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Chemical Equilibrium

Chemical Equilibrium. Systems at Equilibrium & K eq ’s. Review L e Chatelier. What effect does each of the following changes have on the equilibrium position for this reaction? PCl 5(g) + heat PCl 3(g) + Cl 2(g) Addition of Cl 2 Increase in pressure Removal of heat

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Chemical Equilibrium

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  1. Chemical Equilibrium Systems at Equilibrium & Keq’s

  2. Review LeChatelier • What effect does each of the following changes have on the equilibrium position for this reaction? PCl5(g) + heat PCl3(g) + Cl2(g) Addition of Cl2 Increase in pressure Removal of heat Removal of PCl3 as it forms

  3. Our Systems Initially • When reactants initially interact, the rate of the forward reaction is quite large • The rate of the reverse reaction at T=0 is zero (no rxn yet) • As the reaction proceeds, the products begin to get synthesized, increasing the rate of products interacting with one another.

  4. Our Systems after some collisions • This NEW increase in the reverse reaction is caused by the decrease of the forward reaction. • Meaning we are decreasing our [reactants] and increasing our [products] • When the two rates become equal, we have reached _______________________. • This equilibrium means the forward rate is equal to the reverse rate NOT THE CONCENTRATIONS!!

  5. Equilibrium Constants (Keq) • Because reaction rates depend on concentrations, there is a mathematical relationship between the product and reactant concentrations at equilibrium. • The concentrations of the products and the reactants can be used to determine the Keq. Concentration is signified using SQUARE BRACKETS. [HCl] = concentrations of hydrochloric acid • Alternatively, the Keq (if given) can be used to determine concentrations of reactants or products (aka: determining molarities)

  6. Equilibrium Constants (Keq) • The equilibrium constant, denoted Keq, is equal to the concentrations of the products OVER the concentration of the reactants. • Each concentration is raised to its OWN corresponding coefficient from the balanced chemical reaction • These values (all of the products or all of the reactants) are multiplied together • The Keqomits ALL pure solids (s) and liquids (l) • Use only gases and _____________________ solutions.

  7. Equilibrium Constants (Keq) • Consider the following reaction: CaCO3(s) + 2H3O+(aq) Ca2+(aq) + CO2(g) +3H2O(l) • The equilibrium constant, of Keq, (excluding all solids and liquids) is Keq = [Ca2+(aq)] [CO2(g)] [H3O+(aq)]2 Q: What do the square brackets mean??

  8. Rules for writing a Keq • Write out and balance the reaction • Be sure it is at chemical equilibrium (stable) • Write the concentrations of the products in the numerator and the concentrations of the reactants in the denominator (excluding solids and liquids). • Raise each participant to its own coefficient. Keq = [Product #1][Product #2] [Reactant #1]

  9. Practice, Practice, Practice • Write the expression for the equilibrium constant for each reaction 4H2(g) + CS2(g) CH4(g) + 2H2S(g) PCl5(g) PCl3(g) + Cl2(g) 2NO(g) + O2(g) 2NO2(g) CO(g) + H2O(g) H2(g) + CO(g)

  10. Example • An aqueous solution of carbonic acid reacts to reach equilibrium as described below. H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq) • Write out the Keq expression of this equilibrium reaction

  11. Example (continued) • An aqueous solution of carbonic acid reacts to reach equilibrium as described below. H2CO3(aq) + H2O(l) HCO3-(aq) + H3O+(aq) • QUESTION: The solution contains the following solute concentrations: carbonic acid, 3.3x10-2 molL-1; bicarbonate ion, 1.19x10-4 molL-1; and hydronium ion, 1.19x10-4 molL-1. Calculate the Keq using the expression determined in the previous example.

  12. Applying this information • Considering the generic expression for writing out an equilibrium constant, • Do you expect a reaction that does not go very quickly (meaning having more a larger quantity of reactants) to have a small Keq or a large one? • How about if you have more products than reactants? Keq > 1, products favored at equilibrium Keq < 1, reactants favored at equilibrium

  13. Applying this Information • A large Keq represents a lot more products (or complete reaction / dissociation) • A smaller ____________________________________________. • What about a Keq = 1 ?? What would the [prod] / [ react] ratio be? • Keq values can be found in Appendix A in your texts (@ 25°C)

  14. Practice • Consider the following reaction: N2O4(g) 2NO2(g) If at equilibrium, a system contains 0.0045 mol/L of N2O4and 0.030 mol/L NO2 what is the Keq?

  15. Calculating concentration • The Keq for the equilibrium below is 1.8x10-5 at a temperature of 25°C. Calculate [NH4+] when [NH3]=6.82x10-3. NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

  16. In-class practice • Get a book and complete the following: • Page 546-547, Sample Problems #2, 3, and 4 • Then work on Page 548, #14, 17, 18, and 19 • Vocab due!! LAST CALL

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