changing concentrations

Download Changing Concentrations

Post on 03-Feb-2016

29 views

Category:

Documents

0 download

Embed Size (px)

DESCRIPTION

Changing Concentrations. Changing Rates. R  P Equilibrium lies to right (favors products or forward rxn) R  P Equilibrium lies to left (favors reactants or reverse rxn ). AB k f [A] = k r [B] k f /k r = [B]/[A] - PowerPoint PPT Presentation

TRANSCRIPT

  • Changing Concentrations

  • Changing Rates

  • R P Equilibrium lies to right (favors products or forward rxn)

    R P Equilibrium lies to left (favors reactants or reverse rxn )

  • AB kf [A] = kr [B]

    kf/kr= [B]/[A] k(eq) =[B]/[A] k(eq) = equilibrium constant

  • aA+ bB cC + dDR and P are either g or aqEquilibrium expressionK (eq) = [C]c[D]d = [products] [A]a[B]b [reactants]solids and liquids are not included in expressionIf the only product is a solid or liquid- substitute the number 1 for concentration

  • \If K eq > 1000 (103) lot more products equilibrium lies to the rightif K eq is greater than 105 the reaction essentially goes to completionThe larger the K eq, the more the products are favored

    If K eq < 10-3 lot more reactantsequilibrium lies to the leftA very very small Keq indicates that the reaction does not take place.The smaller the Keq the more the reactants are favored

  • 2 SO2(g) + O2(g) 2 SO3(g) 3.0 moles of SO2 and 1.5 mole of O2 are placed in a one liter container and react. At equilibrium 1.8 moles of SO3 are present. Determine the equilibrium constant (Keq).

    substanceMoles InitialMoles ReactMoles at equilibriumMolarity atEquilibriumSO2 O2 SO3

  • 2NO2 N2O4

    ExperimentInitial [NO2] Initial [N2O4] Equilibrium [NO2 ]Equilibrium [N2O4] Keq 10.0200 0.00.01720.00140 2 0.0300 0.0 0.02430.00280 30.0 0.02000.03100.00452

  • PCl5(g) PCl3(g) + Cl2(g)

    K (eq) = 2.2 x 10-2

    If concentrations are: [PCl5 ] = 2.1 x10 -2 [PCl3 ] = 2.0 x10 -1 [Cl2 ] = 2.0 x10 -1Is the reaction at equilibrium?If not, which way does it need to shift?

  • Le Chateliers PrincipleIf a reaction at equilibrium is disturbed, the reaction returns to equilibrium by shifting in such a direction as to partially undo (oppose) the disturbance.New concentrations (equilibrium positions) are established, but the K(eq) remains the sameK(eq) only changes when temperature changes

  • A(aq) + B(g) 2C(aq) + D(aq)

    If D is addedWhat will the system want to do with the added D?______Which way will the equilibrium shift? ________At new equilibrium positions how will concentrations compare to original concentrations? [A] _________[B] _________[C] _________[D] _________What if D was a solid?

  • A(aq) + B(g) 2C(aq) + D(aq)

    If the pressure is increased (for pressure changes look only at the number of gas molecules on each side of the equation)What will the system want to do when the pressure is increased? ___________Which way will the equilibrium shift? ________How could you change the pressure ? ____________

  • A(aq) + B(g) 2C(aq) + D(aq) E = -100 kJ

    If heat is added What will the system want to do with the added heat? ________Which way will the equilibrium shift? ________At new equilibrium positions how will concentrations compare to original concentrations? [A] _________[B] _________[C] _________[D] _________

  • A(aq) + B(g) 2C(aq) + D(aq) E = +100 kJ

    If heat is added What will the system want to do with the added heat? ________Which way will the equilibrium shift? ________At new equilibrium positions how will concentrations compare to original concentrations? [A] _________[B] _________[C] _________[D] _________

  • A(aq) + B(g) 2C(aq) + D(aq)

    If A is added What will the system want to do with the added A? ________Which way will the equilibrium shift? ________At new equilibrium positions how will concentrations compare to original concentrations? [A] _________[B] _________[C] _________[D] _________

  • SOLUBILITY EQUILIBRIAWhen a solution is saturated it is at equilibrium.equilibrium expressions for ionic solids dissolving in water

  • PbF2(s) Pb2+(aq) + 2F-(aq)Keq = equilibrium constantKsp = solubility product constant

    A small Ksp - reactants favored (not much solid dissolves).A large Ksp -products favored (solid dissolves.)

  • From solubility (how much dissolves) can find the solubility product constant. (Ksp)

    From the solubility product constant (Ksp) can find the solubility

  • If the solubility of PbF2 is 1.9 x 10-2 M, what is the solubility product constant (Ksp)?

    Write the equation for the reaction.

    Write the Ksp expression.

    Determine the concentration (M) of each ion.

    Solve

  • If the solubility product constant (Ksp) for BaSO4 is 1.6 x 10 -9,what is the solubility of BaSO4?

    Write the equation for the reaction.

    Write the Ksp expression.

    Let the solubility of BaSO4 equal X.

    Solve

  • Iron(II) hydroxide has a solubility product constant (Ksp) of 1.6 x10 -14Write the equation for the dissolving of iron(II) hydroxide.

    Write the Ksp expression for the reaction

    What is the solubility of iron(II) hydroxide?

  • What is the concentration of each ion?

    How many grams of iron(II) hydroxide can dissolve in 3.5 liters of solution? (total volume)

  • Calcium phosphate has a solubility product constant of1.2 x 10 -26. Determine the solubility of calcium phosphate.

  • Common ion effect and KspKsp for BaSO4 is 1.6 x 10 -9What is the solubility of BaSO4 in a 0.02M Ba(NO3)2 solution?

    **

    *

    *

    ***

    .ANSWER 3.75If the initial moles of the reactants are changed how will that affect the value of the equilibrium constant (Keq) ? _________

    **

    ****************

View more