Chapter 15 – Chemical Equilibrium

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10 15.5 Applications of Equilibrium Constants One use of equilibrium equations is to allow the determination of the concentration or pressure of a reaction component that is difficult to measure. For example, if a reagent were corrosive, it might be easier to measure the concentrations of the other materials and calculate the concentration of the corrosive substance. Prediction of the Direction of Reaction If a system is not at equilibrium, there are times it would be useful to determine in which direction the reaction would shift. For a non-equilibrium system aA + bB Q = c [C] [D] a [A] [B] cC + dD d b where Q is the reaction quotient. This is the same equation as on the third page of the notes for the equilibrium constant. Hence, the reaction quotient, Q, provides a number that is related to the equilibrium constant. When Q equals K eq , the reaction is at equilibrium. When Q ≠ K eq , the reaction is moving towards equilibrium. If Q < K eq , the reaction ‘shifts right.’ If Q > K eq , the reaction ‘shifts left.’ If you find this difficult to remember or are concerned you’ll mix the two up, try the following method to remember these trends. Consider the equilibrium: aA (g) bB (g) Q = P P b B a A Which way will the equilibrium shift if the reaction is begun with A but no B To the right, since there is no B the reaction must try to produce some. In this case, Q = 0, which is less than K eq no matter what the value of K or P A are. Thus if Q < K eq , then the reaction shifts right.
11 The other extreme occurs when the reaction starts with B, but no A. Then the reaction shifts left to produce some A. Here Q = ∞, which is greater than K no matter what the value of K eq or P B are. Thus if Q > K eq , then the reaction shifts left. Calculation of Equilibrium Concentrations A common calculation required for equilibrium mixtures is “What are the concentrations of species in an equilibrium mixture for a given set of initial concentrations” The example below is a variation of Sample Exercise 15.11 (p. 593). Example: What are the equilibrium concentrations of all species for H 2 (g) + I 2 (g) 2HI (g) K eq = 50.5 @ 448 ºC when P H2i = P I2i = 0.100 atm and P HIi = 0 atm 1) Which way does the equilibrium shift Q = 0 < K ⇒ right 2) What are the equilibrium pressures P H2i = 0.100 atm P I2i = 0.100 atm P HIi = 0 atm P H2e = (0.100 – x) atm P I2e = (0.100 – x) atm P HIe = 2x atm The reason why x is subtracted from the H 2 and I 2 pressures, while P HI increases by 2x is reaction stoichiometry. For each molecule of H 2 or I 2 consumed, 2 molecules of HI are produced. Forgetting this is a common mistake when working problems. Now substitute into the equilibrium equation: 2 2 (2x) (2x) 50.5 = = 2 (0.100 - x)(0.100 - x) (0.100 - x) 7.10 = 2x 0.100 - x 2x = (7.10)(0.100 – x)
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Page 3 and 4: 3 The previous equality can be rear
Page 5 and 6: 5 As you can see, the pressures of
Page 7 and 8: 7 Units of Equilibrium Constants As
Page 9: Example: A mixture of 5.000 x 10 -
Page 13 and 14: 13 x 1 = 1.47 x 10 - 3 atm and x 2
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Page 17: 17 The Effect of Catalysts First re

Chapter 15 – Chemical Equilibrium

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