Equilibrium

1
Equilibrium

• Just the Beginning

2
Reactions are reversible

• A B C D ( forward)
• C D A B (reverse)
• Initially there is only A and B so only the
  forward reaction is possible
• As C and D build up, the reverse reaction speeds
  up while the forward reaction slows down.
• Eventually the rates are equal

3
Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time
4
What is equal at Equilibrium?

• Rates are equal
• Concentrations are not.
• Rates are determined by concentrations and
  activation energy.
• The concentrations do not change at equilibrium.
• or if the reaction is verrrry slooooow.

5
Law of Mass Action

• For any reaction
• jA kB lC mD
• K ClDm PRODUCTSpower
  AjBk REACTANTSpower
• K is called the equilibrium constant. It has no
  units.
• is how we indicate a reversible
  reaction

6
Try one together

• 2SO2(g) O2(g) 2SO3(g)
• where the equilibrium concentrations are
• SO2 1.50M O2 1.25M and SO3 3.5M
• K SO32 / SO22O2
• K (3.50)2/(1.50)2(1.25)
• K4.36

7
Try another one together

• 2SO2(g) O2(g) 2SO3(g)
• where the equilibrium concentrations are
• SO2 1.50M O2 1.25M and SO3 3.5M
• Now find the k for the reverse reaction
• K SO22 O2/ SO32
• K (1.50)2(1.25)/(3.50)2
• K0.230

8
Playing with K

• If we write the reaction in reverse.
• Then the new equilibrium constant is
• K 1/K

9
Try one together

• SO2(g) 1/2 O2(g) SO3(g)
• where the equilibrium concentrations are
• SO2 1.50M O2 1.25M and SO3 3.5M
• K SO3 / SO2O21/2
• K (3.50)/(1.50)(1.25)1/2
• K 2.09

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Playing with K

• If we multiply the equation by a constant, n
• Then the equilibrium constant is
• K Kn

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K is CONSTANT

• At any temperature.
• Temperature affects rate.
• The equilibrium concentrations dont have to be
  the same, only K.
• Equilibrium position is a set of concentrations
  at equilibrium.
• There are an unlimited number.

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Equilibrium Constant

• One for each Temperature

13
Calculate K

• N2 3H2 2NH3
• Initial At Equilibrium
• N20 1.000 M N2 0.921M
• H20 1.000 M H2 0.763M
• NH30 0 M NH3 0.157M
• K 6.03 x 10-2

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Calculate K

• N2 3H2 2NH3
• Initial At Equilibrium
• N20 0 M N2 0.399 M
• H20 0 M H2 1.197 M
• NH30 1.000 M NH3 0.203M
• K 6.02 x 10-2

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Equilibrium and Pressure

• Some reactions are gaseous
• PV nRT
• P (n/V)RT
• P CRT
• C is a concentration in moles/Liter
• C P/RT

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Equilibrium and Pressure

• 2SO2(g) O2(g) 2SO3(g)
• Kp (PSO3)2 (PSO2)2 (PO2)
• K SO32 SO22 O2

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Equilibrium and Pressure

• K (PSO3/RT)2 (PSO2/RT)2(PO2/RT)
• K (PSO3)2 (1/RT)2
  (PSO2)2(PO2) (1/RT)3
• K Kp (1/RT)2 Kp RT (1/RT)3

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General Equation

• jA kB lC mD
• Kp (PC)l (PD)m (CCxRT)l (CDxRT)m (PA)j
  (PB)k (CAxRT)j(CBxRT)k
• Kp (CC)l (CD)mx(RT)lm (CA)j(CB)kx(RT)jk
• Kp K (RT)(lm)-(jk) K (RT)Dn
• Dn(lm)-(jk)Change in moles of gas

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Homogeneous Equilibria

• So far every example dealt with reactants and
  products where all were in the same phase.
• We can use K in terms of either concentration or
  pressure.
• Units depend on reaction.

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Heterogeneous Equilibria

• If the reaction involves pure solids or pure
  liquids the concentration of the solid or the
  liquid doesnt change.
• As long as they are not used up we can leave them
  out of the equilibrium expression.
• For example

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For Example

• H2(g) I2(s) 2HI(g)
• K HI2 H2I2
• But the concentration of I2 does not change.
• K HI2 H2

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• Write the equilibrium constant for the
  heterogeneous reaction

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The Reaction Quotient, Q

• Tells you the direction the reaction will go to
  reach equilibrium
• Calculated the same as the equilibrium constant,
  but for a system not at equilibrium
• Q Productscoefficient Reactants
  coefficient
• Compare value to equilibrium constant

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What Q tells us

• If QltK
• Not enough products
• Shift to right
• If QgtK
• Too many products
• Shift to left
• If QK system is at equilibrium

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Example

• for the reaction
• 2NOCl(g) 2NO(g) Cl2(g)
• K 1.55 x 10-5 M at 35ºC
• In an experiment 0.10 mol NOCl, 0.0010 mol NO(g)
  and 0.00010 mol Cl2 are mixed in 2.0 L flask.
• Which direction will the reaction proceed to
  reach equilibrium?

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Solution

• K 1.55 x 10-5 M
• Q NO2Cl / NOCl2
• Q (.001mol/2.0L)2(.0001mol/2.0L) /
  (.1mol/2.0L)2
• Q (5 x 10 -4)2(5 x 10-5) / (5 x 10-2)2
• Q 5 x 10-9
• QltK, therefore more product will need to be
  formed to reach equilibrium

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Solving Equilibrium Problems

• Given the starting concentrations and one
  equilibrium concentration.
• Use stoichiometry to figure out other
  concentrations and K.
• Learn to create a table of initial and final
  conditions.

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• Consider the following reaction at 600ºC
• 2SO2(g) O2(g) 2SO3(g)
• In a certain experiment 2.00 mol of SO2, 1.50 mol
  of O2 and 3.00 mol of SO3 were placed in a 1.00 L
  flask. At equilibrium 3.50 mol of SO3 were found
  to be present. Calculate
• The equilibrium concentrations of O2 and SO2, K
  and KP

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Solution

• K SO32 / SO22O2
• K (3/3.5)2/(2/3.5)2(1.5/3.5)
• K (.857)2 / (.571)2 (.429)
• K 5.25
• Kp k(RT)Dn
• Kp 5.25.0821873-1
• Kp .073