Changing Concentrations
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Transcript of Changing Concentrations
Changing ConcentrationsConcentrations MREACTANTSProductsTIME
Changing Rates
RATErate of forward reactionrate of reverse reactionTIME
• R P• Equilibrium lies to right (favors
products or forward rxn)
• R P
• Equilibrium lies to left (favors
reactants or reverse rxn )
• A<---->B
• kf [A] = kr [B]
• kf/kr= [B]/[A]
• k(eq) =[B]/[A]
• k(eq) = equilibrium constant
aA+ bB <-------> cC + dD• R and P are either g or aq
• Equilibrium expression
• K (eq) = [C]c[D]d = [products]
• [A]a[B]b [reactants]
• solids and liquids are not included in expression• If 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 right• if K eq is greater than 105 the reaction essentially
goes to completion• The larger the K eq, the more the products are favored
• If K eq < 10-3 lot more reactants• equilibrium lies to the left• A 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).
substance Moles Initial Moles React Moles at equilibrium
Molarity atEquilibrium
SO2
O2
SO3
Experiment Initial [NO2]
Initial [N2O4]
Equilibrium [NO2 ]
Equilibrium [N2O4]
Keq
1 0.0200 0.0 0.0172 0.00140
2 0.0300 0.0 0.0243 0.00280
3 0.0 0.0200 0.0310 0.00452
2NO2 <-------->N2O4
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 -1
Is the reaction at equilibrium?If not, which way does it need to shift?
Le Chatelier’s Principle
• If 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 same
• K(eq) only changes when temperature changes
A(aq) + B(g) <-----> 2C(aq) + D(aq)
• If D is added
• What 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) ΔE = -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) ΔE = +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 EQUILIBRIA
• When a solution is saturated it is at equilibrium.
• equilibrium expressions for ionic solids dissolving in water
• PbF2(s) <-->Pb2+(aq) + 2F-(aq)
• Keq = equilibrium constant
• Ksp = 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 -14
• Write 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 Ksp
• Ksp for BaSO4 is 1.6 x 10 -9
• What is the solubility of BaSO4 in a 0.02M Ba(NO3)2 solution?