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N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g)

Chemical Equilibrium. N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g). A + B C + D. Many reactions do not go to completion - under the given conditions it is possible that not all of the reactants are consumed. Instead the extent of the reaction is determined by the equilibrium point .

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N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g)

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  1. Chemical Equilibrium N2(g) + 3 H2(g) --> 2 NH3(g)

  2. A + B C + D Many reactions do not go to completion - under the given conditions it is possible that not all of the reactants are consumed. Instead the extent of the reaction is determined by the equilibrium point. A + B --> C + D As the concentrations of C and D increase, C and D could react to form A and B - the REVERSE reaction. N2(g) + 3H2(g) 2NH3(g)

  3. The forward and reverse directions “oppose” one another. At some point in time, the rate of the forward reaction will equal the rate of the reverse reaction - this point corresponds to EQUILIBRIUM. Hence, when equilibrium has been reached, the concentration of A, B, C and D stay constant, as long as the conditions are held the same. At the equilibrium point: A and B combine to form C and D; C and D combine to form A and B; but both occur at the same rate. There is no NET change in the concentrations of A, B, C & D.

  4. Equilibrium When opposing forces acting on a system are equal in magnitude, the system is said to be in a state of equilibrium. A dynamic equilibrium is one at which changes to the system do occur at the microscopic level, but at the macroscopic level these changes are not observed. In general: Processes not at equilibrium will act or react to reach equilibrium.

  5. A B

  6. Characteristics of equilibrium 1) The attainment of equilibrium is spontaneous; i.e. it is a natural tendency 2) At equilibrium there is no macroscopic evidence of any changes in the system 3) A dynamic balance is established between opposing forces 4) Equilibrium is reached from either direction

  7. For a general reaction: a A + b B c C + d D the ratio: (concentration C)c (concentration D)d (concentration A)a (concentration B)b = K (constant) The Equilibrium Expression K is called the EQUILIBRIUM CONSTANT Note: K has a fixed value for a particular reaction and varies with temperature

  8. PCc PDd [C]c [D]d = K = K PAa PBb [A]a [B]b If all reactants and products are in solution, the relationship between the concentrations of all species at equilibrium is: If all reactants and products are gases, the relationship between the partial pressures of all gases at equilibrium is: Where [X] is the concentration (example molarity) of species X at equilibrium Homogenous reactions; reactants and products in the same phase

  9. [C]c PDd = K [A]a [B]b a A(aq) + b B(aq) c C(aq) + d D(g) Heterogeneous reaction: reactants and products are not in the same phase

  10. K is a dimensionless quantity. 2A(g) B(g) K = PB/P2A is actually PB/Pref K= (PA/Pref)2 Pref is set to 1 atm K is dimensionless For solutions, if concentration is M; [A]ref = 1M

  11. [H+] [CH3COO-] = K [CH3COOH] HCl(aq) H+(aq) + Cl-(aq) [H+] [Cl-] = K K ~ 107 at 25oC [HCl] CH3COOH(aq) H+(aq) + CH3COO-(aq) K ~ 10-5 at 25oC

  12. NH3(g) H2(g) + N2(g)

  13. 2 SO2(g)+ O2(g) 2SO3(g) P2SO3 = K 2 PO2 PSO2 The reaction of SO2(g) and O2(g) forming SO3(g) Equilibrium can be reached for different partial pressures of SO2, O2, and SO3, depending on the starting conditions, but at 25oC, the value of K is the same.

  14. a A + b B c C + d D Reactions that have K values > 1 are favored in the direction written; i.e. forward direction. Reactions that have K values < 1 are favored in the reverse direction Reactions for which K is near I have substantial amounts of both reactants and products when equilibrium is established. The Magnitude of the Equilibrium Constant The magnitude of the equilibrium constant reveals the extent to which the reaction will proceed in the desired direction.

  15. PCl5(g) PCl3(g) + Cl2 (g) K = 2.15 What are the equilibrium partial pressures of all three gases in a closed container containing only PCl5 at 0.100 atm and held at 250oC? According to the ideal gas laws, the partial pressures of gases is proportional to the number of moles of each gas, as long as the volume and temperature are kept fixed. The stoichiometry of this reaction is 1 : 1 : 1 Applying the Equilibrium Expression to Gas Phase Reactions

  16. Initial P (atm) 0.100 0 0 PCl5(g) PCl3(g) + Cl2 (g) Change in P (atm) -x x x Equilibrium P (atm) 0.100-x x x PPCl3 PCl2 = K PPCl5 If the partial pressure of PCl5 decreases by x at equilibrium, the partial pressures of PCl3 and Cl2 increases by x at equilibrium. At equilibrium:

  17. (x) (x) = 2.15 (0.100-x) x = (-b ± √b2 - 4ac) 2a x2 = 2.15 (0.100-x) x2 + 2.15x - 0.215 = 0 This is a quadratic equation of the form ax2+ bx + c = 0 and the solution of this equation is of the form

  18. Using this expression and solving for x, the roots of the equation are x = 0.0957 and -2.25 atm. At equilibrium, the partial pressures of Cl2 and PCl3 are 0.0957 atm, and that of PCl5 is (0.100 - 0.0957) = 0.004atm

  19. 4 NO2(g) N2O(g) + 3 O2 (g) The three gases are introduced into a container at partial pressures of 3.6 atm NO2, 5.1 atm N2O and 8.0atm O2 and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of NO2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.

  20. 4 NO2(g) 2 N2O(g) + 3 O2 (g) K = (5.7)2 (8.9)3 = 6.9 x 102 (2.4)4 Initial P (atm) 3.6 5.1 8.0 Change in P (atm) - x 2x/4 3x/4 Change in P (atm) - 4x 2x 3x Equilibrium P (atm) 2.4 5.1+2x 8.0+3x At equilibrium, the partial pressure of NO2 is 2.4 atm 3.6 - 4x = 2.4 => x = 0.3 atm Hence PN2O at equilibrium = 5.7 atm; PO2 = 8.9 atm

  21. 2 H2(g) + O2 (g) 2H2O(g) 2 2 PH2O PH2O = K1 PH2 PO2 PH2 PO2 2 2 2H2O(g) 2 H2(g) + O2 (g) = K2 -1 K1 = K2 In applying the equilibrium expression the following must be considered. 1) The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the corresponding forward reaction.

  22. H2(g) + O2 (g) H2O(g) 2 H2(g) + O2 (g) 2H2O(g) 2 PH2O PH2O = K1 PH2 PO2 2 K2 = √K1 1 2 = K2 PH2 PO2 1/2 2) When the coefficients in a balanced chemical equation are multiplied by a constant factor, the corresponding equilibrium constant is raised to the power equal to that factor.

  23. 2 BrCl(g) Cl2 (g) + Br2(g) PCl2 PBr2 = K1 = 0.45 at 25oC 2 PBrCl Br2(g) + I2 (g) 2 IBr(g) PIBr 2 = K2 = 0.051 at 25oC PBr2 PI2 3) When chemical equations are added or subtracted to obtain a net equation, the corresponding equilibrium constants are multiplied or divided to obtain the equilibrium constant of the net equation.

  24. 2 BrCl(g) + Br2 (g) + I2 (g) 2 IBr(g) + Cl2 (g) + Br2(g) 2 BrCl(g) + I2 (g) 2 IBr(g) + Cl2 (g) 2 PBrCl PCl2 PIBr 2 = K3 PI2 Adding the two chemical equations gives: Looking at the expressions for K1, K2 and K3 K1 K2 = K3 Hence, K3 = 0.023 at 25oC

  25. PA n [A] = = V R T Concentration of a species A in moles/lit = [A] The Ideal Gas Equation and Chemical Equilibrium For gaseous reactants or products, the concentration may be in moles/liter. The concentrations of the gases in moles/liter must be converted to partial pressures. This equation can be used to convert concentration of a gas in moles/lit to partial pressure of the gas.

  26. a A(g) + b B(g) c C(g) + d D(g) [C]c [D]d (a+b-c-d) R T = K ( ) [A]a [B]b Pref Hence, for a general gas phase reaction where Pref is the reference pressure = 1 atm and ensures that K is unitless.

  27. a A(g) + b B(g) c C(g) + d D(g) PCc PDd = Q PAa PBb Reaction Quotient Define the reaction quotient, Q: For the general reaction: where P is the partial pressure of a species at any point in time.

  28. a A(g) + b B(g) c C(g) + d D(g) If Q = K, the reaction is at equilibrium If Q ≠ K, the reaction is not at equilibrium. If Q > K , reaction proceeds from right to left If Q < K, reaction proceeds from left to right

  29. [CO] [H2 ]3 -2 R T = K ( ) [CH4] [H2O] Pref The equilibrium constant for the reaction CH4(g) + H2O(g) CO(g) + 3 H2(g) Equals 0.172 at 900K. The concentrations of H2(g), CO(g), and H2O(g) in an equilibrium mixture of gases all equal 0.00642 mol/L. Calculate the concentration of CH4(g) in the mixture, assuming that this is the only reaction taking place. [0.00642 mol/L] [0.00642 mol/L]3 [CH4] [0.00642 mol/L] = 0.172 [(0.08206 L atm mol-1) 900 K]-2 [CH4] = 0.00839 mol/L

  30. What happens if a system at equilibrium undergoes a change in conditions? The tendency of a system to achieve equilibrium is spontaneous. Once a system is at equilibrium it will remain at equilibrium. However, if conditions change the system will respond to this change in a way to achieve equilibrium again. Note: the concentrations of species when equilibrium is re-established need not be the same as the ones established at the previous equilibrium.

  31. N2(g) + 3H2(g) 2NH3(g)

  32. a A(g) + b B(g) c C(g) + d D(g) PCc PDd = Q PAa PBb LeChatelier’s Principle If a stress is applied to a system at equilibrium, the system tends to react so that the stress is minimized A) Changing the concentration of a reactant or product If reactant added: Q < K, reaction proceeds from left to right If product added: Q > K, reaction proceeds from right to left If a product is removed from the equilibrium mixture, Q also decreases, and the reaction once again proceeds to the right to increase the concentration of the product.

  33. N2(g) + 3H2(g) 2NH3(g)

  34. 2 NO2(g) N2O4(g) B) Changing the Volume Decreasing the volume, increases the total pressure of the reaction mixture. The reaction will then proceed in the direction which reduces the total pressure. If the volume is decreased, the above reaction will move to the right, to decrease the total number of molecules.

  35. a A(g) + b B(g) c C(g) + d D(g) + heat C) Changing Temperature The effect of changing the temperature of a system at equilibrium depends on whether the reaction proceeds by absorbing energy (endothermic) or by releasing energy (exothermic). An endothermic reaction lowers the temperature of the system and an exothermic reaction raises the temperature of the system. Forward reaction is exothermic; reverse reaction is endothermic. If a reaction is exothermic, raising the temperature causes the equilibrium to shift to the left.

  36. CaCO3(s) + SO3 (g) CaSO3 (s) + CO2(g) [CaSO3] PCO2 = K [CaCO3] PSO3 Hence, PCO2 = K PSO3 In all equilibrium expressions, the concentrations of all pure solids and liquids are set to 1. Heterogeneous Reactions

  37. In general, to write the equilibrium expression for a reaction 1) Concentration of gases are expressed as partial pressures 2) Concentration of dissolved species in solution are expressed as moles/liter 3) Concentrations of pure solids and pure liquids are set to 1 (for a solvent taking part in a reaction, its concentration is also set to 1 providing the solution is dilute)

  38. Problem: The reaction between Ni(s) and CO(g) to form Ni(CO)4(g) is as follows: Ni(s) + 4CO(g) Ni(CO)4(g) A quantity of Ni(s) is added to a vessel containing CO at a partial pressure of 1.282 atm and 354 K. At the equilibrium point of this reaction, the partial pressure of CO is 0.709 atm. Calculate the equilibrium constant of this reaction at 354 K. PNiCO4 4 K= PCO Ni(s) + 4CO(g) Ni(CO)4(g) Initial P (atm) 1.282 0 Change in P (atm) 0.573 0.143 Equilibrium P (atm) 0.709 0.143

  39. 0.143 = 0.567 K= (0.709)4 PNiCO4 4 K= PCO

  40. Applications of Chemical Equilibria Product yields can be increased by adjusting conditions N2(g) + 3H2(g) 2NH3(g) + heat Gas phase reaction: 4 moles of gas --> 2 moles of gas Exothermic reaction Reaction conditions that favor NH3(g) production high pressure (~ 250 atm) low temperature (use a catalyst)

  41. Hb(aq) + O2(g) HbO2 (aq) [HbO2] KO2= [Hb] PO2 Hemoglobin/O2 Equilibrium Hemoglobin (Hb) carries oxygen from the lungs to the body tissue, transporting oxygen from a region of high concentration to low concentration. The oxygen-hemoglobin complex, oxyhemoglobin (HBO2) transports O2 Level of O2 in blood is increased by ~ 70 times because of hemoglobin

  42. Hb(aq) + O2(g) HbO2 (aq) Because of the formation of the HbO2 complex, the amount of O2 in blood is increased by a factor of 70. LeChatelier’s principle predicts that in regions of high O2 partial pressure, the Hb-HbO2 equilibrium is shifted to the right, which is the case in the lungs In regions of low O2 partial pressure, the equilibrium shifts to the left, resulting in a breakup of the HbO2 complex, releasing O2 to the body’s tissues.

  43. Hb(aq) + CO(g) HbCO(aq) [HbCO] KCO= [HbO2] KO2= [Hb] [Hb] PCO PO2 Why is CO lethal?

  44. HbO2 (aq) + CO(g) HbCO(aq) + O2 (g) [HbCO] PO2 Kcompetition = [HbO2] PCO [HbCO] PCO [HbO2] KCO KCO = Kcompetition = = PO2 KO2 KO2 When Hb is exposed to both O2 and CO, there is competition for the Hb, and the following reaction takes place: The Kcompetition for this reaction is:

  45. HbO2 (aq) + CO(g) HbCO(aq) + O2 (g) Since KCO > KO2 Kcompetition is >1 At 38oC, the value of Kcompetition is 210 strongly favoring the formation of the HbCO complex CO displaces O2 from the HbO2 complex, resulting in asphyxiation. The process is reversible - from LeChatelier’s principle, a large partial pressure of O2 will shift the reaction above from right to left.

  46. A solute dissolved in a solvent A can be extracted using another solvent B. Condition: The solute must dissolve in both solvents A & B and solvent B must be immiscible with the solvent A. CCl4 and H2O are immiscible. I2(s) dissolves in both solvents. Extraction and Separation

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