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Chapter 15. Chemical Equilibrium - Weeb · PDF fileAP Chemistry Chapter 15 Equilibrium - 1 - Chapter 15. Chemical Equilibrium Common Student Misconceptions • Many students need to

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  • AP Chemistry Chapter 15 Equilibrium

    - 1 -

    Chapter 15. Chemical Equilibrium

    Common Student Misconceptions Many students need to see how the numerical problems in this chapter are solved. Students confuse the arrows used for resonance ()and equilibrium (). Students often have problems distinguishing between K and Q. Students who have difficulty with some of the mathematical manipulations in this chapter should be

    directed to Appendix A of the text. Students often do not know (or check) whether an approximate equilibrium calculation is valid. Lecture Outline 15.1 The Concept of Equilibrium Consider colorless frozen N2O4. At room temperature, it decomposes to brown NO2.

    N2O4(g) 2NO2(g) At some time, the color stops changing and we have a mixture of N2O4 and NO2. Chemical equilibrium is the point at which the concentrations of all species are constant.

    The N2O4(g) 2 NO2(g) equilibrium. (a) The concentration of N2O4 decreases while the concentration of NO2 increases during the course of the reaction. Equilibrium is indicated when the concentrations no longer change with time. (b) The rate of disappearance of N2O4 decreases with time as the concentration of N2O4 decreases. At the same time, the rate of formation of NO2 also decreases with time. Equilibrium occurs when these two rates are equal.

    (a) The concentration of N2O4 decreases while the concentration of NO2 increases during the course of the reaction. Equilibrium is indicated when the concentrations no longer change with time. (b) The rate of disappearance of N2O4 decreases with time as the concentration of N2O4 decreases. At the same time, the rate of formation of NO2 also decreases with time. Equilibrium occurs when these two rates are equal.

  • AP Chemistry Chapter 15 Equilibrium

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    At equilibrium the concentrations of N2O4 and NO2 do not change. This mixture is called an equilibrium mixture. This is an example of a dynamic equilibrium. A dynamic equilibrium exists when the rates of the forward and reverse reactions

    are equal. No further net change in reactant or product concentration occurs. The double arrow implies that the process is dynamic. Assume that both the forward and reverse reactions are elementary processes. We can write rate expressions for each reaction. Forward reaction: N2O4(g) 2 2(g) Ratef = kf[N2O4] kf = rate constant (forward reaction) Reverse reaction: 2 NO2(g) N2O4(g) Rater = kr[NO2]

    2 kr = rate constant (reverse reaction)

    15.2 The Equilibrium Constant

    As N2O4 reacts to form NO2, the concentration of N2O4 will decrease and the concentration of NO2 will increase.

    Thus we expect the forward reaction rate to slow and the reverse reaction rate to increase. Eventually we get to equilibrium where the forward and reverse rates are equal. At equilibrium: Ratef = Rater

    kf[N2O4] = kr[NO2]2

    Rearranging, we get: kf = [NO2]2 kr [N2O4]

    The ratio of the rate constants is a constant at a specific temperature, and the expression becomes

    Keq = kf = [NO2]2 kr [N2O4] Sidebar: The Haber Process, used for the preparation of ammonia from nitrogen and hydrogen. Consider the reaction: N2(g) + 3H2(g) 2NH3(g) If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach

    equilibrium with constant concentrations of nitrogen, hydrogen and ammonia. However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2

    and H2 will be produced until equilibrium is achieved.

  • AP Chemistry Chapter 15 Equilibrium

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    No matter the starting composition of reactants and products is, the equilibrium mixture contains the same relative concentrations of reactants and products.

    Equilibrium can be reached from either direction.

    Equilibrium-Constant Expression We can write an expression for the relationship between the concentration of the reactants and

    products at equilibrium. This expression is based on the law of mass action. For a general reaction aA + bB dD + eE The equilibrium-constant expression is given by:

    Where Kc is the equilibrium constant. The subscript c indicates that molar concentrations were used to evaluate the constant. Note that the equilibrium constant expression has products in the numerator and reactants in the

    denominator.

    Equilibrium Constants in Terms of Pressure, Kp When the reactants and products are gases we can write an equilibrium expression using partial

    pressures rather than molar concentrations. The equilibrium constant is Kp where p stands for pressure. For the reaction:

    aA + bB dD + eE

    They can be interconverted using the ideal gas equation and our definition of molarity: PV = nRT thus P = (n/V)RT

    If we express volume in liters the quantity (n/V) is equivalent to molarity. Thus the partial pressure of a substance, A, is given as:

    PA =(nA/V)RT = [A]RT We can use this to obtain a general expression relating Kc and Kp:

    Kp = Kc(RT)n

    Where n = (moles of gaseous products) (moles of gaseous reactants). The numerical values of Kc and Kp will not differ if n = 0. (Note: See p. 635 in textbook for info about why we do not use units for K)

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  • AP Chemistry Chapter 15 Equilibrium

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    Sample Exercise 15.1 (p. 634)

    Write the equilibrium expression for Keq for these three reactions:

    a) 2 O3(g) 3 O2(g)

    b) 2 NO(g) + Cl2(g) 2 NOCl(g)

    c) Ag+(aq) + 2 NH3(g) Ag(NH3)2+(aq)

    Practice Exercise 1 (15.1)

    For the reaction 2 SO2(g) + O2(g) 2 SO3(g) which of the following is the correct equilibrium-constant expression:

    a) Kp = PSO22 PO2 PSO32

    b) Kp = 2 PSO2 PO2 2 PSO3

    c) Kp = PSO32 PSO22 PO2

    d) Kc = 2 PSO3 2 PSO2 PO2

    Practice Exercise 2 (15.1)

    Write the equilibrium expression for Keq for these two reactions:

    a) H2(g) + I2(g) 2 HI(g)

    b) Cd2+(aq) + 4 Br-(aq) CdBr42-(aq

  • AP Chemistry Chapter 15 Equilibrium

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    Evaluating Kc The value of Keq does not depend on initial concentrations of products or reactants. Consider the reaction: N2O4(g) 2NO2(g) The equilibrium constant is given by: Kc = [NO2]2 [N2O4]

    The equilibrium expression depends on stoichiometry. It does not depend on the reaction mechanism. The value of Keq varies with temperature. We generally omit the units of the equilibrium constant.

    Concentration changes approaching equilibrium.

    As seen in Table 15.1, the same equilibrium mixture is produced starting with either 0.0400 M NO2 (Experiment 3) or 0.0200 M N2O4 (Experiment 4).

  • AP Chemistry Chapter 15 Equilibrium

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    15.3 Interpreting and Working with Equilibrium Constants The Magnitude of Equilibrium Constants The equilibrium constant, Keq, is the ratio of products to reactants. Therefore, the larger Keq the more products are present at equilibrium. Conversely, the smaller Keq the more reactants are present at equilibrium. If Keq >> 1, then products dominate at equilibrium and equilibrium lies to the right. If Keq

  • AP Chemistry Chapter 15 Equilibrium

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    The Direction of the Chemical Equation and K An equilibrium can be approached from either direction. Consider the reaction: N2O4(g) 2 NO2(g)

    The equilibrium constant for this reaction (at 100oC) is: Kc = [NO2]2 = 0.212

    [N2O4]

    However, when we write the equilibrium expression for the reverse reaction, 2 NO2(g) N2O4(g) The equilibrium constant for this reaction (at 100oC) is:

    Kc = [N2O4] = 4.72 [NO2]2

    The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of the reaction in the opposite direction. Relating Chemical Equations and Equilibrium Constants The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium

    constant of the reaction in the forward direction. The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium

    constant raised to a power equal to that number. The equilibrium constant for a net reaction made up of two or more steps is the product of the

    equilibrium constants for the individual steps.

    Sample Exercise 15.4 (p. 640)

    Given the following information, HF(aq) H+(aq) + F-(aq) Kc = 6.8 x 10-4 H2C2O4(aq) 2 H+(aq) + C2O42-(aq) Kc = 3.8 x 10-6

    determine the value of Kc for the following reaction: 2 HF(aq) + C2O42-(aq) 2 F-(aq) + H2C2O4(aq)

    (0.12)

  • AP Chemistry Chapter 15 Equilibrium

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    Practice Exercise 1 (15.4)

    Given the equilibrium constants for the following two reactions in aqueous solution at 25oC,

    HNO2(aq) H+(aq) + NO2-(aq) Kc = 4.5 x 10-4

    H2SO3(aq) 2 H+(aq) + SO32-(aq) Kc = 1.1 x 10-9

    What is the value of Kc for the following reaction?

    2 HNO2(aq) + SO32-(aq) H2SO3(aq) + 2 NO2-(aq)

    a) 4.9 x 10-13 b) 4.1 x 105 c) 8.2 x 105 d) 1.8 x 102 e) 5.4 x 10-3

    Practice Exercis

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