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Section 8.2— Equilibrium Constant ow can we describe a reaction at equilibrium?

Section 8.2—Equilibrium Constant

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Section 8.2—Equilibrium Constant. How can we describe a reaction at equilibrium?. Equilibrium Constant Expression. Equilibrium Constant Expression – Equation showing the ratio of the concentrations of products to reactants at equilibrium. We use brackets, [ ], to symbolize concentration!!!. - PowerPoint PPT Presentation

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Page 1: Section 8.2—Equilibrium Constant

Section 8.2—Equilibrium Constant

How can we describe a reaction at equilibrium?

Page 2: Section 8.2—Equilibrium Constant

Equilibrium Constant Expression

Equilibrium Constant Expression – Equation showing the ratio of the concentrations of products to reactants at equilibrium

We use brackets, [ ], to symbolize concentration!!!

Page 3: Section 8.2—Equilibrium Constant

Example:Write the equilibrium constant expression

for the following:2 H2 (g) + O2 (g) 2 H2O (g)

Write the product of the product concentrations on the top—take each one to a power of the coefficient from the balanced equation.

Write the product of the reactant concentrations on the bottom—also take each to the power of the balanced equation coefficient.

1

2

Writing Equilibrium Constant Expressions

Page 4: Section 8.2—Equilibrium Constant

Example:Write the equilibrium constant expression

for the following:2 H2 (g) + O2 (g) 2 H2O (g)

[H2O]

[H2] [O2]

Write the product of the product concentrations on the top—take each one to a power of the coefficient from the balanced equation.

Write the product of the reactant concentrations on the bottom—also take each to the power of the balanced equation coefficient.

1

2

K =

22

Writing Equilibrium Constant Expressions

Page 5: Section 8.2—Equilibrium Constant

Homogeneous Equilibrium – All of the species are the same state of matter

Heterogeneous Equilibrium – There are at least 2 states of matter

2 H2 (g) + O2 (g) 2 H2O (g)

2 H2 (g) + O2 (g) 2 H2O (l)

Heterogeneous Equilibrium

Page 6: Section 8.2—Equilibrium Constant

Concentrations of Solids and Liquids

Solids and Liquids are PURE – they are not described using concentration terms.

•Concentration means a part (solute) in the whole (solvent) If a substance is pure it can’t be a part of a whole

THEREFORE – WE DO NOT INCLUDE SOLIDS OR LIQUIDS IN THE EXPRESSION!!!!

Page 7: Section 8.2—Equilibrium Constant

If the “concentration” of a pure solid or liquid is constant, then it will not change during equilibrium and it is not written in the “K” expression.

2 H2 (g) + O2 (g) 2 H2O (g)

2 H2 (g) + O2 (g) 2 H2O (l)

][][

][

22

2

22

OH

OHK

][][

1

22

2 OHK

H2O is not included in this “K” expression because it’s a liquid.

Only gases and aqueous solutions are included in “K” expressions!

“K” Expressions with Solids or Liquids

Page 8: Section 8.2—Equilibrium Constant

Example #1—Writing K expression

Example:Write the equilibrium constant expression for Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g)

Page 9: Section 8.2—Equilibrium Constant

32

32

][

][

H

OHK

Fe2O3 and Fe were not included in the K expression as they are solids!

Example #1—Writing K expression

Example:Write the equilibrium constant expression

for Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g)

Page 10: Section 8.2—Equilibrium Constant

Equilibrium Constant

Equilibrium Constant (K)– The number calculated from the equilibrium constant expression

“K” is different for every reaction at every temperature!

Page 11: Section 8.2—Equilibrium Constant

Example #2—Calculating K

Example:Solve for the equilibrium constant for

Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g) If at equilibrium [H2] = 0.45 M and [H2O] = 0.18 M

Page 12: Section 8.2—Equilibrium Constant

32

32

][

][

H

OHK

3

3

]45.0[

]18.0[K K = 0.064

Most instructors and textbooks do not require units for “K” as each one would be different

Example #2—Calculating K

Example:Solve for the equilibrium constant for

Fe2O3 (s) + 3 H2 (g) 2 Fe (s) + 3 H2O (g) If at equilibrium [H2] = 0.45 M and [H2O] = 0.18 M

Page 13: Section 8.2—Equilibrium Constant

In general….If K >1 or large… [Products]

[Reactants]

There is a much larger ratio of products to reactants at equilibrium

The reaction is said to “lie to the right” (products are on the right)

If K < 1 or small…. [Products]

[Reactants]

There is a much smaller ratio of products to reactants at equilibrium

The reaction is said to “lie to the left”

Meaning of Equilibrium Constant

If K = 1 it means we have similar amounts of reactants and products at equililbrium.

Page 14: Section 8.2—Equilibrium Constant

Let’s Practice #1

Write the equilibrium constant expression for N2 (g) + O2 (g) 2 NO (g)

Page 15: Section 8.2—Equilibrium Constant

]][[

][

22

2

ON

NOK

Let’s Practice #1

Write the equilibrium constant expression for N2 (g) + O2 (g) 2 NO (g)

Page 16: Section 8.2—Equilibrium Constant

Let’s Practice #2

If the equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is 1.24 × 10-4, what can be said in general about this reaction at

equilibrium?

Page 17: Section 8.2—Equilibrium Constant

Let’s Practice #2

If the equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is 1.24 × 10-4, what can be said in general about this reaction at

equilibrium?

The equilibrium constant is very small, so at equilibrium the concentration of products is much lower than reactants. The reaction lies to the left.

Page 18: Section 8.2—Equilibrium Constant

Let’s Practice #3

If the equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is 1.24 × 10-4 and the equilibrium concentration of [N2] = 0.166 M and

[O2] = 0.145 M, what is the equilibrium concentration of NO?

Page 19: Section 8.2—Equilibrium Constant

]][[

][

22

2

ON

NOK

[NO]eq = 0.00172 M

)145.0)(166.0(

][1024.1

24

MM

NO

][145.0166.01024.1 4 NOMM

Let’s Practice #3

If the equilibrium constant for N2 (g) + O2 (g) 2 NO (g) is 1.24 × 10-4 and the equilibrium concentration of [N2] = 0.166 M and

[O2] = 0.145 M, what is the equilibrium concentration of NO?