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Quantitative Changes in Equilibrium Systems

Quantitative Changes in Equilibrium Systems. Chapter 7. Testing to see if an equilibrium has been established. The reaction quotient ( Q ), uses the equilibrium expression with the available concentrations to determine if the reaction has attained equilibrium or not.

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Quantitative Changes in Equilibrium Systems

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  1. Quantitative Changes in Equilibrium Systems Chapter 7

  2. Testing to see if an equilibrium has been established. • The reaction quotient (Q), uses the equilibrium expression with the available concentrations to determine if the reaction has attained equilibrium or not. For the generalchemical reaction: aA + bB DcC + dD

  3. Predicting the direction of an equilibrium

  4. Quantitative analysis of equilibria • Using Le Châtelier’s Principle & the Equilibrium Law one can quantitatively assess; • The equilibrium constant (Keq) • The position or progress of the equilibrium system with the reaction quotient (Q) • The equilibrium concentrations

  5. Determining the position or status of an equilibrium system. • Calculate the reaction quotient using the equilibrium concentrations. • Compare the reaction quotient value to the equilibrium constant value. • Determine whether the reaction is reactant or product rich and adjust the equilibrium to establish the equilibrium.

  6. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? • Balanced equation for the reaction N2 (g) + 3 H2 (g) 2 NH3 (g) • Generate the equilibrium expression and reaction quotient expression

  7. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? • Substitute the supplied concentrations into the expression to solve for Q. • Compare the Q value to the known Keq value. Q < Keq14.8 < 25.0

  8. At 500o C, the value of Keq for reaction between nitrogen and hydrogen to produce ammonia is 25.0. The equilibrium concentrations for the chemicals are; [N2] = 0.10 mol/L, [H2] = 0.30 mol/L & [NH3]=0.20 mol/L. Is the mixture at equilibrium? If not, which direction must it be adjusted to acquire an equilibrium? • Use the Q and Keq values to determine the position of the equilibrium system. Since Q < Keq, the system has fewer products than the equilibrium state. Therefore the reaction needs to progress toward the products to attain equilibrium. (i.e.- Shift to the right.)

  9. Calculating equilibrium concentrations • To determine the equilibrium concentrations, one must use: • A balanced equation for the equilibrium reaction • An equilibrium expression and constant, Keq • An Initial concentration, Change in concentration, and Equilibrium concentration (ICE) table.

  10. Calculating equilibrium concentrations • In cases where the Keq value is very small the addition and subtraction of the change value (x) may be insignificant and thereby omitted. • Use the ratio of the smallest initial concentration and the Keq to assess the affect of the change.

  11. Calculating equilibrium concentrations • If clue > 500, the addition or subtraction of “x” is insignificant and may be ignored. • If 100<clue<500, the addition or subtraction of “x” is probably insignificant, but should be checked. • If clue<100, the addition or subtraction of “x” is significant and must be included in the calculations.

  12. Calculating equilibrium concentrations • In cases where the Keq expression results in the resolution of a quadratic equation, the use of the quadratic formula my prove helpful. • When the quadratic formula produces two answers select the one that is viable. Remembering that you can not have a negative concentration!

  13. Example – At high temperatures, as with lightning, nitrogen and oxygen will react to produce nitrogen monoxide. A chemist puts 0.085 moles of N2(g) and 0.038 mol of O2(g) in a 1.0 L flask at high temperature, where the Keq= 4.2 x 10-8. What is the concentration of the NO(g) in the mixture at equilibrium? Strategy • Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored. • Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. • Set up an ICE table letting “x” represent the change in concentrations. • Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. • Calculate the required value(s).

  14. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored. The clue is far greater than 500 so we can ignore the changes in N2(g) and O2(g).

  15. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. N2(g) + O2(g)  NO(g) N2(g) + O2(g)  2 NO(g)

  16. Set up an ICE table letting “x” represent the change in concentrations.

  17. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. The negative value is impossible as one can not have a negative concentration.

  18. Calculate the required value(s). What is the NO(g) concentration at equilibrium? The NO(g) concentration at equilibrium is 1.2 x 10-5 mol/L.

  19. Example – In a 1.00 L flask, 2.00 mol of H2(g) is combined with 3.00 mol of I2(g) which produces HI(g). The Keq for the reaction is 25 at 1100 K. What is the What is the concentration of each gas in the mixture at equilibrium? Strategy • Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored. • Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. • Set up an ICE table letting “x” represent the change in concentrations. • Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. • Calculate the required value(s).

  20. Divide the smallest initial concentration by the Keq to determine if the change in concentration can be ignored. The clue is much less than 500 so we can not ignore the changes in H2(g), I2(g) and HI(g).

  21. Write a balanced chemical reaction for the equilibrium reaction and determine the equilibrium expression. H2(g) + I2(g)  HI(g) H2(g) + I2(g)  2 HI(g)

  22. Set up an ICE table letting “x” represent the change in concentrations.

  23. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

  24. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”.

  25. Substitute the equilibrium concentrations (ICE Table) into the expression and solve for “x”. 4.3 and 1.7 are determined through the quadratic formula calculation, but only one is correct! 3.00 – 4.3 = negative value so 1.7 mol/L is the correct value. (A negative concentration is impossible!) Therefore the final equilibrium concentrations are: [H2]eq = 2.00 – 1.7 = 0.3 mol/L[I2]eq = 3.00 – 1.7 = 1.3 mol/L[HI]eq = 2( 1.7) = 3.4 mol/L

  26. SUMMARY OF LESSONS 1,2 AND 3 EQUILIRIUM EXPRESSION Solids and liquids are not included in equilibrium expression ONLY gases and aqueous. units: use M for Kc and or atm for Kp SIZE OF K K >1 → products favored in equilibrium system K < 1 → reactants favored in equilibrium system ENDOTHERMIC K increases with increase temperature. K decreases with decrease temperature EXOTHERMIC K decreases with increase temperature. K increases with decrease temperature

  27. SUMMARY OF LESSONS 1,2 AND 3 • HOW “K” CHANGES AS THE REACTION CHANGES • Reverse rxn: K →1/K • Multiply rxn by n: K → Kn • Add rxs: K3 = K1 x K2 • REACTION QUOTIENT (Q): a snapshot of the reaction; shows which way reaction proceeds (L, R, no shift) • Q > K → left • Q < K → right • Q = K → at equilibrium (no shift) • CALCULATIONS • given [ ]eq → find K • given [ ]o + other information → find [ ]eq → K • given K → [ ]eq using one of three methods • perfect squares • Approximation: 100 rule • Quadratic equation

  28. SUMMARY OF LESSON 1,2 AND 3 • Le CHATELIER’S PRINCIPLE: reaction shifts to restore equilibrium (opposes stress on the system) • FACTORS AFFECTING EQUILIBRIUM • Concentration; Temperature; Pressure • Concentration: • add/remove reactants/products part of K expression; pure liquids or solids usually do not affect equilibrium • increase concentration of A, move away from A ( A ) • Decrease concentration of A, move towards A ( A) • Temperature: • endothermic: heat written as a reactant; • exothermic: heat written as a product • Increase temp always favours the endothermic reaction • Pressure: Must be gaseous and must have unequal number of moles. • P ↑ move in a direction with less gas mol • if possible; convert V changes to P • FACTORS NOT AFFECTING EQUILIBRIUM • Catalyst, Pure liquid or solid (water sometimes can shift rxn). Inert gases does not affect equilbrium

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