1. CHEMICAL
EQUILIBRIUM
notes

2. EQUILIBRIUM
In many chemical reactions both a forward and reverse reaction occur simultaneously.
When the rate of forward and reverse reactions are equalized, the system is at equilibrium.
Not all reactions can reach equilibrium. Consider a burning log. The products of the combustion (the smoke and ashes) can never reunite to form the reactants (the log and the oxygen)
By contrast some common processes are equilibrium systems, such as the formation of a saturated solution or the vaporization of a liquid in a sealed container.

3. RECOGNIZING EQUILIBRIUM
When equilibrium systems do exist they may be recognized by their apparent static nature (it looks as if nothing is happening). Additionally, all equilibrium systems must be closed, that is, nothing let in and nothing let out including energy.
While equilibrium appear unchanging when observed on a large scale (macroscopically) they are in constant change on the molecular level (microscopically). Reactants are continually forming products and products are continually forming reactants at equal rates.
Simply stated, equilibrium systems are macroscopically static and microscopically dynamic(changing).

4. SOLUBILITY EQUILIBRIUM
RATE OF DISSOLVING = RATE OF CRYSTALIZATION
RATE OF FORWARD PROCESS = RATE OF REVERSE PROCESS
The amount of substance dissolved remains constant
The system appears macroscopically static.

5. CHEMICAL EQUILIBRIUM
When the rates of opposing processes are equal, equilibrium has been established whether the system is physical or chemical.
Given the reaction: a A + b B  c C + d D when the rate at which A reacts with B equals the rate at which C reacts with D the system is at equilibrium.
Recall that the rate equation for a reaction is:
Rate = k x [reactants]n
At equilibrium Rate forward = Rate reverse or:
kf [reactants] = kr[products] and rearranging the equation:
kf / kr = [products] / [reactants] and a constant divided by a constant gives another constant so we get:
Ke = [PRODUCTS] / [REACTANTS] at equilibrium for the given reaction then is:
Ke = ([C]c x [D]d) / ([A]a x [B]b) notice that in this equation called the equilibrium expression, the coefficients of the balanced equation serve as powers.

6. THE EQUILIBRIUM EXPRESSION
a A + b B c C d D c d
K = [C] x [D] e
_________ b a
[A] x [B]

7. THE EQUILIBRIUM EXPRESSION
For a system at equilibrium, the value of the equilibrium constant (Ke) remains constant unless the temperature is changed. The concentrations of products and reactants may change as a result of “equilibrium shifts” but the ratio of products to reactants as calculated using the equilibrium expression remains unchanged.
Let’s find the equilibrium expression for the following reaction: 2 H2(g) + O2(g)  2 H2O(g)
(the double arrow  means equilibrium)
Ke = [H2O] 2 / ([H2]2 x [O2])
How about H2(g) + 1/2 O2(g)   H2O(g) ?
The reaction is the same but it is balanced with a different set of coefficients and therefore the equilibrium expression is different: Ke = [H2O] / ([H2] x [O2]1/2)

8. THE EQUILIBRIUM EXPRESSION
The coefficients used in balancing the reaction alters the equilibrium expression and also the value of the equilibrium constant.
For example, using the equilibrium equation H2(g) + I2(g)  2 HI(g) the equilibrium expression is
Ke = [HI] 2 / ([H2] x [I2])
Let us assume the following equilibrium concentrations [H2] = M, [I2] = and [HI] = M
Ke = (0.012)2 / ((0.094) x (0.094)) = 0.016
If the equation was balanced as ½ H2(g) + ½ I2(g)  HI(g) the equilibrium expression would be:
Ke = [HI] / ([H2]1/2 x [I2] 1/2 ) and
Ke = (0.012) / ((0.094)1/2 x (0.094)1/2 ) = 0.127
The numerical value of the equilibrium constant depends on the coefficients used to balance the equation!

9. HOW ARE Ke VALUES RELATED?
In our previous example when the coefficients of the balanced equation were halved the Ke value changed. But how?
H2(g) + I2(g)  2 HI(g) ,
Ke = [HI] 2 / ([H2] x [I2]) = 0.016
½ H2(g) + ½ I2(g)  HI(g) ,
Ke = [HI] / ([H2]1/2 x [I2] 1/2 ) = 0.127
Notice that the square root (half power) of = 0.127
If for some reason the equation was balanced as: H2(g) + 2 I2(g)  4 HI(g), the constant would change to the square of the original value or and Ke would equal 0.064
When the balancing coefficients of an equation are changed by a factor, the Ke value changes by that factor as a power of the original Ke

10. HOW ARE Ke VALUES RELATED?
How else may the original balanced equation be altered?
Often equations are reversed (interchanging products with reactants. For example:
H2(g) + I2(g)  2 HI(g) may be rewritten as:
2 HI(g)  H2(g) + I2(g)
The equilibrium expression for the first equation is:
Ke = [HI] 2 / ([H2] x [I2]) , for the second it is:
Ke = ([H2] x [I2]) / [HI] 2 , the reciprocal of the first
Therefore when an equation is reversed, the Ke becomes the reciprocal of the original Ke value.
Ke (reverse) = 1 / K (forward)

11. HETEROGENEOUS EQUILIBRIUM SYSTEMS
Heterogeneous and homogeneous are terms that are applied to the physical state (phases) of the components in equilibrium systems. A system consisting of all gases is heterogeneous while one including, for example, both solids and gases is heterogeneous.
H2(g) + I2(g)  2 HI(g) (homogeneous – all gases)
H2(g) + ½ O2(g)  H2O(l) (heterogeneous – gas & liquid)
Solids and liquids have concentrations which can vary only slightly due to temperature changes. For all practical proposes the concentrations of solids and liquids are fixed. Only gases and dissolved substances can change concentration. Since the equilibrium expression involves only components which change concentration, solids and liquids are not included in its format.

12. HETEROGENEOUS EQUILIBRIUM SYSTEMS
When solids or liquids are encountered in an equilibrium equation, a “1” is substituted in the equilibrium expression for that component.
2 H2(g) + O2(g)  2 H2O(l)
Ke = 1 / ([H2]2 x [O2])
For a system such as:
CaCO3(s)  CaO(s) + CO2(g)
The equilibrium expression is:
Ke = 1 x [CO2] / 1 or simply Ke = [CO2]
notice that since both CaCO3(s) and CaO(s) are solids they have been replaced by ones in the equilibrium expression.

13. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS
Equilibrium means that the rate of the forward reaction
and the rate of the reverse reaction are equal.
The concentrations of the products and the reactants need
not be equal and rarely are equal at equilibrium
When the equilibrium constant is small, the concentrations
of the reactants are large as compared to the products at
equilibrium. (The equilibrium favors the reactants).
[ PRODUCTS ]
Ke =
= a small value
__________
[ REACTANTS ]

14. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS
When the equilibrium constant is large, the concentrations
of the reactants are small as compared to the products at
equilibrium. (The equilibrium favors the products).
[ PRODUCTS ]
= a large value
Ke =
__________
[ REACTANTS ]

15. THE MAGNITUDE OF Ke AND EQUILIBRIUM CONCENTRATIONS
When the equilibrium constant is about one, the
Concentrations of the reactants and the products are
about equal at equilibrium.
[ PRODUCTS ]
Ke =
= about one (1)
__________
[ REACTANTS ]

16. MEASURING Ke VALUES *
Since concentration may be measured for solutions in moles per liter and for gases in terms of pressures (atms) or moles per liters two unit systems are possible for Ke.
When the concentration values used in calculating Ke are in terms of molar units the constant is referred to as Kc (for concentration units – moles per liter). When concentration of gases is measured in pressure units (atms) the constant is referred to as Kp (for pressure units – atms).
The equilibrium constant in concentration terms (Kc) can be converted to an equivalent value in pressure units (Kp).
Kp = Kc(R x T) n , in this equation:
R = atm l / moles K, T = Kelvin temperature
n = moles of gaseous products – moles of gaseous reactants in the balanced equation.

17. CONVERTING KP AND KC VALUES*
Problem: For the reaction:
N2(g) H2(g)  2 NH3(g) , Kc = at 472 0C. Find Kp at this temperature.
Solution:
Kp = Kc(R x T) n , R = atm x l / mole x K
T= = 745 K, n = 2 – 4 = (two moles of NH3(g) product gases) – (one mole of N2(g) and three moles of H2(g) a total of four moles of reactant gases)
Kp = x ( x 745)-2 = 2.81 x 10-5

18. CALCULATING THE EQUILIBRIUM CONSTANT
Problem: If 0.20 moles of PCl3(g) and 0.10 moles of Cl2(g) are placed in a 1.0 liter container at 200 0C the equilibrium concentration of PCl5 is found to be molar. What is Kc for the reaction:
PCl3(g) + Cl2(g)  PCl5(g)
Solution: Since one PCl3 and one Cl2 must be consumed to form one PCl5, (0.20 M, the starting molarity of PCl3 – M, the molarity of PCl5 formed) = M PCl3 remains and
(0.10 M, the starting molarity of Cl2 – M the molarity of PCl5 formed) = M Cl2 remaining..
Kc = [PCl5] / ([PCl3] x [Cl2])
Kc = (0.012) / (0.188 x 0.088) = 0.725
NOTE: The temperature given in the problem is not used in the calculation. It is refer data only. Temperatures are not used in equilibrium calculations of this type!

19. USING THE EQUILIBRIUM CONSTANT
Problem: For the reaction: H2(g) + I2(g)  2 HI(g) , Kc = at 793 K. If 1.00 mole of hydrogen and 1.00 mole of iodine are mixed in a 10.0 liter container, what is the equilibrium concentration of all components?
Solution: H2(g) + I2(g)  2 HI(g) and therefore,
Ke = [HI] 2 / ([H2] x [I2])
The starting [H2] and [I2] both equal 1.00 mole / 10.0 liters or M.
The concentration of HI at equilibrium is unknown (X).
In order to form 2 HI, only 1 H2 and 1 I2 (half as many of each) are needed. The equilibrium [H2] and [I2] are both then ( X). This represents the starting concentrations of each minus the concentration used to form the HI.

20. USING THE EQUILIBRIUM CONSTANT (cont’d)
Substituting the equilibrium concentrations into the equilibrium expression we get:
Kc = X2 / (0.10 – 0.5 X)(0.10 – 0.5X) = or
X2 / (0.10 – 0.5X)2 = 0.016, taking the square root of both sides of the equation we get:
X / (0.10 – 0.5X) = and X = – 0.008X
Rearranging the equation to solve for X we get:
1.008X = , X = / =
[HI]e = M,
[H2]e = [I2]e = 0.10 – .5(0.0125) = M
CHECK – Placing the equilibrium concentrations back into the equilibrium expression should give the correct constant!
(0.0125)2 / (0.938)2 = a very close approximation of Kc (0.016) considering the rounding errors!

21. USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT*
Equilibrium calculations often require the use of the quadratic equation. Its function is to solve mathematical equations involving squared, first power and zero power terms in the same equation such as:
4X X – 3.2 x = 0 this equation cannot be solved easily by inspection and requires the quadratic formula:
Using the form aX2 + bX + c = 0 the formula is:
( - b + b2 – 4ac )/ 2a
In the given equation: a = 4, b = and c = – 3.2 x 10-4
( (0.0048)2 – 4(4)(– 3.2 x 10-4 )) / 2(4) =
Note: every quadratic has two answers.

22. USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT (cont’d)*
Problem: Given the reaction
H2(g) + I2(g)  2 HI(g) , assume the original [H2] = M and [I2] = M. The Kc value is at 793 K. Find all equilibrium concentrations.
Solution: Let X = [H2] consumed to reach equilibrium.
For every H2 consumed so is one I2 therefore X also = [I2] consumed.
For every H2 and I2 consumed 2 HI are formed. Therefore [HI]e = 2X, and [H2]e = (0.200 –X) and [I2]e = (0.100 – X)
Ke = [HI] 2 / ([H2] x [I2]) = (2X)2 / (0.200 – X)(0.100 –X)
Kc = 4X2/ ( X + X2) = 0.016
4X2 = 0.016( X + X2)
4X2 = X X2

23. USING THE QUADRATIC EQUATION WITH THE EQUILIBRIUM CONSTANT (cont’d)*
3.984X X = 0
a = , b = , c =
(- b + b2 – 4ac) / 2a
( (0.0048)2– 4(3.984)( )) /2(3.984)
X = ( ) / =
NOTE: A negative X value means less than no H2 was consumed which is of course impossible. Therefore the positive value must be correct!
[HI]e = 2 x = M
[H2]e = – = M
[I2]e = – = M
CHECK: (0.017)2 / (0.192)(0.092) = ~ Kc = 0.016

24. REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM*
It is often important to determine whether a reacting system is at equilibrium, and if it is not, how will it change as it moves towards the equilibrium state.
In order to make this determination, a value called the reaction quotient. The reaction quotient is a mathematical setup just like the equilibrium expression except equilibrium concentrations are no necessarily used in its calculation.
Q = [products] / [reactants] but not necessarily at equilibrium!
The value of Q is compared the the equilibrium constant Ke to decide if the reaction is already at equilibrium and in which direction the system will move to attain equilibrium.

25. REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM*
If the value of Q > Ke the reaction is not at equilibrium and will favor the reverse reaction to reach equilibrium.
If the value of Q < Ke the reaction is not at equilibrium and will favor the forward reaction to reach equilibrium.
If the value of Q = Ke the reaction is already at equilibrium and no net reaction will occur to reach equilibrium.

26. REACTION QUOTIENTS AND THE DIRECTION OF EQUILIBRIUM*
Problem: Given the following reaction:
2 HF(g)  H2(g) + F2(g) , Ke = 1 x at 1000 oC
[HF] = 0.50 M, [H2] = M, [F2] = M
Is the system at equilibrium? If not in which direction will it move to attain equilibrium?
Solution: Q = [HF] 2 / ([H2] x [F2])
Q = (0.50)2 / ( x ) = 1.6 x 10-5
Since Q does not equal Ke the system is not at equilibrium
Q > Ke therefore the system will move in reverse until equilibrium is reached. HF will be formed and H2 and F2 will be consumed!

27. LeChatelier’s Principle and Chemical Equilibrium
LeChatelier’s Principle how equilibrium systems change when equilibrium conditions are changed.
The changes imposed on an equilibrium are called stresses. The response to these changes (stresses) are called shifts.
Stresses consist of changes in concentrations, pressures and temperatures. Shifts that occur are characterized as “shift towards products” (shift right) which means increased rate of forward reaction occurs or “shift towards reactants” (shift left) which means increased rate of reverse reaction occurs.
LeChatelier’s Principle defines the relationship between applied stresses and equilibrium shifts which result.

28. LeChatelier’s Principle and Chemical Equilibrium
In the reaction of A + B  C + D if the concentration of a reactant (A or B) increases, the collision rate between reactant molecules increases and the forward rate therefore increases. As the reaction continues and reactants are consumed the forward rate slows. At the same time more product is formed and the reverse reaction rate increases. At some point the rates forward and reverse again become equal and a new equilibrium is reached. At this new equilibrium however the concentrations of products are larger and the concentrations reactants are lower than in the original reaction. This is an equilibrium shift toward products.

29. LeChatelier’s Principle and Chemical Equilibrium
C + D A
A + B C D
A stress is applied, the concentration
of reactant A is increased
Equilibrium shifts forward and concentrations of products
(C and D) increases. Concentration of reactant B is reduced
because it must be consumed to allow the shift forward.
The concentration of A is of course larger than the original
because it was added although some of it is consumed in
the forward shift.The equilibrium constant value remains
unchanged despite the shift!

30. LeChatelier’s Principle and Chemical Equilibrium
In the reaction of A + B  C + D if the concentration of a product (C or D) increases, the collision rate between product molecules increases and the reverse rate therefore increases. As the reaction continues and products are consumed the reverse rate slows. At the same time more reactant is formed and the forward reaction rate increases. At some point the rates forward and reverse again become equal and a new equilibrium is reached. At this new equilibrium however the concentrations of reactants are larger and the concentrations products are lower than in the original reaction. This is an equilibrium shift toward reactants.

31. A + B C A + B C + D LeChatelier’s Principle and Chemical Equilibrium
A stress is applied, the concentration
of product C is increased
Equilibrium shifts reverse and concentrations of reactants
(A and B) increases. Concentration of reactant D is reduced
because it must be consumed to allow the shift reverse.
The concentration of C is of course larger than the original
because it was added although some of it is consumed in
the reverse shift.The equilibrium constant value remains
unchanged despite the shift!

32. D A + B C + D C C A B LeChatelier’s Principle and Chemical Equilibrium
A stress is applied, the concentration
of product C is decreased
The reaction rate in the forward direction remains unchanged
because reactant concentrations are not changed. The rate of
the reverse reaction slows due to the reduced concentration
of C and the net reaction shifts forward making more D
and also increasing C while consuming reactants A and B.
The value of the equilibrium constant remains unchanged. D
A B C D C C A B

33. LeChatelier’s Principle and Chemical Equilibrium
When changing temperature is the applied stress, the exothermic or endothermic nature of the reaction must be considered.
When heat is added to endothermic reactions they shift forward. When heat is removed from endothermic reactions they shift in reverse.
When heat is added to an exothermic reactions they shift reverse. When heat is removed from exothermic reactions they shift forward.
Endothermic reactions are identified with a positive H or heat written into the equation as a reactant. Exothermic reactions are identified with a negative H or heat written into the equation as a product.

34. LeChatelier’s Principle and Chemical Equilibrium
A + B
heat
A + B C + D + heat
Exothermic reactions shift reverse when heat is added
Concentrations of A and B increase, C and D decreases
C + D
A + B C + D + heat
A + B
heat
Endothermic reactions shift forward when heat is removed
Concentrations of C and D increases, A and B decreases.

35. LeChatelier’s Principle and Chemical Equilibrium
In general, Le Chatelier’s Principle tells us that equilibrium systems when disturbed by an applied stress, shift away from increased concentrations and towards reduced concentrations.
Similarly, systems shift away from added heat (increased temperatures) and towards removed heat (lower temperatures).
It is important to note that altering concentrations shift equilibriums but DO NOT CHANGE the value of the equilibrium constant. Changing temperatures cause equilibrium shifts but DO CHANGE the value of the equilibrium constant.

36. LeChatelier’s Principle and Chemical Equilibrium
In addition to concentrations and temperatures often pressures are changed in an equilibrium system.
LeChatelier’s Principle can be applied to these system also. When gaseous systems are compressed the equilibrium shifts towards the side of the reaction containing the fewer number of gas phase molecules.
When gaseous systems are expanded, the equilibrium shifts towards the side of the reaction containing the greater number of gas phase molecules
For example: N2(g) H2(g)  2 NH3(g)
compressing this system would cause a shift to the right (products) which contains only two gas phase molecules (2 ammonia NH3).
Expanding this system would cause a shift to the left (reactants) which contains four gas phase molecules (1 nitrogen and 3 hydrogen)

37. LeChatelier’s Principle and Chemical Equilibrium
Problem: Given the following reaction:
2 H2(g) + O2(g)  2 H2O(l) + heat
How can the amount of water formed be increased?
Solution:
Increasing the concentration of hydrogen will cause an equilibrium shift forward.(more water formed)
Increasing the concentration of oxygen will cause an equilibrium shift forward. (more water formed)
The reaction is exothermic. Removing heat (lowering the temperature) will cause a shift forward.
Compressing the system will cause a shift forward (two gas phase molecules convert to zero gas phase molecules)
Removing liquid water from the system will not cause a equilibrium shift. Remember that liquids and solids are not involved in equilibrium systems!

38. SUMMARY OF KEY IDEAS (1) At equilibrium
Rate forward Reaction = Rate reverse Reaction
(2) For the reaction: a A + b B  c C + d D
Ke = ([C]c x [D]d) / ([A]a x [B]b)
(3) The numerical value of the equilibrium constant depends on the coefficients used to balance the equation!
(4) When the balancing coefficients of an equation are changed by a factor, the Ke value changes by that factor as a power of the original Ke
(5) Therefore when an equation is reversed, the Ke becomes the reciprocal of the original Ke value.
(6) Since the equilibrium expression involves only components which change concentration, solids and liquids are not included in its format.

39. SUMMARY OF KEY IDEAS
(7) Large equilibrium constants mean products are favored at equilibrium. Small values mean reactants are favored. Constants of about one mean products and reactants are present in approximately equal concentrations at equilibrium
(8) Kp = Kc(R x T) n (converting Kp to Kc)
(9) The quadratic formula is sometimes required for problem solution: aX2 + bX + c = 0
( - b + b2 – 4ac )/ 2a
(10) Q = [products] / [reactants] but not necessarily at equilibrium!
(11) Q> Ke (reactants favored), Q< Ke (products favored), Q = Ke (the system is at equilibrium)

40. SUMMARY OF KEY IDEAS (12) LeChatelier’s Principle
Concentration of reactants increases, shift towards the products
Concentration of products increases, shift towards the reactants
Exothermic reactions, add heat, shift toward reactants
Endothermic reactions, add heat, shift toward products
Compress a gaseous system, shift toward fewer gas molecules
Expand a gaseous system, shift toward more gas molecules

41. THE END