Equilibrium Constant Kp

8 Key excerpts on "Equilibrium Constant Kp"

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  eBook - PDF

  A Textbook of Physical Chemistry

  • Arther Adamson(Author)
  • 2012(Publication Date)
  • Academic Press

    (Publisher)

  Alternatively, K P is determined by AG 0 , which depends only on the standard-state free energies of the reactants and products and is a constant for the system. However, AG 0 and hence Κ P will vary with temperature, as discussed in Section 7-4. The value of AG 0 is a measure of the tendency of the reaction to occur. If it is large and negative, K P is general, however, and is applied to solution equilibria in Chapter 12 with only minor modifications in terminology. 230 CHAPTER 7: CHEMICAL EQUILIBRIUM a large number, and at equilibrium the concentrations of products will be large compared to those of the reactants. Conversely, if AG 0 is a large positive number, K P will be small, and the degree of reaction will be small at equilibrium. Finally, Eq. (7-11) relates equilibrium constants to the general body of thermodynamic data. Free energies for individual substances may be obtained from spectroscopic information through the methods of statistical thermodynamics so that equilibrium constants can be calculated indirectly (Section 7-8) or measured equilibrium constants can be given a thermodynamic interpretation. The equilibrium constant for a reaction involving ideal gases may be expressed in several alternative forms. Since P t = (ni/N)P, where Ν and Ρ are, respectively, the total moles present and the total pressure at equilibrium, it follows that Κ Ρ ~ « Α ° ν - Ν) -Η ΊΓ) 9 (7 ' 12) where Δη denotes the number of moles of products minus those of reactants. Since K P is independent of pressure, it is evident that K n must be a function of pressure. Therefore K n is not a true equilibrium constant; it is useful, however, as an inter-mediate quantity in equilibrium constant calculations. Similarly, we have Υ τη,γ η . . . K P = ?αΛ... = K * P * n > ( 7 13 ) χ Α χ Β where χ denotes mole fraction; Κ χ , like Κ η , is a function of pressure. We may also express K P in terms of concentration since C = n/v = P/RT.

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  Chemistry, 5th Edition

  • Allan Blackman, Steven E. Bottle, Siegbert Schmid, Mauro Mocerino, Uta Wille(Authors)
  • 2022(Publication Date)
  • Wiley

    (Publisher)

  The thermo- dynamic equilibrium constant is obtained when activities, rather than pressures or concentrations, are used in the equilibrium constant expression. The equilibrium constant is constant at constant temperature. The magnitude of the equilibrium constant reflects the extent to which the forward reaction has proceeded to completion when equilibrium is reached. The reaction quotient, Q, has the same form as K, but the concentrations or pressures are nonequilibrium values. Pure solids and pure liquids do not appear in the expressions for K and Q. When an equation is multiplied by a factor n to obtain a new equation, we raise its value of K to the power of n to obtain K for the new equation. When two equations are added, we multiply their values of K to obtain the new K. When an equation is reversed, we take the reciprocal of its value of K to obtain the new K. The values of K p and K c are equal only if the same number of moles of gas appear on each side of the chemical equation. When the numbers of moles of gas are different, K p is related to K c by the following equation where R = 8.314 J mol -1 K -1 , T = absolute temperature and Δn gas is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants in the balanced equation. K p = K c ( 1000 RT p o ) Δn gas An equilibrium where all substances are in the same phase is called a homogeneous equilibrium, while one with substances in more than one phase is a heterogeneous equilibrium. The equilibrium constant expression for a heterogeneous equilibrium omits concentration terms for pure liquids and pure solids. 9.3 Show how equilibrium and Gibbs energy are related. Gibbs energy diagrams for chemical reactions show the variation in Gibbs energy of a system as a reaction mixture proceeds from pure reactants to pure products.

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  eBook - ePub

  Rapid Review of Chemistry for the Life Sciences and Engineering

  With Select Applications

  • Armen S. Casparian, Gergely Sirokman, Ann Omollo(Authors)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  c is discussed in more detail later in this section.

  Four points of caution regarding the use of this expression are noteworthy:

  • Only concentrations in units of molarity are permitted.
  • Only species that have concentrations can appear. Thus, species in the solution state (aq) or in the gaseous state (g) are included, but species that are pure solids (s) or liquids (l) are not.
  • Kc is a function of temperature (check the van’t Hoff Equation, Equation 4.5, later in this chapter).
  • All concentrations substituted into this expression must be taken at equilibrium.

  A reaction in which all species are present in the same physical state (e.g., all gases or all solutions) is called a homogeneous reaction, while a reaction in which species are present in more than one physical state (i.e., mixed physical states) is called a heterogeneous reaction. If all species in a reaction are present in the gaseous state, an alternative form of the equilibrium constant expression can be written as Kp :

  K p

    =  

  P G

  g

  P H

  h

  P A

  a

  P B

  b

  (4.3)

  Here, P represents the partial pressure of each gas present at equilibrium. Any pressure unit is acceptable as long as it is consistent with the others and with the units of Kp , but atmospheres and torr are common.

  A formula to convert Kc to Kp , and vice versa, is as follows:

  K p

    =

  K c

  (

  R T

  )

  Δ n

  (4.4)

  where

  1. T = the absolute temperature in Kelvins
  2. R = the gas constant to be used in units consistent with partial pressure
  3. ∆n = the change in the number of moles of gas, i.e., the total moles of product minus the total moles of reactant in the balanced reaction

  Example 4.1: Expressing and Evaluating Kc and Kp

  Write the equilibrium constant expression Kc

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  Physical Chemistry

  Thermodynamics

  • Horia Metiu(Author)
  • 2006(Publication Date)
  • Taylor & Francis

    (Publisher)

  20 CHEMICAL EQUILIBRIUM: THE DEPENDENCE OF THE EQUILIBRIUM CONSTANT ON TEMPERATURE AND PRESSURE §1. Introduction. Of all the quantities calculated here, the equilibrium composi-tion is the most interesting to the practicing chemist. To calculate it, the equilibrium constant K must be known. This puts K at the center of the stage: G 0 and S 0 are of interest only as intermediate quantities for computing K . Given this central role of K , it would be useful to have equations that give the vari-ation of K with temperature and pressure. In this chapter, I derive such equations. They have three uses. 1. They lead to Le Chatelier’s principle, which allows us to predict the direction in which the equilibrium composition changes when we change T or p . These predictions are qualitative (e.g. if you raise the temperature, you make more ammonia) and require no calculations. 409 410 Dependence of Equilibrium Constant on T and p 2. The equations allow me to calculate the equilibrium constant at any temperature and pressure, if I know its value at one temperature and pressure. 3. The equations allow me to calculate the heat of reaction if I measure the equilib-rium composition at several temperatures. Since concentration measurements are accurate and heat measurements are not, this is a good method for obtaining heats of reaction. The Change of the Equilibrium Constant K with Temperature and Pressure: the Equations §2. The Change of Equilibrium Constant with Temperature. In Chapter 17, §19, I derived the equation − RT ln K = c i = 1 ν ( i ) µ 0 ( i ; T , p ) ≡ G 0 ( T , p ) (20.1) This is Eq. 17.35 for A = 0, which is the equilibrium condition. µ 0 ( i ; T , p ) is the chemical potential of the pure component i . To proceed I need to remind you of the equation (see Chapter 14, §12, Eq. 14.24) ∂ ∂ T µ 0 ( i ) T p , n = − h 0 ( i ) T 2 (20.2) Here h 0 ( i ) is the molar enthalpy of the pure component i .

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  Physical Chemistry

  • Robert J. Silbey, Robert A. Alberty, George A. Papadantonakis, Moungi G. Bawendi(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  5.8 Heterogeneous Reactions 151 To define a dimensionless equilibrium constant in terms of concentration, we introduce the standard concentration c ∘ , which represents one mole per liter. Introducing this standard concentration into each term of equation 5.55 yields K P = Ns ∏ i=1 [ ( c i c ∘ ) ( c ∘ RT P ∘ )] v i = ( c ∘ RT P ∘ ) Σv i ∏ i ( c i c ∘ ) v i = ( c ∘ RT P ∘ ) Σv i K c (5.56) where the equilibrium constant expressed in terms of concentration, K c = ∏ i ( c ∘ c ∘ ) v i (5.57) is a function only of temperature for a mixture of ideal gases. If c ∘ = 1 mol L −1 and P ∘ = 1 bar, then c ∘ RT/P ∘ = 24.79 at 298.15 K. Example 5.15 A gas equilibrium constant expressed in concentrations What is the value of the equilibrium constant K c for the dissociation of ethane into methyl radicals at 1000 K? C 2 H 6 (g) = 2CH 3 (g) Δ r G ∘ = 2Δ f G ∘ (CH 3 ) − Δ f G ∘ (C 2 H 6 ) = 2(159.82) − 109.55 = 210.09 kJ mol −1 K P = exp ( −Δ r G ∘ RT ) = exp (−210.09) (8.3145 × 10 −3 )(1000) = 1.062 × 10 −11 K c = ([CH 3 ]∕c ∘ ) 2 [C 2 H 6 ]∕c ∘ = K P P ∘ c ∘ RT = (1.062 × 10 −11 ) (1 bar) (1mol L −1 )(0.083145)(1000 K) = 1.278 × 10 −13 Thus, at equilibrium [CH 3 ] 2 /[C 2 H 6 ] = 1.278 × 10 −13 mol L −1 , where the brackets indicate concentrations in moles per liter. 5.8 HETERO- GENEOUS REACTIONS A reaction involving more than one phase that does not involve equilibria of a species between phases is referred to as a heterogeneous reaction. Examples are CaCO 3 (s) = CaO(s) + CO 2 (g) (5.58) CH 4 (g) = C(s) + 2H 2 (g) (5.59) Depending on the initial conditions, reactions of this type can go to completion, whereas reactions in a sin- gle phase do not go to completion because of the entropy of mixing (Section 3.5). The equilibrium constant for reaction 5.58 is equal to the partial pressure of CO 2 gas that is measured at equilibrium when all three phases are present. If the pressure applied to the system is less than K, reaction 5.58 will go to completion to the right.

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  Student Solutions Manual for Chemistry, Third Canadian Edition

  • John A. Olmsted, Gregory M. Williams, Robert C. Burk(Authors)
  • 2016(Publication Date)
  • Wiley

    (Publisher)

  Equilibrium constant expressions have the concentrations of the products in the numerator and the concentrations of the reactants in the denominator, with each concentration raised to the power of its stoichiometric coefficient: 2 2 3 2 2 3 2 4 2 2 3 eq CO eq H O eq 4 3 NH eq O eq eq 2 N eq 2 CH CHO eq eq 2 C H eq O eq 2 2 eq eq 4 eq eq H S eq NH eq (a) ( ) ( ) ( ) ( ) (b) ( ) ( ) (c) ( ) ( ) (d) [Ag ] [SO ] (e) ( ) ( ) K p p p p K p p K p p K K p p + − = = = = = 14.51 This problem describes an equilibrium reaction. We are asked to determine the equilibrium pressures of all the gases. To calculate pressures at equilibrium from initial conditions, set up a concentration table, write the K eq expression, and solve for the pressures. For a gas phase reaction, concentrations must be expressed in bar. Chapter 14 Chemistry, 3ce Student Solutions Manual 276 © John Wiley & Sons Canada, Ltd. 1 2 1 2 2 1 2 1 (3.00 bar)(273 352 K) 6.29 bar 298 K p p p T p T T T + = = = = Reaction: COCl 2  CO + Cl 2 Initial pressure (bar) 6.29 0 0 Change in pressure (bar) – x + x + x Equilibrium pressure (bar) 6.29 – x x x Now substitute into the equilibrium constant expression and solve for x: 2 2 2 2 2 CO eq Cl eq 4 eq COCl eq 2 2 4 3 2 CO eq Cl eq COCl eq ( ) ( ) ( ) 8.3 10 ; assume that 6.29: ( ) 6.29 8.3 (6.29)(8.3 10 ) 5.22 10 6.29 7.2 10 bar ( ) ( ) ( ) 6.29 ( p p x K x p x x x x p p p − −4 − − − = × = = << − × 10 = = × = × = × = = = − 2 2 7.2 10 ) 6.22 bar 7.2 10 is 1.1% of 6.29, so is << 6.29 and the approximation is valid. x − − × = × 14.53 At equilibrium, a molecular picture should show the presence of both reactants and products, in relative amounts that are determined by the value of the equilibrium constant.

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  Physical Chemistry

  • Robert J. Silbey, Robert A. Alberty, Moungi G. Bawendi(Authors)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  It is essential to know that equilibrium has been reached before the analysis of the mixture can be used for calculating the equilibrium constant. The following criteria for the attainment of equilibrium at constant temperature are useful: The same value of the equilibrium constant should be obtained when the equilibrium is approached from either side. The same value of the equilibrium constant should be obtained when the initial concentrations of reacting material are varied over a wide range. The determination of the density of a partially dissociated gas provides one of the simplest methods for measuring the extent to which the gas is dissociated. When a gas dissociates, more molecules are produced, and at constant tempera- ture and pressure the volume increases. If the equilibrium extent of reaction has been determined, we need the ex- pression for the equilibrium constant in terms of the equilibrium extent of reac- tion and the pressure. We will assume that the gases are ideal. The first step is to express the mole fractions of reactants and products, in terms of the extent of reaction . 1 0 0.6 0.4 0.2 0.0 0.8 1.0 2 3 4 5 ξ P/bar K = 10 K = 1 K = 0.1 n n n n g P P K P P P P P P P P . P K P . K P P n 140 P P K                  2 2 4  0 2 4 2 4 2 2 4 0 2 4 2 0 2 4 2 4 2 0 2 4 2 4 2 2 NO N O 2 2 2 2 4 2 1/2 0 2 4 Extent of reaction at equilibrium for the reaction A(g) 2B(g) as a function of / , for various values of , us- ing equation 5.31. (See Computer Problem 5.H.)  dimensionless extent of reaction Chapter 5 Chemical Equilibrium                                                    Figure 5.3 For example, consider the dissociation of an initial amount (N O ) of N O to NO at a given temperature and pressure. If the equilibrium extent of reac- tion is , the equilibrium amount of N O is (N O ) and the equilibrium amount of NO is 2 .

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  Engineering and Chemical Thermodynamics

  • Milo D. Koretsky(Author)
  • 2012(Publication Date)
  • Wiley

    (Publisher)

  584 ► Chapter 9. Chemical Reaction Equilibria The equilibrium constant is calculated in the usual manner. First, we use the Gibbs energy of reaction to calculate K at 25ºC: K 298 5 exp ¢ 2 Dg rxn o RT ≤ 5 exp a2 232,900 3 8.314 4 3 298.2 4 b 5 5.81 3 10 5 (a) For Dh rxn o 5 constant, we again use Equation (9.21): ln K T K 298 5 Dh rxn o R a 1 T 2 1 298 b 5 92,220 8.314 a 1 773 2 1 298 b 5 222.88 and solving: K T 5 6.754 3 10 25 Plugging this into Equation (E9.7) with P 5 1 bar and solving for the extent of reaction gives: j 5 0.005 (b) For Dh rxn o 5 D rxn o 1 T 2 , we can use Equation (9.24): ln ¢ K T K 298 ≤ 5 h B 2 Dh rxn,298 o R 1 DA 1 298 2 1 DB 2 1 298 2 2 1 DC 3 1 298 3 2 2 DD 298 R a 1 T 2 1 298 b 1DA ln a T 298 b 1 DB 2 1 T 2 298 2 1 DC 6 1 T 2 2 298 2 2 1 DD 2 ¢ 1 T 2 2 1 298 2 ≤ x 5 224.37 and solving: K T 5 1.51 3 10 25 Plugging this into Equation (E9.7) with P 5 1 bar and solving for the extent of reaction gives j 5 0.003 In summary, under these conditions we do not expect to produce an appreciable amount of ammonia. Consider the production of ammonia from the catalytic reaction of a stoichiometric feed of nitrogen and hydrogen as in Example 9.7. As we saw, when the reaction temperature is 500°C and the reactor pressure is 1 bar, the conversion is very low. We wish to change the reactor conditions to increase conversion. Would you pick T or P? Which way would you change? SOLUTION This reaction is exothermic 1 Dh rxn o , 0 2 . As can be deduced from Equation (9.18), lower temperature will lead to higher equilibrium conversions. However, reducing the tem- perature also reduces the rate of reaction, which is also a major industrial concern. In this case, the rates become too slow, so T cannot be reduced to increase conversion. If we next consider P, the primary effect is determined by n, as shown in Equation (E9.5B). Since n is negative, an increase in pressure will increase conversion.

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