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    will the equilibrium constant for the reaction increase, decrease, or stay the same?

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    equilibrium constants and changing conditions

    A look at the relationship between equilibrium constants and Le Chatelier's Principle.

    Explanation in terms of the constancy of the equilibrium constant

    The equilibrium constant, Kc for this reaction looks like this:

    If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?

    This is actually the wrong question to ask! We need to look at it the other way round.

    Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

    If you decrease the concentration of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.

    The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant.

    If you decrease the concentration of C:

    Changing pressure

    This only applies to systems involving at least one gas.

    The facts aren't changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.

    That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.

    Explanation

    Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right.

    Because this is an all-gas equilibriium, it is much easier to use Kp:

    Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure, Kp will increase as well. Not so!

    To understand why, you need to modify the Kp expression.

    Remember the relationship between partial pressure, mole fraction and total pressure?

    Source : www.chemguide.co.uk

    The Effect of Changing Conditions

    This page looks at the relationship between equilibrium constants and Le Châtelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as …

    The Effect of Changing Conditions

    Last updated Aug 15, 2020 The Contact Process The Haber Process picture_as_pdf Readability Cite this page Donate Jim Clark

    Former Head of Chemistry and Head of Science at Truro School in Cornwall

    This page looks at the relationship between equilibrium constants and Le Chatelier's Principle. Students often get confused about how it is possible for the position of equilibrium to change as you change the conditions of a reaction, although the equilibrium constant may remain the same.

    Changing concentrations

    Equilibrium constants are not changed if you change the concentrations of things present in the equilibrium. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.

    Suppose you have an equilibrium established between four substances A, B, C and D.

    A+2B⇌C+D (1) (1)A+2B⇌C+D

    According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again.

    Explanation in terms of the constancy of the equilibrium constant

    The equilibrium constant,

    K c Kc

    for this reaction looks like this:

    K c = [C][D] [A][B ] 2 (2) (2)Kc=[C][D][A][B]2

    If you have moved the position of the equilibrium to the right (and so increased the amount of

    C C and D D

    ), why hasn't the equilibrium constant increased? This is actually the wrong question to ask! We need to look at it the other way round.

    Let's assume that the equilibrium constant must not change if you decrease the concentration of

    C C

    - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?

    If you decrease the concentration of

    C C , the top of the K c Kc

    expression gets smaller. That would change the value of

    K c Kc

    . In order for that not to happen, the concentrations of

    C C and D D

    will have to increase again, and those of

    A A and B B

    must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before. The position of equilibrium moves - not because Le Chatelier says it must - but because of the need to keep a constant value for the equilibrium constant.

    If you decrease the concentration of

    C C :

    Changing pressure

    This only applies to systems involving at least one gas. Equilibrium constants are not changed if you change the pressure of the system. The only thing that changes an equilibrium constant is a change of temperature. The position of equilibrium may be changed if you change the pressure. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change that you have made.

    That means that if you increase the pressure, the position of equilibrium will move in such a way as to decrease the pressure again - if that is possible. It can do this by favoring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.

    Case 1: Differing Numbers of Gaseous Species on each side of the Equation

    Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are three molecules on the left, but only two on the right. An increase in pressure would move the position of equilibrium to the right.

    A (g) +2 B (g) ⇌ C (g) + D (g) (3)

    (3)A(g)+2B(g)⇌C(g)+D(g)

    Because this is an all-gas equilibrium, it is much easier to use

    K p Kp : K p = P c P D P A P 2 B (4) (4)Kp=PcPDPAPB2

    Once again, it is easy to suppose that, because the position of equilibrium will move to the right if you increase the pressure,

    K p Kp

    will increase as well. Not so! To understand why, you need to modify the

    K p Kp

    expr ession. Remember the relationship between partial pressure, mole fraction and total pressure?

    P A

    =(mole fraction of A)(total pressure)

    (5)

    (5)PA=(mole fraction of A)(total pressure)

    P A = χ A P tot (6) (6)PA=χAPtot

    Replacing all the partial pressure terms in

    4 4 by mole fractions χ A χA

    and total pressure (

    P tot Ptot ) g ives you this: K p ( χ C P tot

    Source : chem.libretexts.org

    Chem 1102 Exam 3 Chapter 17 Flashcards

    Memorize flashcards and build a practice test to quiz yourself before your exam. Start studying the Chem 1102 Exam 3 Chapter 17 flashcards containing study terms like Which of the following would occur if solid NH4Cl was added to an aqueous solution of NH3?, Which of the following conjugate acid-base pairs will function as a buffer?, The pH of the equivalence point of a weak acid-strong base titration is _________. and more.

    Chem 1102 Exam 3 Chapter 17

    31 studiers in the last day

    Which of the following would occur if solid NH4Cl was added to an aqueous solution of NH3?

    Click card to see definition 👆

    The pH decreases.

    Click again to see term 👆

    Which of the following conjugate acid-base pairs will function as a buffer?

    Click card to see definition 👆

    HF/F-

    Click again to see term 👆

    1/13 Created by zaitawii

    Terms in this set (13)

    Which of the following would occur if solid NH4Cl was added to an aqueous solution of NH3?

    The pH decreases.

    Which of the following conjugate acid-base pairs will function as a buffer?

    HF/F-

    The pH of the equivalence point of a weak acid-strong base titration is _________.

    greater than 7

    Consider the following equilibrium:

    B(aq)+H2O(l)⇌HB+(aq)+OH−(aq)

    Suppose that a salt of HB+ is added to a solution of B at equilibrium.

    Will the equilibrium constant for the reaction increase, decrease, or stay the same?

    The equilibrium constant for the reaction will stay the same.

    Consider the following equilibrium:

    B(aq)+H2O(l)⇌HB+(aq)+OH−(aq)

    Suppose that a salt of HB+ is added to a solution of B at equilibrium.

    Will the concentration of B(aq) increase, decrease, or stay the same?

    The concentration of B(aq) will increase.

    Consider the following equilibrium:

    B(aq)+H2O(l)⇌HB+(aq)+OH−(aq)

    Suppose that a salt of HB+ is added to a solution of B at equilibrium.

    Will the pH of the solution increase, decrease, or stay the same?

    The pH of the solution will decrease.

    A buffer is made with sodium acetate (CH3COONa) and acetic acid (CH3COOH); the Ka for acetic acid is 1.80×10−5. The pH of the buffer is 3.98. What is the ratio of the equilibrium concentration of sodium acetate to that of acetic acid if you assume that the Henderson-Hasselbalch equation is accurate?

    0.174

    Calculate the pH of a solution that is 0.065 M in potassium propionate (C2H5COOK or KC3H5O2) and 0.080 M in propionic acid (C2H5COOH or HC3H5O2).

    pH = 4.80

    Calculate the pH of a solution that is 0.075 M in trimethylamine, (CH3)3N, and 0.11 M in trimethylammonium chloride, ((CH3)3NHCl).

    pH = 9.64 pH= 14−pOH = 14+log[OH−]

    = 14+log(Kb[(CH3)3N]/[(CH3)3NH+])

    = 14+log((6.4×10−5)(0.075)/0.11)

    Calculate the pH of a solution that is made by mixing 50.0 mL of 0.14 M acetic acid and 50.0 mL of 0.22 M sodium acetate.

    pH = 4.94

    Which of the following solutions is a buffer?

    0.10 MCH3COOH and 0.10 MCH3COONa

    The conjugate base of the weak acid is the majority species in solution at the equivalence point, and this conjugate base reacts with water to produce O H −.

    Therefore, the p H at the equivalence point for a weak acid/strong base titration is greater than 7.00.

    than 7.00.

    Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water?

    [Ag+]^3 [PO4^3−]

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