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Equilibrium. Chapter 12. 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.

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equilibrium

Equilibrium

Chapter 12

reactions are reversible
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
slide3

Forward Reaction

Reaction Rate

Equilibrium

Reverse reaction

Time

what is equal at equilibrium
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 very slow, no significant change will occur.
law of mass action
Law of Mass Action
  • For any reaction
  • aA + bB cC + dD
  • K = [C]c[D]d PRODUCTSpower [A]a[B]b REACTANTSpower
  • K is called the equilibrium constant.
  • is how we indicate a reversible reaction
calculating k
calculating K
  • If we write the reaction in reverse.
  • cC + dD aA + bB
  • Then the new equilibrium constant is
  • K’ = [A]a[B]b = 1/K[C]c[D]d
calculating with k
Calculating with K
  • If we multiply the equation by a constant
  • naA + nbB ncC + ndD
  • Then the equilibrium constant is
  • K’ = [A]na[B]nb= ([A] a[B]b)n= Kn[C]nc[D]nd ([C]c[D]d)n
adding k
Adding K
  • When we have two equations which we are adding to get a resulting equation
  • We will need to multiply the K from each equation.
  • K1 X K2 = K3 (sum of equations 1 +2)
the units for k
The units for K
  • Are determined by the various powers and units of concentrations.
  • They depend on the reaction.
  • Will usually cancel out to leave no units.
k is a constant
K is a CONSTANT
  • At each temp, the reaction has a unique K that won’t change for that temp..
  • Temperature affects rate and K.
  • The equilibrium concentrations won’t be the same, only K.
  • Equilibrium position is a set of concentrations at equilibrium.
  • There are an unlimited number equilibrium positions.
calculate k
Calculate K
  • N2 + 3H2 2NH3
  • Initially At Equilibrium
  • [N2]0 =1.000 M [N2] = 0.921M
  • [H2]0 =1.000 M [H2] = 0.763M
  • [NH3]0 =0 M [NH3] = 0.157M
calculate k1
Calculate K
  • N2 + 3H2 2NH3
  • Initial At Equilibrium
  • [N2]0 = 0 M [N2] = 0.399 M
  • [H2]0 = 0 M [H2] = 1.197 M
  • [NH3]0 = 1.000 M [NH3] = 0.203M
  • K is the same no matter what the amount of starting materials
equilibrium and pressure
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
equilibrium and pressure1
Equilibrium and Pressure
  • 2SO2(g) + O2(g) 2SO3(g)
  • Kp = (PSO3)2 (PSO2)2 (PO2)
  • K = [SO3]2 [SO2]2 [O2]
equilibrium and pressure2
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
general equation
General Equation
  • aA + bB cC + dD
  • Kp= (PC)c(PD)d=(CCxRT)c(CDxRT)d(PA)a(PB)b(CAxRT)a(CBxRT)b
  • Kp=(CC)c(CD)dx(RT)c+d(CA)a (CB)bx(RT)a+b
  • Kp =K (RT)(c+d)-(a+b) = K (RT)Dn
  • Dn=(c+d)-(a+b)=Change in moles of gas
homogeneous equilibria
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.
heterogeneous equilibria
Heterogeneous Equilibria
  • If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesn’t change.
  • So, we can leave them out of the equilibrium expression.
for example
For Example
  • H2(g) + I2(s) 2HI(g)
  • K = [HI]2 [H2][I2]
  • But the concentration of I2 does not change.
  • K= [HI]2 [H2]
the reaction quotient q
The Reaction Quotient (Q)
  • Tells you the directing the reaction will go to reach equilibrium
  • Calculated the same as the equilibrium constant, but for a system not at equilibrium
  • Q = [Products]coefficient [Reactants] coefficient
  • Compare value to equilibrium constant
what q tells us
What Q tells us
  • If K>Q

Not enough products

Shift to right

  • If K<Q

Too many products

Shift to left

  • If Q=K system is at equilibrium
example
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?
solving equilibrium problems
Solving Equilibrium Problems
  • Given the starting concentrations and one equilibrium concentration.
  • Use stoichiometry to figure out other concentrations and K.
slide24
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
slide25
Consider the same reaction at 600ºC
  • In a different experiment .500 mol SO2 and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present.
  • Calculate the final concentrations of SO2 and SO3 and K
what if you re not given an equilibrium concentration
What if you’re not given an equilibrium concentration?
  • The size of K will determine what approach to take.
  • First let’s look at the case of a LARGE value of K ( >100).
  • Allows us to make simplifying assumptions.
example1
Example
  • H2(g) + I2(g) 2HI(g)
  • K = 7.1 x 102 at 25ºC
  • Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.9 g of H2 294 g I2 .
  • [H2]0 = (15.7g/2.02)/5.00 L = 1.56 M
  • [I2]0 = (294g/253.8)/5.00L = 0.232 M
  • [HI]0 = 0
slide28
Q= 0, which is <K so more product will be formed.
  • Assume that since K is large, then the reaction will go to completion.
  • Set up table of initial, final and equilibrium in concentrations.
slide29

H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2Xequil.

  • Using to stoichiometry we can find
  • Change in H2 = 1.56M -X
  • Change in HI = twice change in H2
  • Change in HI = 0 +2X
slide30

H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M change -X -X +2X

equil 1.56-X 0.232-X +2X

  • Now plug these values into the equilibrium expression
  • K = (2X)2 = 7.1 x 102 (1.56-X)(0.232-X)
slide31
When we solve for X we get 0.232
  • So we can find the other concentrations
  • I2 = 0 M
  • H2 = 1.328 M
  • HI = 0.464 M
practice
Practice
  • For the reaction Cl2 + O2 2ClO(g) K = 156
  • In an experiment 0.100 mol ClO, 1.00 mol O2 and 0.0100 mol Cl2 are mixed in a 4.00 L flask.
  • If the reaction is not at equilibrium, which way will it shift?
  • Calculate the equilibrium concentrations.
for example1
For example
  • For the reaction 2NOCl 2NO +Cl2
  • K= 1.6 x 10-5
  • If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl2 are mixed in a 1 L container
  • What are the equilibrium concentrations
  • Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 M [NOCl]2 (1.20)2
slide34

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change

Equil

slide35

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change -2X +2X +X

Equil 1.20-2X 0.45+2X 0.87+X

slide36

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change -2X +2X +X

Equil 1.20-2X 0.45+2X 0.87+X

  • Now we can write equilibrium constant
  • K = (.45 + 2X)2 (0.87 + X) (1.20 – 2X)2 = 1.6E-5
practice problem
Practice Problem
  • For the reaction 2ClO(g) Cl2 (g) + O2 (g)
  • K = 6.4 x 10-3
  • In an experiment 0.100 mol ClO(g), 1.00 mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container.
  • What are the equilibrium concentrations.
example2
Example
  • H2(g) + I2 (g) 2 HI(g) K=38.6
  • What is the equilibrium concentrations if 1.800 mol H2, 1.600 mol I2 and 2.600 mol HI are mixed in a 2.000 L container?
problems involving pressure
Problems Involving Pressure
  • Solved exactly the same, with same rules for choosing X depending on KP
  • For the reaction N2O4(g) 2NO2(g) KP = .131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N2O4?
le chatelier s principle
Le Chatelier’s Principle
  • If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress.
  • 3 Types of stress
change amounts of reactants and or products
Change amounts of reactants and/or products
  • Adding product makes K<Q
  • Removing reactant makes K<Q
  • Adding reactant makes K>Q
  • Removing product makes K>Q
  • Determine the effect on Q, will tell you the direction of shift
change pressure
Change Pressure
  • By changing volume
  • System will move in the direction that has the least moles of gas.
  • Because partial pressures (and concentrations) change a new equilibrium must be reached.
  • System tries to minimize the moles of gas.
change in pressure
Change in Pressure
  • By adding an inert gas
  • Partial pressures of reactants and product are not changed
  • No effect on equilibrium position
change in temperature
Change in Temperature
  • Affects the rates of both the forward and reverse reactions.
  • Doesn’t just change the equilibrium position, changes the equilibrium constant.
  • The direction of the shift depends on whether it is exo- or endothermic
exothermic
Exothermic
  • DH<0
  • Releases heat
  • Think of heat as a product
  • Raising temperature push toward reactants.
  • Shifts to left.
endothermic
Endothermic
  • DH>0
  • Produces heat
  • Think of heat as a reactant
  • Raising temperature push toward products.
  • Shifts to right.
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