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Predicting the Direction of Reaction

Predicting the Direction of Reaction. N 2 (g) + 3H 2 (g) 2NH 3 (g). K P (472°C)= (P NH3 ) 2 (P N2 )(P H2 ) 3.

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Predicting the Direction of Reaction

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  1. Predicting the Direction of Reaction N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 Starting with 1.00 mol N2, 2.00 mol H2, and 2.00 mol NH3 in a 1.00-L container at 472°C, which way will the reaction proceed? KP(472°C) = 2.79 x 10-5. Substituting these values into the equilibrium expression gives a value known as the reaction quotient Q. (2.00 mol RT)2 Q = 1.00 L2 = 124 2 (1.00 mol RT)(2.00 mol RT)3 62.0 1243 1.00 L 1.00 L3 R = 0.083145 L*bar/mol*K

  2. Predicting the Direction of Reaction Using the Reaction Quotient Q N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 Starting with 1.00 mol N2, 2.00 mol H2, and 2.00 mol NH3 in a 1.00-L container at 472°C, which way will the reaction proceed? Q = 1.30 x 10-4 KP(472°C) = 2.79 x 10-5 Q > KP This means there are more products than at equilibrium and that some of the NH3, therefore, will decompose to N2 and H2. When Q > K, the system will react to use up products and form reactants.

  3. Predicting the Direction of Reaction Using the Reaction Quotient Q When Q > K, the system will react to use up products and form reactants. When Q < K, the system will react to use up reactants and form products. When Q = K, the system is at equilibrium and the relative amounts of reactants and products will not change.

  4. Heterogeneous Equilibria PbCl2(s) Pb2+(aq) + 2Cl-(aq) K = [Pb2+] [Cl-]2 [PbCl2] Ksp = [Pb2+] [Cl-]2 What is the molarity (moles per volume) of a pure substance? Or, more accurately, what is the ratio of the concentration of the pure substance to the standard value? The ratio is constant and can be incorporated into K. This applies when the pure substance is a solid or a liquid. Ksp is called the solubility product constant.

  5. Heterogeneous Equilibria PbCl2(s) Pb2+(aq) + 2Cl-(aq) K = [Pb2+] [Cl-]2 [PbCl2] Ksp = [Pb2+] [Cl-]2 If a pure solid or liquid is involved in an equilibrium, its concentration is not included in the equilibrium expression.

  6. Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) What is the equilibrium expression for this reaction? KP = PCO2 Although CaCO3(s) and CaO(s) are not present in the equilibrium expression for the reaction, THEY MUST BE PRESENT IN THE REACTION SYSTEM AT EQUILIBRIUM FOR THE EQUILIBRIUM TO APPLY.

  7. Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) KP = PCO2 In any equilibrium, ALL substances must be present at equilibrium. A SOLVENT IS CONSIDERED A PURE SUBSTANCE.

  8. Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) KP = PCO2 1. Starting with only CaCO3, could equilibrium be reached? Yes. CaO and CO2 will form as CaCO3 decomposes. However, there must be enough CaCO3 to allow the equilibrium PCO2 to be reached.

  9. Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) KP = PCO2 2. Starting with CaO and PCO2 > KP, could equilibrium be reached? Yes. PCO2 > KP means there is more than the equilibrium amount of CO2 (Q > KP). Therefore some CO2 will react with the CaO to form CaCO3.

  10. Heterogeneous Equilibria CaCO3(s) CaO(s) + CO2(g) KP = PCO2 3. Starting with CaCO3 and PCO2 > KP, could equilibrium be reached? No. PCO2 > KP (Q > KP) means the reaction favors the formation of reactants and so no CaO will be formed. 4. Starting with CaCO3 and CaO, could equilibrium be reached? Yes.

  11. Calculation of Equilibrium Concentrations Calculate KC for the reaction NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) using the following information: Enough ammonia is dissolved in 5.00 L of water at 25°C to produce a solution that is 0.0124 M in ammonia. The solution is then allowed to come to equilibrium. At that time, the concentration of hydroxide is found to be 4.64 x 10-4 M. KC = [NH4+] [OH-] [NH3]

  12. Calculating Equilibrium Concentrations Enough ammonia is dissolved in 5.00 L of water at 25°C to produce a solution that is 0.0124 M in ammonia. The solution is then allowed to come to equilibrium. At that time, the concentration of hydroxide is found to be 4.64 x 10-4 M. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Initial: 0.0124 M 0 M 0 M Change: - 4.64 x 10-4 M 4.64 x 10-4 M 4.64 x 10-4M Equilibrium: 0.011936 M 4.64 x 10-4 M 4.64 x 10-4M KC = [NH4+] [OH-] = (4.64 x 10-4)2 = 1.80 x 10-5 [NH3] 0.011936

  13. Calculating Equilibrium Concentrations At 500°C an equilibrium mixture of gases in the Haber process was found to contain a hydrogen partial pressure of 0.928 bar and a nitrogen partial pressure of 0.432 bar. What is the partial pressure of ammonia? KP(500°C) = 1.45 x 10-5 N2(g) + 3H2(g) 2NH3(g) Equilibrium: 0.432 bar 0.928 bar x bar KP = (PNH3)2 = x2 = x2 (PN2)(PH2)3 (0.432)(0.928)3 0.34525 x2 = (0.34525)(1.45 x 10-5) = 5.006 x 10-6 x = √(5.006 x 10-6) = 2.24 x 10-3bar

  14. Calculating Equilibrium Concentrations A gas cylinder at 500 K is charged with phosphorus pentachloride at an initial pressure of 1.66 bar. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature? KP(500 K) = 0.497 PCl5(g) PCl3(g) + Cl2(g) Initial: 1.66 bar 0 bar 0 bar Change: -x bar x bar x bar Equilibrium: 1.66 - x bar x bar x bar KP = (PPCl3)(PCl2) = x2x2 = 0.497(1.66 – x) (PPCl5) (1.66-x) x2 + 0.497x – 0.82502 = 0 x = 0.693 bar

  15. Calculating Equilibrium Concentrations A gas cylinder at 500 K is charged with phosphorus pentachloride at an initial pressure of 1.66 bar. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature? KP(500 K) = 0.497 PCl5(g) PCl3(g) + Cl2(g) Equilibrium 0.97 bar 0.693 bar 0.693 bar Total pressure at equilibriumPtot = 0.97 + 0.693 + 0.693 = 2.35 bar

  16. Le Châtelier’s Principle The conclusions we draw from the size of the reaction quotient Q relative to the equilibrium constant K tell us which way the system will move to reach equilibrium. We are, in effect, using Le Châtelier’s Principle. Le Châtelier’s Principle If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.

  17. Le Châtelier’s Principle Changing reactant or product concentrations N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 How would adding nitrogen to the equilibrium mixture change the partial pressures of the gases? At equilibrium, Q = KP. Adding N2 would cause Q < KP. The equilibrium would shift to the right, using up some of the N2 and H2 and creating more NH3. So the partial pressure of H2 would decrease and the partial pressure of NH3 would increase. We would have to calculate the effect on the partial pressure of N2.

  18. Le Châtelier’s Principle Changing reactant or product concentrations N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 How would removing ammonia from the equilibrium mixture change the partial pressures of the gases? At equilibrium, Q = KP. Removing NH3 would cause Q < KP. The equilibrium would shift to the right, using up some of the N2 and H2 and creating more NH3. Removing product, in this case, is a way to generate more product.

  19. Le Châtelier’s Principle Changing reactant or product concentrations 2H+(aq) + 2CrO42-(aq) Cr2O72-(aq) + H2O(l) How would adding HCl(aq) affect this equilibrium? How would adding NaOH(aq) affect this equilibrium?

  20. Le Châtelier’s Principle Decreasing the volume of a system (this has the effect of increasing the pressure). N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 The reaction will shift toward the side with fewer moles of gas.

  21. Le Châtelier’s Principle Decreasing the volume of a system (this has the effect of increasing the pressure). N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3 In this reaction, the product side has fewer moles of gas. So decreasing the volume will increase PNH3. OR At equilibrium, Q = KP . If the volume were halved, the partial pressures would all double, resulting in Q < KP . The equilibrium would shift to the right.

  22. Le Châtelier’s Principle Adding an inert gas to a reaction system Since an inert gas does not participate in the reaction, the equilibrium will not change. The only thing that changes is the total pressure. N2(g) + 3H2(g) 2NH3(g) KP (472°C)= (PNH3)2 (PN2)(PH2)3

  23. Le Châtelier’s Principle Changing the temperature of a reaction system This is a change that affects the value of K itself. Increasing T effectively adds heat to the system. PCl5(g) PCl3(g) + Cl2(g)ΔH° = 87.9 kJ PCl5(g) + 87.9 kJ PCl3(g) + Cl2(g) If the reaction is endothermic, increasing T will increase K, shifting the equilibrium to the right.

  24. Le Châtelier’s Principle Changing the temperature of a reaction system This is a change that affects the value of K itself. Look up in Appendix C N2(g) + 3H2(g) 2NH3(g)ΔH°(25°C) = ? kJ N2(g) + 3H2(g) 2NH3(g) + 92.38 kJ If the reaction is exothermic, increasing T will decrease K, shifting the equilibrium to the left.

  25. Le Châtelier’s Principle The effect of a catalyst Catalysts change reaction rates, not equilibrium mixtures. The use of a catalyst will not change K or the equilibrium mixture.

  26. Integrating Concepts C(s) + H2O(g) CO(g) + H2(g) KP(800°C) = 14.1 • If we start with carbon and and 0.100 mol water vapor in a 1.00-L vessel, what will the partial pressures of the gases be at equilibrium? • What is the minimum amount of carbon needed to achieve equilibrium at 800°C? • What will the total pressure inside the vessel be at equilibrium? • At 25°C, KP for the reaction is 1.7 x 10-21. Is the reaction endothermic or exothermic? • To maximize yield at 800°C, should the volume of the system be increased or decreased?

  27. Integrating Concepts • If we start with carbon and and 0.100 mol water vapor in a 1.00-L vessel, what will the partial pressures of the gases be at equilibrium? C(s) + H2O(g) CO(g) + H2(g) Initial: present 0.100 mol 0 mol 0 mol Change: -x mol -x mol x mol x mol Equilibrium: ? 0.100-x mol x mol x mol KP (800°C) = 14.1 = (PCO)(PH2) = (xRT/V)2 = x2RT (PH2O) (0.100-x)RT/V (0.100-x)V 1.41 - 14.1x = x2(0.083145 L*bar/mol*K)(1073 K)/(1.00 L) 89.215x2+ 14.1x - 1.41 = 0

  28. Integrating Concepts • If we start with carbon and and 0.100 mol water vapor in a 1.00-L vessel, what will the partial pressures of the gases be at equilibrium? C(s) + H2O(g) CO(g) + H2(g) Initial: present 0.100 mol 0 mol 0 mol Change: -x mol -x mol x mol x mol Equilibrium: ? 0.100-x mol x mol x mol x = 0.0695 mol, -0.228 mol P(steam) = 2.72 bar P(CO) = P(H2) = 6.20 bar

  29. Integrating Concepts 2. What is the minimum amount of carbon needed to achieve equilibrium at 800°C? C(s) + H2O(g) CO(g) + H2(g) Equilibrium: present 0.0303 mol 0.0697 mol 0.0697 mol 0.0695 mol CO x 1 mol C x 12.01 g C = 0.835 g 1 mol CO 1 mol C There must be a little more than 0.835 g C present at the start of the reaction.

  30. Integrating Concepts 3. What will the total pressure inside the vessel be at equilibrium? C(s) + H2O(g) CO(g) + H2(g) Equilibrium: 0 bar2.72 bar6.20 bar 6.20 bar Ptot = 2.72 + 6.20 + 6.20 = 15.12 bar

  31. Integrating Concepts C(s) + H2O(g) CO(g) + H2(g) KP(800°C) = 14.1 4. At 25°C, KP for the reaction is 1.7 x 10-21. Is the reaction endothermic or exothermic? KP increases with increasing T, so the reaction is endothermic.

  32. Integrating Concepts C(s) + H2O(g) CO(g) + H2(g) KP(800°C) = 14.1 5. To maximize yield at 800°C, should the volume of the system be increased or decreased? Increasing the volume favors formation of products (because it decreases the partial pressures of all of the gases and causes Q < KP. The reaction shifts to the right.)

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