Approximations and Extent of Reaction

Transcription

1 Approximations and Extent of Reaction Model 1: Driving Forces for Equilibrium Processes Interpreting the Reaction Quotient Review: In an earlier activity, we examined the following system: Consider the following reaction. Suppose the initial concentrations of H 2 S, H 2 and S 2 are 1.00 M, 0.00 M, and 0.00 M, respectively: 2 H 2 S(g) 2 H 2 (g) + S 2 (g) K C = Determine the equilibrium concentrations of all species. 1. How will the concentrations of each species change (increase or decrease) to achieve equilibrium? 2. Do you expect a large or small change from the initial concentrations? Explain. 3. Without doing a calculation, will the relative change in the concentration of dihydrogen sulfide be a high or low percentage of the original concentration? The answer to CTQ#3 suggested we could make a simple approximation which made the calculation much simpler, as explored earlier Now, let's change the starting concentrations: New initial concentrations: H 2 S, H 2, and S 2 are 1.00 M, each. 4. Is the system at equilibrium? How can you tell? Approximations and the Extent of Reaction page 1 of 9

2 5. How will the concentration of the each species change (increase or decrease) to achieve equilibrium? 6. Given the values of K and the calculated value of Q, do you expect a large or small change in the initial concentrations? Explain. 7. From these initial concentrations, will the assumption "the change is small" be valid? If the concentrations change extensively, the determination of the equilibrium concentrations may involve more challenging calculational methods. A better understanding of the energy changes associated with a reaction may provide some insight into approaching the calculation. Model 2: Energy changes associated with reactions: Equilibrium reactions are related to the free energy changes of the process. The free energy of the process can be thought of as a combination of the relative potential energy of the components and the kinetic energies of the atoms in the molecules. A. Potential energy (enthalpy) Earlier, we examined the energy changes for chemical processes from the measurable heat flow which coincided with the reaction. For example, the combustion of methane, CH 4, proceeds by the following reaction: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) ΔH = 890 kj/mol 8. From the sign in the change in enthalpy, was energy absorbed or released in this reaction? Approximations and the Extent of Reaction page 2 of 9

3 9. If we describe reactions as the breaking and forming of bonds, are the stronger bonds on the reactant or the product side of the above methane reaction? 10. If bonds are a measure of potential energy of the molecules, which side of the methane combustion reaction is at lower potential energy? For the reaction 2 H 2 S(g) 2 H 2 (g) + S 2 (g), we can determine the enthalpy change from the enthalpies of formation. H 2 S(g) H 2 (g) S 2 (g) DH f (kj/mol) Confirm this reaction has a enthalpy change of kj/mol. 12. On which side of the reaction (reactants or products) are the bonds stronger? Approximations and the Extent of Reaction page 3 of 9

4 B. Kinetic energy As molecules move in space, they possess kinetic energy. As defined in the model for Kinetic Molecular Theory of a Gas, the average kinetic energy of the molecule's movement in space is directly related to the temperature (measured in Kelvin). K.E. = 3 2 R T The Kinetic Energy is associated with the molecules moving in space, but also the motions of the atoms within the molecule. It is the sum of: the energy of molecule moving in some direction (translation); the energy of the molecule tumbling in space (rotation); and, the energies of the atoms moving within the molecule vibrations stretching bonds and deforming the bond angles. The combination of the translations, vibrations, and rotations reflect the various ways the molecules distributes the kinetic energy present; chemists refer to these as the degrees of freedom of the molecules. While each individual atom's kinetic energy may be small, the energies add up. The kinetic energy of a mole of a simple monoatomic gas (e.g., He) at room temperature is approximately 3.7 kj; for water, the total kinetic energy12 kj Why is the value of the kinetic energy for water higher than that of Helium if they are at the same temperature? C. Gibbs energy When examining reactions and the drive toward equilibrium, both the changes in potential energy associated with the bonds broken and formed and changes in how the kinetic energy is distributed among translations, vibrations, and rotations are important. Either one or both can serve as a driving force to process. A new state function, the Gibbs Energy (G), is dependent on both potential and kinetic energies. As a reaction proceeds, the value of the Gibbs energy must decrease; once the reaction arrives at it's minimum energy, the system is at equilibrium. Qualitatively, we can graph the Gibbs energy as a function of extent of the reaction (symbol: ξ). According to the IUPAC, the Extent of reaction is an extensive quantity describing the progress of a chemical reaction. as the reaction proceeds from reactants to products. 1 In general terms, total kinetic energy is (½ RT) (number of atoms in molecule, i.e., degrees of freedom) the larger, more complex molecules have additional ways to distribute energy through vibrations and rotations. Approximations and the Extent of Reaction page 4 of 9

5 The graph of the Gibbs energy vs. the extent for the reaction N 2 O 4 (g) 2 NO 2 (g) starting from initial concentration of N 2 O 4 and NO 2 of 1.00 M and 0.00 M, respectively, at 298 K (0% Extent: all reactants) to complete conversion to NO 2 (100% extent: all products) is shown to the right: 14. From the information on the graph, is the reaction product favored or reactant favored? 15. Will the value of K be greater or less than 1 for this reaction? From an earlier activity (CA39), we examined the following data: Y(g) Z(g) Set Trial [Y] 0 [Z] 0 k Y k Z [Y] eq M Is this reaction Y(g) Z(g) product favored or reactant favored? 17. Sketch the graph of Gibbs energy vs. Extent. Approximations and the Extent of Reaction page 5 of 9

6 18. Complete the row for Trial 3. Where is the initial point of the third trial on the Gibbs energy vs. Extent curve from CTQ 17?: 19. Are the equilibrium concentrations of Y and Z affected by where you start on the graph of G vs. extent? Model 3: Using the Gibbs/extent curve with Equilibrium Calculations/Approximations Return to the initial problem in Model 1: Suppose the initial concentrations of H 2 S, H 2, and S 2 are 1.00 M, 1.00 M, and 1.00 M, respectively: 2 H 2 S(g) 2 H 2 (g) + S 2 (g) K C = We know from CTQ 4 7, for this initial set of concentrations: Q>>K; the concentrations will change significantly from these initial values; and, the simplifying assumption ( the change from the initial conditions is insignificant ) won't work. But, it doesn't matter where you start on the graph of G vs. extent you start, you end at the same place (the lowest value of G). The value of K indicates this system will favor the reactants at equilibrium, so applying that logic to this system, we can redefine the starting point for the calculations using simple stoichiometry: 2 H 2 S(g) 2 H 2 (g) + S 2 (g) K C = Initial 1.00 M 1.00M 1.00M Q >> K stochiometric reaction M 1.00 M 0.50 M H 2 is Limiting Reagent Redefined Initial 2.00 M 0.00 M 0.50M Q closer to K Approximations and the Extent of Reaction page 6 of 9

7 2 H 2 S(g) 2 H 2 (g) + S 2 (g) K C = Initial Q >> K Redefined Initial Q closer to K 20. Verify the values in the above table. 21. How does the value of K lead you to redefine the set of initial conditions? Using the redefined initial concentrations: 2 H 2 S(g) 2 H 2 (g) + S 2 (g) K C = Redefined Initial Change 2X +2X +X Equilibrium X 0 + 2X X 22. Do you expect a large or small change in these redefined initial concentrations? Explain. 23. From these redefined initial concentrations, will the assumption "the change is small" be valid? Approximations and the Extent of Reaction page 7 of 9

8 24. Make the appropriate approximation(s) from the redefined initial concentrations and substitute into the mass action expression. 25. Determine the equilibrium concentrations of all species. 26. Were the approximations valid (less than 5%)? Exercises 27. Many "insoluble" ionic solids are actually dissolve to a very small extent which can be described by an equilibrium constant; for example, silver chloride dissolves by the following equilibrium: AgCl(s) Ag + (aq) + Cl (aq) K SP = = [Ag + ][Cl ] How many grams of solid silver chloride can be prepared by the reaction of ml of M silver nitrate with ml of M calcium chloride? Calculate the concentration of all species remaining in solution after the precipitation is complete. Approximations and the Extent of Reaction page 8 of 9

9 Problems: 1. A solution is prepared by mixing moles of acetic acid (CH 3 CO 2 H) and moles of sodium acetate (NaCH 3 CO 2 ) in 750. ml of water. Acetic acid in water undergoes the following net ionic equilibrium: CH 3 CO 2 H(aq) + H 2 O(l) H 3 O + (aq) + CH 3 CO 2 (aq) K A = ⁵ Determine the concentration of all species in solution at equilibrium. 2. One of the equilibrium processes examined in an upcoming experiment is the formation of CoCl 4 ² ; the process is described by the following equilibrium: Co 2+ (aq) + 4 Cl (aq) CoCl 4 ² (aq) K f = If 2.00 mmoles of Co(NO 3 ) 2 (s) are dissolved in 150. ml of M sodium chloride, what are the equilibrium concentrations of all species present?