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Chemical Equilibrium. The reversibility of reactions. Equilibrium. Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

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chemical equilibrium

Chemical Equilibrium

The reversibility of reactions

equilibrium
Equilibrium

Many chemical reactions do not go to completion. Initially, when reactants are present, the forward reaction predominates. As the concentration of products increases, the reverse reaction begins to become significant.

equilibrium1
Equilibrium

The forward reaction rate slows down since the concentration of reactants has decreased. Since the concentration of products is significant, the reverse reaction rate gets faster.

Eventually, the

forward reaction rate = reverse reaction rate

equilibrium2
Equilibrium

forward reaction rate = reverse reaction rate

At this point, the reaction is in equilibrium. The equilibrium is dynamic. Both the forward and reverse reactions occur, but there is no net change in concentration.

equilibrium 2no 2 n 2 o 4
Equilibrium: 2NO2↔ N2O4

The concentrations of N2O4 and NO2 level off when the system reaches equilibrium.

equilibrium 2no 2 n 2 o 41
Equilibrium: 2NO2↔ N2O4

The system will reach equilibrium starting with either NO2 , N2O4, or a mixture of the two.

equilibrium3
Equilibrium

Reactions that can reach equilibrium are indicated by double arrows (↔ or ).

The extent to which a reaction proceeds in a particular direction depends upon the reaction, temperature, initial concentrations, pressure (if gases), etc.

equilibrium4
Equilibrium

Scientists studying many reactions atequilibrium determined that for a general reaction with coefficients a, b, c and d :

a A + b B ↔ c C + d D

K is the equilibrium constant for the reaction.

[C]c[D]d

K=

[A]a[B]b

equilibrium5
Equilibrium

a A + b B ↔ c C + d D

This is the equilibrium constant expression for the reaction. K is the equilibrium constant. The square brackets indicate the concentrations of products and reactants at equilibrium.

[C]c[D]d

K=

[A]a[B]b

equilibrium constants
Equilibrium Constants

The value of the equilibrium constant depends upon temperature, but does not depend upon the initial concentrations of reactants and products. It is determined experimentally.

equilibrium constants2
Equilibrium Constants

The units of K are usually omitted. The square brackets in the equilibrium constant expression indicate concentration in moles/liter.

Some texts will use the symbol Kc or Keq for equilibrium constants that use concentrations or molarity.

equilibrium constants3
Equilibrium Constants

The size of the equilibrium constant gives an indication of whether a reaction proceeds to the right.

equilibrium constants4
Equilibrium Constants

If a reaction has a small value of K, the reverse reaction will have a large value of K.

At a given temperature, K = 1.3 x 10-2 for:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

If the reaction is reversed,

2 NH3(g) ↔ N2(g) + 3 H2(g)

K = 1/(1.3 x 10-2 ) = 7.7 x 101

equilibrium constants5
Equilibrium Constants

Equilibrium constants for gas phase reactions are sometimes determined using pressures rather than concentrations. The symbol used is Kp rather than K (or Kc).

For the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

Kp = (Pammonia)2

(PN2)(PH2)3

equilibrium constants6
Equilibrium Constants

The numerical value of Kp is usually different than that for K. The values are related, since P= nRT = MRT, where M = mol/liter.

The method for solving equilibrium problems with Kp or K is the same. With Kp, you use pressure in atmospheres. With K or Kc, you use concentration in moles/liter.

V

heterogeneous equilibria
Heterogeneous Equilibria

Many equilibrium reactions involve reactants and products where more than one phase is present. These are called heterogeneous equilibria.

An example is the equilibrium between solid calcium carbonate and calcium oxide and carbon dioxide.

CaCO3(s) ↔ CaO(s) + CO2(g)

heterogeneous equilibria1
Heterogeneous Equilibria

CaCO3(s) ↔ CaO(s) + CO2(g)

Experiments show that the position of the equilibrium does not depend upon the amounts of calcium carbonate or calcium oxide. That is, adding or removing some of the CaCO3(s) or CaO(s) does not disrupt or alter the concentration of carbon dioxide.

heterogeneous equilibria2
Heterogeneous Equilibria

CaCO3(s) ↔ CaO(s) + CO2(g)

heterogeneous equilibria3
Heterogeneous Equilibria

CaCO3(s) ↔ CaO(s) + CO2(g)

This is because the concentrations of pure solids (or liquids) cannot change. As a result, the equilibrium constant expression for the above reaction is:

K=[CO2] or Kp = PCO2

heterogeneous equilibria4
Heterogeneous Equilibria

The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.

problem calculation of k
Problem: Calculation of K
  • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

problem calculation of k1
Problem: Calculation of K
  • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

1. Write the equilibrium constant expression.

problem calculation of k2
Problem: Calculation of K
  • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

1. Write the equilibrium constant expression.

K = [NH3]2/([N2][H2]3)

problem calculation of k3
Problem: Calculation of K
  • At a certain temperature, 3.00 moles of ammonia were placed in a 2.00 L vessel. At equilibrium, 2.50 moles remain. Calculate K for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

2. Make a table of concentrations.

problem calculation of k4
Problem: Calculation of K

N2(g) + 3 H2(g) ↔ 2NH3(g)

problem calculation of k5
Problem: Calculation of K

N2(g) + 3 H2(g) ↔ 2NH3(g)

3. Complete the table.

problem calculation of k6
Problem: Calculation of K

N2(g) + 3 H2(g) ↔ 2NH3(g)

3. Complete the table.

problem calculation of k7
Problem: Calculation of K

N2(g) + 3 H2(g) ↔ 2NH3(g)

3. Complete the table.

slide31

N2(g) + 3H2(g) ↔2NH3(g)

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

slide32

N2(g) + 3H2(g) ↔2NH3(g)

The equilibrium values are the sum of the initial concentrations plus any changes that occur.

slide33

N2(g) + 3H2(g) ↔2NH3(g)

4. Substitute and solve for K.

K = [NH3]2/([N2][H2]3)

slide34

N2(g) + 3H2(g) ↔2NH3(g)

4. Substitute and solve for K.

K = [NH3]2/([N2][H2]3)

K=(1.25)2/[(.13)(.38)3] = 2.2 x 102

the reaction quotient
The Reaction Quotient

If reactants and products of a reaction are mixed together, it is possible to determine whether the reaction will proceed to the right or left. This is accomplished by comparing the composition of the initial mixture to that at equilibrium.

the reaction quotient1
The Reaction Quotient

If the concentration of one of the reactants or products is zero, the reaction will proceed so as to make the missing component.

If all components are present initially, the reaction quotient, Q, is compared to K to determine which way the reaction will go.

the reaction quotient2
The Reaction Quotient

Q has the same form as the equilibrium constant expression, but we use initial concentrations instead of equilibrium concentrations. Initial concentrations are usually indicated with a subscript zero.

the reaction quotient3
The Reaction Quotient

In the previous problem, we determined that K = 2.2 x 102 for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

  • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?
the reaction quotient4
The Reaction Quotient

K = 2.2 x 102 for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

  • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?

1. Calculate Q and compare it to K.

the reaction quotient5
The Reaction Quotient

A comparison of Q with K will indicate the direction the reaction will go.

the reaction quotient6
The Reaction Quotient

K = 2.2 x 102 for the reaction:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

  • Suppose 1.00 mol N2, 2.0 mol H2 and 1.5 mol of NH3 are placed in a 1.00L vessel. What will happen? In which direction with the reaction proceed?

1. Calculate Q and compare it to K.

problem
Problem
  • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102

Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

problem1
Problem
  • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102

Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

1. Write the Kpexpression.

problem2
Problem
  • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102

Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

1. Write the Kpexpression.

Kp = (PHI)2

(PH2)(PI2)

problem3
Problem
  • H2(g) + I2(g) ↔ 2HI(g) Kp = 1.00 x 102

Initially, Phydrogen = Piodine = 0.500 atm. Calculate the equilibrium partial pressures of all three gases.

2. Make a table of initial, change and equilibrium pressures.

problem h 2 g i 2 g 2hi g3
Problem: H2(g) + I2(g) ↔ 2HI(g)

3. Substitute and solve.

problem h 2 g i 2 g 2hi g4
Problem: H2(g) + I2(g) ↔ 2HI(g)
  • Substitute and solve.
  • Kp= (PHI)2/(PH2)(PI2) = 1.00 x 102
  • (2x)2= 1.00 x 102
  • (.500-x) (.500-x)
problem h 2 g i 2 g 2hi g5
Problem: H2(g) + I2(g) ↔ 2HI(g)

3. Substitute and solve.

Kp= (PHI)2/(PH2)(PI2) = 1.00 x 102

(2x)2= 1.00 x 102

(.500-x) (.500-x)

Take the square root of both sides:

(2x) = 10.0

(.500-x)

problem h 2 g i 2 g 2hi g6
Problem: H2(g) + I2(g) ↔ 2HI(g)

3. Substitute and solve.

(2x) = 10.0

(.500-x)

2x = 10.0 (.500-x) = 5.00 -10.0x

12.0 x = 5.00

x = .417

problem h 2 g i 2 g 2hi g7
Problem: H2(g) + I2(g) ↔ 2HI(g)

4. Answer the question: Calculate the equilibrium partial pressures of all three gases.

x = .417

problem h 2 g i 2 g 2hi g8
Problem: H2(g) + I2(g) ↔ 2HI(g)

4. Answer the question: Calculate the equilibrium partial pressures of all three gases.

x = .417

problem h 2 g i 2 g 2hi g9
Problem: H2(g) + I2(g) ↔ 2HI(g)

5. Check your answer, if possible. Does (PHI)2/(PH2)(PI2) = 1.00 x 102 ?

(.834)2/(.083)(.083) = (.696)/(.0069) = 1.0 x 102

problem n 2 g o 2 g 2no g
Problem: N2(g) + O2(g) ↔ 2NO(g)
  • At 2200 oC, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.
problem n 2 g o 2 g 2no g1
Problem: N2(g) + O2(g) ↔ 2NO(g)
  • At 2200 oC, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.

1. Write the equilibrium constant expression.

K = 0.050 = [NO]2/[N2][O2]

problem n 2 g o 2 g 2no g2
Problem: N2(g) + O2(g) ↔ 2NO(g)
  • At 2200 oC, K = 0.050 for the above reaction. Initially, 1.60 mol of nitrogen and 0.400 mol of oxygen are sealed in a 2.00 liter vessel. Calculate the concentration of all species at equilibrium.

2. Make a table of initial, change and equilibrium concentrations.

problem n 2 g o 2 g 2no g4
Problem: N2(g) + O2(g) ↔ 2NO(g)
  • Substitute and solve.

K = 0.050 = [NO]2/[N2][O2]

approaches to solving problems
Approaches to Solving Problems

Some equilibrium problems require use of the quadratic equation to obtain an accurate solution. In cases where the equilibrium constant is quite small (roughly 10-5 or smaller), it is often possible to make assumptions that will simplify the mathematics.

problem 2nocl g 2no g cl 2 g
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • At 35oC, K = 1.6 x 10-5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl2 are placed in a 1.0 liter flask.

1. K = 1.6 x 10-5 =[NO]2[Cl2]/[NOCl]2

problem 2nocl g 2no g cl 2 g1
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • At 35oC, K = 1.6 x 10-5 for the above reaction. Calculate the equilibrium concentrations of all species present if 2.0 moles of NOCl and 1.0 mole of Cl2 are placed in a 1.0 liter flask.

2. Make a table of concentrations.

problem 2nocl g 2no g cl 2 g3
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Substitute and solve.
  • K = 1.6 x 10-5 =[NO]2[Cl2]/[NOCl]2
  • 1.6 x 10-5 =[2x]2[1.0 + x]/[2.0-2x]2
problem 2nocl g 2no g cl 2 g4
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Substitute and solve.
  • 1.6 x 10-5 = [2x]2 [1.0 + x]
  • [2.0-2x]2
  • Since K is small, x, the amount of product formed, will be a small number. As a result, 1.0 + x ≈1.0, and 2.0-2x≈2.0
problem 2nocl g 2no g cl 2 g5
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Substitute and solve.
  • 1.6 x 10-5 = [2x]2 [1.0 + x]
  • [2.0-2x]2
  • The expression simplifies to:
  • 1.6 x 10-5 = [2x]2 [1.0]
  • [2.0]2

0

0

problem 2nocl g 2no g cl 2 g6
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Solve for x.
  • 1.6 x 10-5 = [2x]2 [1.0]
  • [2.0]2
  • 4x2 = 1.6 x 10-5 (4)/1 = 6.4 x 10-5
  • x2 = 1.6 x 10-5 ; x =0.0040
problem 2nocl g 2no g cl 2 g7
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)

4. Check your assumption.

x =0.0040

Does 1.0 + x ≈1.0, and 2.0-2x≈2.0?

1.0 + .004 = 1.0

2.0 – 2(.004) = 2.0 - .008=2.0

validity of the assumption
Validity of the Assumption

These assumptions are generally considered valid if they cause an insignificant error. In this case, due to significant figures, no error is introduced. Usually, if the difference is within 5%, the error is considered acceptable.

problem 2nocl g 2no g cl 2 g8
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Answer the question.
  • [NOCl] = 2.0M; [NO]= 0.0080M;
  • [Cl2] = 1.0M
problem 2nocl g 2no g cl 2 g9
Problem: 2NOCl(g)↔2NO(g) + Cl2(g)
  • Check your answer.
  • [NOCl] = 2.0M; [NO]= 0.0080M;
  • [Cl2] = 1.0M
  • Does [NO]2[Cl2]/[NOCl]2 = 1.6 x 10-5 ?
  • (.0080)2(1.0)/(2.0)2 = 1.6 x 10-5
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 so as to counteract the stress.

le chatelier s principle1
Le Chatelier’s Principle

If a stress is applied to a system at equilibrium, the position of the equilibrium will shift so as to counteract the stress.

Types of stresses:

  • Addition or removal of reactants or products
  • Changes in pressure or volume (for gases)
  • Changes in temperature
le chatelier s principle2
Le Chatelier’s Principle

For reactions involving gaseous products or reactants, changes in pressure of volume may shift the equilibrium. The equilibrium will shift only if there is an unequal number of moles of gas on either side of the reaction. Reactions involving liquids or solids are not significantly affected by changes in pressure or volume.

le chatelier s principle4
Le Chatelier’s Principle
  • For the reaction

C(s) + 2 H2(g) ↔ CH4(g) ∆Ho=-75kJ

Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes.

a) add carbon

b) remove methane

c) increase the temperature

le chatelier s principle5
Le Chatelier’s Principle

The value of K will change with temperature. The effect of a change in temperature on K depends on whether the reaction is exothermic or endothermic.

The easiest way to predict the result when temperature is changed is to write in heat as a product (for exothermic reactions) or a reactant (for endothermic reactions).

le chatelier s principle6
Le Chatelier’s Principle
  • For the reaction

C(s) + 2 H2(g) ↔ CH4(g) ∆Ho=-75kJ

Assuming the reaction is initially at equilibrium, predict the shift in equilibrium (if any) for each of the following changes.

d) add a catalyst

e) decrease the volume

f) remove some carbon

g) add hydrogen