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Chemical Equilibrium
Equilibrium
Fast Initial Step
• Because Ratef = Rater ,
Solving for [NOBr2] gives us
Can also write as:
Equilibrium
Equilibrium
The Concept of Equilibrium
Chemical equilibrium occurs when a
reaction and its reverse reaction proceed at
the same rate.
The Concept of Equilibrium
• As a system
approaches equilibrium,
both the forward and
reverse reactions are
occurring.
• At equilibrium, the
forward and reverse
reactions are
proceeding at the same
rate.
Equilibrium
A System at Equilibrium
Once equilibrium is
achieved, the
amount of each
reactant and product
remains constant.
Equilibrium
A System at Equilibrium
Rates become equal Concentrations become constant
Equilibrium
Depicting Equilibrium
In a system at equilibrium, both the
forward and reverse reactions are
running simultaneously. We write the
chemical equation with a double arrow:
Equilibrium
Equilibrium
Equilibrium
The Equilibrium
Constant
Equilibrium
The Equilibrium Constant
Forward reaction: Reverse reaction:
Rate Law
Rate law
Equilibrium
The Equilibrium Constant
At equilibrium
Rearranging gives:
Equilibrium
The Equilibrium Constant
The ratio of the rate constants is a
constant (as long as T is constant).
The expression becomes
Equilibrium
The Equilibrium Constant
To generalize, the reaction:
Has the equilibrium expression:
This expression is true even if you don’t know the elementary
reaction mechanism. Equilibrium
SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions
Write the equilibrium expression for Kc for the following reactions:
Equilibrium
PRS here…
Equilibrium
SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions
Write the equilibrium expression for Kc for the following reactions:
Solution
Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each.
Plan: Using the law of mass action, we write each expression as a quotient having the product concentration
terms in the numerator and the reactant concentration terms in the denominator. Each term is raised to the power
of its coefficient in the balanced chemical equation.
Solve:
PRACTICE EXERCISE
Write the equilibrium-constant expression, Kc for
Equilibrium
Equilibrium Can Be Reached from Either
Direction
Kc, the final ratio of [NO2]2 to [N2O4], reaches a
constant no matter what the initial concentrations
of NO2 and N2O4 are (with constT).
Equilibrium
Equilibrium Can Be Reached from
Either Direction
This graph shows data
from the last two trials
from the table.
QuickTime™ and
TIFF (LZW) decompre
e needed to see this pi
Equilibrium
Equilibrium
Equilibrium Can Be Reached from Either
Direction
It does not matter whether we start with N2 and H2 or
whether we start with NH3. We will have the same
proportions of all three substances at equilibrium.
What is the equilibrium expression?
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
Equilibrium
What Does the Value of K Mean?
• If K >> 1, the reaction
is product-favored;
product predominates
at equilibrium.
• If K << 1, the reaction is
reactant-favored;
reactant predominates
at equilibrium.
Equilibrium
Manipulating Equilibrium Constants
The equilibrium constant of a reaction in the
reverse reaction is the reciprocal of the
equilibrium constant of the forward reaction.
Equilibrium
Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant
raised to a power that is equal to that number.
Equilibrium
Manipulating Equilibrium Constants
The equilibrium constant for a net reaction made up
of two or more steps can be found from the
equilibrium constants for the individual steps.`
At 1565 K we have these equilibrium constants:
Equilibrium
ans=2.9
The Equilibrium Constant
Because pressure is proportional to
concentration for gases, the equilibrium
expression can also be written in terms of
partial pressures (instead of concentration):
Mixed versions are also used sometimes:
Equilibrium
Relationship between Kc and Kp
• From the ideal gas law we know that
= Pressure in terms of concentration
Equilibrium
Relationship between Kc and Kp
Equilibrium
Substituting P=[A]RT into the expression for
Kp for each substance, the relationship
between Kc and Kp becomes
Kp = Kc (RT)n
Where:
n = (moles of gaseous product) – (moles of gaseous reactant)
Hint: ‗products–reactants‘ and ‗products over reactants‘
is a common theme in chemistry.
Heterogeneous
Equilibrium
Equilibrium
The Concentrations of Solids and
Liquids Are Essentially Constant
Concentrations of liquids and solids can be
obtained by dividing the density of the
substance by its molar mass—and both of
these are constants at constant temperature.
Equilibrium
The Concentrations of Solids and
Liquids Are Essentially Constant
Therefore, the concentrations of solids
and liquids do not appear in the
equilibrium expression
Pb2+
(aq) + 2 Cl−
(aq)
Kc = [Pb2+] [Cl−]2
PbCl2 (s)
Equilibrium
As long as some CaCO3 or CaO remain in
the system, the amount of CO2 above the
solid will remain the same.
CaCO3 (s) CO2 (g) + CaO(s)
Equilibrium
What Are the Equilibrium
Expressions for These Equilibria?
Equilibrium
The real scoop: units of equilibrium constants
Equilibrium constants are really defined in terms
of activity, not concentration.
becomes:
Activity is unitless, so K is unitless.
Equilibrium
Equilibrium
Calculations
Equilibrium
Equilibrium Calculations
A closed system initially containing
1.000 x 10-3 M H2 and 2.000 x 10-3 M I2
At 448C is allowed to reach equilibrium.
Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10-3 M.
Calculate Kc at 448C for the reaction:
H2 (g) + I2 (g) 2 HI (g)
Equilibrium
What Do We Know?
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change
At
Equilibrium
1.87 x 10-3
ICE method:
[HI] Increases by 1.87 x 10-3 M
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change +1.87 x 10-3
At
equilibrium
1.87 x 10-3
Stoichiometry tells us [H2] and [I2]
decrease by half as much
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At
equilibrium
1.87 x 10-3
We can now calculate the equilibrium
concentrations of all three compounds…
Equilibrium
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At
equilibrium
6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
…and, therefore, the equilibrium constant
c
K =
[HI]2
[H2] [I2]
= 51
=
(1.87 x 10-3)2
(6.5 x 10-5)(1.065 x 10-3)
Equilibrium
The Reaction Quotient (Q)
Equilibrium
• To calculate Q, one substitutes the
initial concentrations on reactants and
products into the equilibrium
expression.
• Q gives the same ratio the equilibrium
expression gives, but for a system that
is not at equilibrium.
If Q = K,
the system is at equilibrium.
Equilibrium
If Q > K,
there is too much product and the
equilibrium shifts to the left.
Equilibrium
If Q < K,
there is too much reactant, and the
equilibrium shifts to the right.
Equilibrium
Le Châtelier‘s
Principle
Equilibrium
Le Châtelier‘s Principle
Equilibrium
―
I
fa system at equilibrium is disturbed by
a change in temperature, pressure, or the
concentration of one of the components,
the system will shift its equilibrium
position so as to counteract the effect of
the disturbance.‖
Systems shift from ―
Q
‖towards ―
K
‖
.
What Happens When More of a
Reactant Is Added to a System?
Equilibrium
The Haber Process
The transformation of nitrogen and hydrogen into
ammonia (NH3) is of tremendous significance in
agriculture, where ammonia-based fertilizers are of
utmost importance.
Equilibrium
The Haber Process
If H2 is added to the
system, N2 will be
consumed and the
two reagents will
form more NH3.
Equilibrium
The Haber Process
This apparatus
helps push the
equilibrium to the
right by removing
the ammonia (NH3)
from the system as
a liquid.
Equilibrium
The Effect of Changes in Pressure
Equilibrium
The Effect of Changes in
Temperature
Co(H2O)6 (aq) + 4 Cl(aq)
2+ CoCl4 (aq) + 6 H2O (l)
Equilibrium
The Effect of Changes in
Temperature
Equilibrium
Catalysts increase the rate of both the
forward and reverse reactions.
Equilibrium
Equilibrium is achieved faster, but the
equilibrium composition remains unaltered.
Equilibrium
Equilibrium

More Related Content

chemical-equilibria ppt.pptx

  • 2. Fast Initial Step • Because Ratef = Rater , Solving for [NOBr2] gives us Can also write as: Equilibrium
  • 3. Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.
  • 4. The Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate. Equilibrium
  • 5. A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. Equilibrium
  • 6. A System at Equilibrium Rates become equal Concentrations become constant Equilibrium
  • 7. Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are running simultaneously. We write the chemical equation with a double arrow: Equilibrium
  • 11. The Equilibrium Constant Forward reaction: Reverse reaction: Rate Law Rate law Equilibrium
  • 12. The Equilibrium Constant At equilibrium Rearranging gives: Equilibrium
  • 13. The Equilibrium Constant The ratio of the rate constants is a constant (as long as T is constant). The expression becomes Equilibrium
  • 14. The Equilibrium Constant To generalize, the reaction: Has the equilibrium expression: This expression is true even if you don’t know the elementary reaction mechanism. Equilibrium
  • 15. SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions: Equilibrium
  • 17. SAMPLE EXERCISE 15.1 Writing Equilibrium-Constant Expressions Write the equilibrium expression for Kc for the following reactions: Solution Analyze: We are given three equations and are asked to write an equilibrium-constant expression for each. Plan: Using the law of mass action, we write each expression as a quotient having the product concentration terms in the numerator and the reactant concentration terms in the denominator. Each term is raised to the power of its coefficient in the balanced chemical equation. Solve: PRACTICE EXERCISE Write the equilibrium-constant expression, Kc for Equilibrium
  • 18. Equilibrium Can Be Reached from Either Direction Kc, the final ratio of [NO2]2 to [N2O4], reaches a constant no matter what the initial concentrations of NO2 and N2O4 are (with constT). Equilibrium
  • 19. Equilibrium Can Be Reached from Either Direction This graph shows data from the last two trials from the table. QuickTime™ and TIFF (LZW) decompre e needed to see this pi Equilibrium
  • 20. Equilibrium Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N2 and H2 or whether we start with NH3. We will have the same proportions of all three substances at equilibrium. What is the equilibrium expression?
  • 21. What Does the Value of K Mean? • If K >> 1, the reaction is product-favored; product predominates at equilibrium. Equilibrium
  • 22. What Does the Value of K Mean? • If K >> 1, the reaction is product-favored; product predominates at equilibrium. • If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium. Equilibrium
  • 23. Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. Equilibrium
  • 24. Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. Equilibrium
  • 25. Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps can be found from the equilibrium constants for the individual steps.` At 1565 K we have these equilibrium constants: Equilibrium ans=2.9
  • 26. The Equilibrium Constant Because pressure is proportional to concentration for gases, the equilibrium expression can also be written in terms of partial pressures (instead of concentration): Mixed versions are also used sometimes: Equilibrium
  • 27. Relationship between Kc and Kp • From the ideal gas law we know that = Pressure in terms of concentration Equilibrium
  • 28. Relationship between Kc and Kp Equilibrium Substituting P=[A]RT into the expression for Kp for each substance, the relationship between Kc and Kp becomes Kp = Kc (RT)n Where: n = (moles of gaseous product) – (moles of gaseous reactant) Hint: ‗products–reactants‘ and ‗products over reactants‘ is a common theme in chemistry.
  • 30. The Concentrations of Solids and Liquids Are Essentially Constant Concentrations of liquids and solids can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature. Equilibrium
  • 31. The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression Pb2+ (aq) + 2 Cl− (aq) Kc = [Pb2+] [Cl−]2 PbCl2 (s) Equilibrium
  • 32. As long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same. CaCO3 (s) CO2 (g) + CaO(s) Equilibrium
  • 33. What Are the Equilibrium Expressions for These Equilibria? Equilibrium
  • 34. The real scoop: units of equilibrium constants Equilibrium constants are really defined in terms of activity, not concentration. becomes: Activity is unitless, so K is unitless. Equilibrium
  • 36. Equilibrium Calculations A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 At 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448C for the reaction: H2 (g) + I2 (g) 2 HI (g) Equilibrium
  • 37. What Do We Know? Equilibrium [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change At Equilibrium 1.87 x 10-3 ICE method:
  • 38. [HI] Increases by 1.87 x 10-3 M Equilibrium [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change +1.87 x 10-3 At equilibrium 1.87 x 10-3
  • 39. Stoichiometry tells us [H2] and [I2] decrease by half as much Equilibrium [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 At equilibrium 1.87 x 10-3
  • 40. We can now calculate the equilibrium concentrations of all three compounds… Equilibrium [H2], M [I2], M [HI], M Initially 1.000 x 10-3 2.000 x 10-3 0 Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3 At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
  • 41. …and, therefore, the equilibrium constant c K = [HI]2 [H2] [I2] = 51 = (1.87 x 10-3)2 (6.5 x 10-5)(1.065 x 10-3) Equilibrium
  • 42. The Reaction Quotient (Q) Equilibrium • To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. • Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.
  • 43. If Q = K, the system is at equilibrium. Equilibrium
  • 44. If Q > K, there is too much product and the equilibrium shifts to the left. Equilibrium
  • 45. If Q < K, there is too much reactant, and the equilibrium shifts to the right. Equilibrium
  • 47. Le Châtelier‘s Principle Equilibrium ― I fa system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.‖ Systems shift from ― Q ‖towards ― K ‖ .
  • 48. What Happens When More of a Reactant Is Added to a System? Equilibrium
  • 49. The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH3) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance. Equilibrium
  • 50. The Haber Process If H2 is added to the system, N2 will be consumed and the two reagents will form more NH3. Equilibrium
  • 51. The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid. Equilibrium
  • 52. The Effect of Changes in Pressure Equilibrium
  • 53. The Effect of Changes in Temperature Co(H2O)6 (aq) + 4 Cl(aq) 2+ CoCl4 (aq) + 6 H2O (l) Equilibrium
  • 54. The Effect of Changes in Temperature Equilibrium
  • 55. Catalysts increase the rate of both the forward and reverse reactions. Equilibrium
  • 56. Equilibrium is achieved faster, but the equilibrium composition remains unaltered. Equilibrium