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Reaction Rates & Equilibrium

Reaction Rates & Equilibrium. Rates of Reaction (rxn.) Reversible Reactions and Equilibrium Determining Whether a Reaction Will Occur Calculating Entropy and Free Energy The Progress of Chemical Reactions. Chemical Equilibria. Reactions going to completion ( ↓ )

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Reaction Rates & Equilibrium

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  1. Reaction Rates & Equilibrium Rates of Reaction (rxn.) Reversible Reactions and Equilibrium Determining Whether a Reaction Will Occur Calculating Entropy and Free Energy The Progress of Chemical Reactions

  2. Chemical Equilibria Reactions going to completion ( ↓ ) All reactants are converted to products Yields arrow; full-headed arrow in a single direction Equilibrium ( ∏ ) The reaction reaches a steady-state of forward and reverse reactions Products and Reactants coexist in a closed system Equally sized, half-headed arrows heading in opposite directions

  3. Chemical Equilibria Reaction Quotient (Q) the relative proportion of products to reactants at any given point during a rxn., at times other than, but including equilibrium. aA + bB ∏ cC + dD Q can be solved using the molarities (M) of the reactants and products present. Note: Q is without unit. Molarities numeric values should be inputted without units.

  4. Chemical Equilibria Equilibrium Constant (Keq, Kc, sometimes Q) the reaction quotient (Q) at equilibrium for a specific rxn. basically a very specific reaction quotient. dependent on temperature, pressure, and volume. Keq can be solved using the molarities (M) of the reactants and products present at equilibrium. Note: Keq is without unit.

  5. Chemical Equilibria Equilibrium Constant (Keq) If Keq is greater than 1 ( Keq› 1 ) then products are favored. Concentrations of products are greater than reactants. More products are present! If Keq is less than 1 ( Keq‹ 1 ) then reactants are favored. Concentrations of reactants are greater than products. More reactants are present! Equilibrium Constant (Keq) vs. Reaction Quotient (Q) If Keq > Q : The rxn. will shift to the right If Keq < Q  : The rxn. will shift to the left If Keq = Q  : The rxn. is at equilibrium

  6. Calculating Keq Given the balanced equation and conc. at equilibrium: CO2 + CF4∏ 2COF2 Equilibrium conc. CO2: 0.116MCF4: 0.0844MCOF2: 0.305M What is the Keq?

  7. Calculating Keq First, determine the proper Keq equation. Second, put things where they belong and then solve: Products in numerator; on top Coefficients become exponents! Reactants in denominator; on bottom

  8. Calculating Keq Given the balanced equation and conc. at equilibrium: CO2 + CF4∏ 2COF2 Equilibrium conc. CO2: 0.116MCF4: 0.0844MCOF2: 0.305M What is the Keq? 9.50 × 100

  9. Calculating Keq - Advanced Given the balanced equation, initial conc., and one (1) conc. at equilibrium: 2NO2∏ NO3 + NO Initial conc.Equilibrium conc. NO2: 0.100M NO2: 0. 0370M NO3: 0.000M NO3: ???? M NO: 0.000M NO: ???? M What is the Keq?

  10. Calculating Keq - Advanced First, determine how much reactant was used… this will tell you how much reactant contributed to the products. In this case 0.100M – 0.037M = 0.0630 M NO2 USED Products are only produced from something—it’s not magic, it’s chemistry. 0.0630 M NO2 USED to make these, but how much? conc. initially: 0.100M 0.00M 0.00M conc. equilibrium: 0.037M 2NO2∏ NO3 + NO

  11. Calculating Keq - Advanced Second, use that value to determine how much product was produced or how much of other reactants were used. You will use the mole ratios from the balanced equation. conc. initially: 0.100M 0.00M 0.00M 2NO2∏ NO3 + NO conc. equilibrium: 0.037M 0.0315M 0.0315M 0.0630 mol NO2 1 mol NO3 1 L 2 mol NO2 = 0.0315 M NO3 0.0630 mol NO2 1 mol NO 1 L 2 mol NO2 = 0.0315 M NO

  12. Calculating Keq - Advanced Third, determine the proper Keq equation. Then, put things where they belong and solve: Products in numerator; on top Coefficients become exponents! Reactants in denominator; on bottom

  13. Calculating Keq - Advanced Given the balanced equation, initial conc., and one (1) conc. at equilibrium: 2NO2∏ NO3 + NO Equilibrium conc. NO2: 0.0370M NO3: 0.0315M NO: 0.0315M What is the Keq? 7.25 × 10-1

  14. Calculating an Equilibrium Concentration Given the balanced equation, the Keq value and the partial set of conc. at equilibrium: PCl3 + Cl2∏ PCl5 Equilibrium conc.Keq PCl3: 0.0482M 1.81 × 101Cl2: ????MPCl5: 0.400M What is the conc. Cl2?

  15. Calculating an Equilibrium Concentration • First, determine the proper Keq equation: • Second, isolate or solve for the desired variable, before you put in numbers! Products in numerator; on top Reactants in denominator; on bottom

  16. Calculating an Equilibrium Concentration Third, put things where they belong and then solve: [Cl2] =4.58 × 10-1

  17. The LeChâtelier-Braun Principle Equilibria are dynamic and so, controllable. If you change the conditions then you are changing the rxn. Equilibria shifts (making more reactants or more products) occur in response to changes in volume, pressure, conc., and temperature as a method of partially counteracting the change.

  18. The LeChâtelier-Braun Principle Note: Solids and liquids can do very little to affect pressure and volume of a closed system—they only take up a particular amount of space. Gases however fill whatever container you put them in. ConditionShift ↑ Pressure Toward the lower total pressure ↑ Volume Toward higher gaseous volume Remember Boyle’s Law (P1V1=P2V2). Pressure and volume are inversely related.

  19. The LeChâtelier-Braun Principle is heat needed for the rxn. or produced by the rxn. ConditionShift ↑ Conc. Toward lower conc. reagents ↑ Temperature Depends… Endothermic rxn.: Since heat is required as a reactant an increase would shift the rxn. towards products. Exothermic rxn.: Since heat is produced as a product an increase would shift the rxn. towards reactants.

  20. The LeChâtelier-Braun Principle Assume that the reaction below is at equilibrium. A stress is then applied as described. How will the system respond to the stress. CS2(g) + 4H2(g)∏ CH4(g) + 2H2S(g) H=10KJ The pressure is decreased. The reaction will shift: To the LEFT because there is now more room. Increased pressure forces gases to minimize the space they occupy. The above reactants take up a 5 moles of space, the products 3 moles.

  21. The LeChâtelier-Braun Principle Assume that the reaction below is at equilibrium. A stress is then applied as described. How will the system respond to the stress. CS2(g) + 4H2(g)∏ CH4(g) + 2H2S(g) H=10KJ The volume is decreased. The reaction will shift: To the RIGHT because there is less room. Decreased volume forces gases to minimize the space they occupy. The above reactants take up a 5 moles of space, the products 3 moles.

  22. The LeChâtelier-Braun Principle Assume that the reaction below is at equilibrium. A stress is then applied as described. How will the system respond to the stress. CS2(g) + 4H2(g)∏ CH4(g) + 2H2S(g) H=10KJ The H2 is removed. The reaction will shift: To the LEFT because when reactants are removed the balance of atoms on each side of the rxn. is upset. To readjust more reactants must be made.

  23. The LeChâtelier-Braun Principle Assume that the reaction below is at equilibrium. A stress is then applied as described. How will the system respond to the stress. CS2(g) + 4H2(g)∏ CH4(g) + 2H2S(g) H=10KJ The temperature is increased. The reaction will shift: To the RIGHT because heat (enthalpy) is a reactant; adding more of any reactant will shift the reactants toward the products.

  24. The LeChâtelier-Braun Principle & Keq Remember Rxn. Quotient (Q)? Of course you do. It is the conc. of each substance you have at any point during rxn.; conc. at times other than, but including, equilibrium. Comparing the rxn. quotient (Q) to the equilibrium constant (Keq) is a quantitative method of applying the LeChâtelier-Braun principle.

  25. The LeChâtelier-Braun Principle & Keq Rxn. Quotient (Q) can be solved for just like Keq which is why we can compare them. If a Q value is larger than the Keq value for a given rxn. it means that the conc. of products is greater than expected at equilibrium. The rxn. will shift LEFT to make more reactants before reaching equilibrium

  26. The LeChâtelier-Braun Principle & Keq If a Q value is smaller than the Keq value for a given rxn. it means that the conc. of products is less than expected at equilibrium. The rxn. will shift RIGHT to make more products before reaching equilibrium. If a Q value is EQUAL to the Keq value for a given rxn. it means that the rxn. is at equilibrium.

  27. The LeChâtelier-Braun Principle & Keq Given the balanced equation, the Keq value, and the complete set of conc. to find a rxn. quotient: H2 + 2ICl Ý I2 + 2HCl Current conc.Keq H2: 0.413M 1.39 × 10-1 ICl: 0.398MI2: 0.470M HCl: 0.309M Which is favored, the production of reactants or products?

  28. The LeChâtelier-Braun Principle & Keq First, determine the proper Q equation: Second, put things where they belong and then solve: Products in numerator; on top Coefficients become exponents! Reactants in denominator; on bottom

  29. The LeChâtelier-Braun Principle & Keq Third, compare the Q and Keqvalues then correctly choose which direction the rxn. will proceed: Q is larger which means there is a higher conc. of products than anticipated at equilibrium therefore the rxn. will favor producing reactants. Q = 6.86 × 10-1 vs. Keq = 1.39 × 10-1

  30. Reaction Rates The speed of a reaction, the rate at which it occurs, can be affected by changing conditions. The LeChâtelier-Braun Principle clearly shows that we can change the rxn. (producing products or reactants) by changing the conditions of the rxn. It doesn’t mention rate. So, what factors influence the rate of rxn.?

  31. Rate Influencing Factors Temperature More kinetic energy means more collisions Endothermic rxn. speeds up with heat up. Exothermic rxn. speeds up with cool down. (within reason, if activation energy is not met the rxn. will not occur) Concentration More substance, more collisions, more rxn. Surface Area (Particle Size) More substance exposed, more collisions that generate rxn. Agitation/Stirring More kinetic energy means more collisions. Catalysts (accelerants & inhibitors) Reduced or increased activation energy.

  32. Getting a Rxn. Started • Activation Energy (Ea) • The minimum amount of energy required to cause a rxn. • This energy can be lowered or raised by using catalysts. • Increasing temperature usually helps a rxn. obtain the necessary energy for rxn. • This is not lowering the Ea only meeting it. • Agitation/Stirring will also increase energy needed for rxn. • This is not lowering the Ea only meeting it.

  33. Activation Energy of Exothermic Rxn. Uncatalyzed Rxn. Ea Ea Free Energy Reactants ΔH Catalyzed Rxn. Products Course of Rxn.

  34. Activation Energy of Endothermic Rxn. Uncatalyzed Rxn. Ea Products Ea ΔH Free Energy Reactants Catalyzed Rxn. Course of Rxn.

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