phase transition.

30 Temperature dependence of the vapor pressure We have seen that the vapor pressure depends on temperature. We can determine that dependence from the van t Hoff equation, Now we use the fact that K is P for vapor pressure.

31 Colorimetric humidity sensor Cobalt chloride hexahydrate is a pink solid. Under vacuum it loses H 2 O to become anhydrous cobalt chloride. This equilibrium is given by Given the following free energies of formation calculate the vapor pressure of water above cobalt chloride hexahydrate. Molecule

32 Colorimetric humidity sensor The colorimetric reaction is In this case the stoichiometry for the balanced reaction requires that Inserting the value for D rxn G o from the previous slide and therefore we find that the equilibrium vapor pressure of P H2 O for this equlibrium is:

33 The reaction Colorimetric humidity sensor Can be used to make a colorimetric sensor. We can consider two important cases: 1. If the reaction is in a closed container, it will come to equilibrium, and the equilibrium vapor pressure of P H2 O will reach atm. This is not a sensor. 2. If this mixture is open to air, then it the balance of the equilibrium Will be shifted to the left if the vapor pressure, P H2 O, in the atmosphere is greater than This powder will be pink. However, if the vapor pressure is less than then the equilibrium will move to the right and the powder will be blue.

34 Definition of humidity Recall that the carrying capacity (i.e. the theoretical maximum vapor pressure) of the atmosphere is P H2 O = atm at 298 K. We have Discussed the fact that this vapor pressure can be calculated using The temperature dependence of the equilibrium constant: However, in a dry climate the atmosphere will not reach that value. There is not enough water around to get to the equilibrium value.

35 Equilibrium of a gas solution

36 Equilibrium of a gas solution The equilibrium constant for a gas above a solution is: This equilibrium is referred to as the Henry s law equilibrium. It is crucial for determining the concentration of O 2 in water, so that fish can breathe. It also accounts for the fact that CO 2 dissolves in the ocean. For example, K O = 774. What is the 2 concentration of O 2 in the ocean?

37 Heterogeneous equilibria CaCO 3 (s) CaO (s) + CO 2 (g) Calculate the equilibrium pressure of CO 2 in a closed container at 298 K. (D rxn H o = +178 kj/mol ; D rxn S o = +222 J/mol-K)

38 Heterogeneous equilibria CaCO 3 (s) CaO (s) + CO 2 (g) Calculate the equilibrium pressure of CO 2 in a closed container at 825 K. (D rxn H o = +178 kj/mol ; D rxn S o = 222 J/mol-K)

39 Solubility product equilibria When a solid (usually ionic) compound is in equilibrium with its ions in solution we refer to this as a solubility product. Since we do not include the solid (here MX) in the equilibrium constant, we have

40 Illustration of solvation of NaCl Illustration of the process of solution of a salt

41 Examples of enthalpy of solution (kj/mol) are tabulated below and refer to an infinitely dilute solution cation \anion OH - F - Cl - Br - I- CO 32 - NO 3 - SO 4 2- Li Na K NH Mg Ca Sr Al

42 Solution of ammonium nitrate Thermodynamic data We have seen that (NH 4 ) 2 NO 3 solvation is endothermic. Therefore, its solution must be entropy driven. Using these values estimate the equilibrium constant for solvation.

43 Solution of ammonium nitrate Thermodynamic data We have seen that (NH 4 ) 2 NO 3 solvation is endothermic. Therefore, its solution must be entropy driven. A. Using these values estimate the equilibrium constant for solvation.

44 Solution of ammonium nitrate Equilibrium constant calculation B. Estimate the solubility of (NH 4 ) 2 NO 3 based on these data.

45 Phase distribution equilibria Phase distribution equilibria are measurements of relative solubility. In the example shown on the right iodine is shown in carbon disulfide and water. The relative concentrations in the two solutions are given by: Example: Calculate the concentration of I 2 remaining in the aqueous phase after 50.0 ml of 0.10M I 2 in water is shaken with 10.0 ml of CS 2.

46 Phase distribution equilibria Let n 1 and n 2 represent the numbers of millimoles of solute in the water and CS 2 layers, respectively. K d can then be written as The number of moles of solute is (50 ml)(0.10 mmol/ml) = 5.0 mmol and mass conservation requires that n 1 + n 2 = 5.00 mmol, so n 2 = (5.00 n 1 ) mmol and we now have only the single unknown n 1.

47 Phase distribution equilibria The equilibrium constant then becomes Simplifying and solving for n 1 yields ( )n 1 / (0.02 m 1 ) = 650, with n 1 = mmol. The concentration of solute in the water layer is ( mmol)/(50 ml) = M, showing that almost all of the iodine has moved into the CS 2 layer.

48 Log P is a partitioning for the pharmaceutical industry

49 Log P partitioning Component log P OW T ( C) Acetamide [26] Methanol [27] Formic acid [28] Diethyl ether [27] p- Dichlorobenzene [2 9] Hexamethylbenze ne [29] 2,2',4,4',5- Pentachlorobiphe nyl [30] Ambient

50 Additional problem H 2 (g) + I 2 (g) 2 HI(g) K = 14 Start with 1.0 atm of H 2 How much I 2 needs to be added to form 1.5 atm of HI?

51 Additional problem H 2 (g) + I 2 (g) 2 HI(g) K = 14 Start with 1.0 atm of H 2 How much I 2 needs to be added to form 1.5 atm of HI? H 2 I 2 HI 1.0 x x

52 Additional problem H 2 (g) + I 2 (g) 2 HI(g) K = 14 How much I 2 needs to be added to form 1.5 atm of HI?

53 Additional problem N 2 O 4 (g) 2 NO 2 (g) K = 0.25 What pressure of NO 2 do you need to start with to form 1.0 atm of N 2 O 4?

54 Additional problem N 2 O 4 (g) 2 NO 2 (g) K = 0.25 What pressure of NO 2 do you need to start with to form 1.0 atm of N 2 O 4? N 2 O 4 NO 2 0 X x-2.0

55 Additional problem N 2 O 4 (g) 2 NO 2 (g) K = 0.25 What pressure of NO 2 do you need to start with to form1.0 atm of N 2 O 4? Answer: 2.5 atm

56 Additional problem N 2 O 4 (g) 2 NO 2 (g) K = 0.25 What pressure of NO 2 do you need to start with to form1.0 atm of N 2 O 4? We choose the answer 2.5 instead of 1.5 since the pressure of NO 2 would be negative for x = 1.5. The quadratic formula gives you two answers, but only one of them will be physically meaningful.

57 Goals Manipulate K with scale, direction and adding reactants or products. Calculate pressures or K from equilibrium data

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