- 114 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' CHEMICAL EQUILIBRIUM' - imelda

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### CHEMICAL EQUILIBRIUM

Dr. Saleha Shamsudin

What is chemical equilibrium?

- A dynamic state in which the rates of forward (f) and reverse (r) reaction are identical.

Chemical Equilibrium – A Dynamic Equilibrium

- Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium.
- Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

1. Chemical Reactions

1.1 Introduction

- Chemical reactions
Involves chemical change.

Chemical change: the atoms rearrange themselves in such a way that new substances are formed.

Components decrease in quantity are called reactants and those increase in quantity are called products.

The chemical reaction is:

aA + bB ↔ cC + dD

The rate of reaction = constant x concentration each species raised to the power of the number of molecules participating in the reaction.

Forward reaction:

kf= rate constant

[A] and [B]= molar concentrations of A and B

Forward and reverse rates are equal

kf[A]a[B]b = kb[C]c[D]d

Molar equilibrium constant, K, raised to the power of the number of molecules participating in the reaction.

kb[C]c[D]d = kf = K

[A]a[B]b kb

- Rearranging these equation gives molar equilibrium constant (K) :
[C]c[D]d = kf = K

[A]a[B]b kb

When the two rates become equal, the system is in a state of equilibrium.

Forward and reverse rates are equal

kf[A]a[B]b = kb[C]c[D]d

Type of equilibria raised to the power of the number of molecules participating in the reaction.

- We can write equilibrium constants for many types of chemical processes. The equilibria may represent :
- Dissociation (acid/base, solubility)
- Formation of products (complexes)
- Reaction (redox)
- Distribution between 2 phases (water and non-aqueous solvent)

Types of equilibria raised to the power of the number of molecules participating in the reaction.

Equilibrium constant of reverse reaction: raised to the power of the number of molecules participating in the reaction.

aA + bB cC + dD

cC + dD aA + bB

Equilibrium constants for dissociating and associating species

- For weak electrolyte
- Slightly soluble substances
- Example:
1) AB A + B

- Equilibrium constant written as:
[A][B] = Keq

[AB]

- Stepwise dissociation species
A2B A + AB K1 = [A][AB]

[A2B]

AB A + B K2 = [A][B]

[AB]

Overall dissociation:

A2B 2A + B Keq = [A]2[B]

[A2B]

3) Heterogeneous equilibria species

- Involving more than one physial phase, example, solids, water and pure liquids.
- Concentrations of any solid reactants and products are omitted from the equi. cons. expression.
- Molar concentrations of water (or of any liquid reactant or product) is omitted from the equi. cons. expression.

E.g. speciesWrite out the equilibrium expression for K using the reaction below:

N2(g) + 3H2(g) 2NH3(g) Kc = ?

Calculating Equilibrium Constants species

Try this example:

An aqueous solution of ethanol and acetic acid, each with a concentration of 0.810 M, is heated to 100oC. At equilibrium, the acetic acid concentration is 0.748 M. Calculate K at 100oC for the reaction

C2H5OH(aq) + CH3CO2H(aq) CH3CO2C2H3(aq) + H2O(l)

ethanol acetic acid ethyl acetate

- Write the equilibrium constant expressions in terms of concentrations:
- Calculate K at 100oC for the reaction:

- The equilibrium constant expressions is: species
- K = [CH3CO2C2H3] notice that liquid
- [C2H5OH][CH3CO2H] water does not
- 2. The calculation of K is: appear in equi.
- = 0.062 = 0.11 expressions.
- (0.748)(0.748)

Calculating an Equilibrium from an Equilibrium Constant species

Problem: K= 55.64 for H2(g) + I2(g) 2HI(g)

Has been determined at 425oC. If 1.00 mol

Each of H2 and I2 are placed in 0.5 L flask at 425oC, what are the concentrations of H2, I2 and HI when equilibrium has been achieved?

Solution: species

- Write the equi. Cons. Expression.
- Set up an ICE table

- Assume x is equal to the quantity of H2 or I2
- consumed in the reaction

- Substitute the equi. Con. Into K expression. species
55.64 = (2x)2 = (2x)2

(2.0 – x)(2.0 – x) (2.0 – x)2

55.64 = 7.459 = 2x

2.0 – x

x = 1.58

H = I = 2.0 – x = 0.42 M

HI = 2x = 3.16 M

Calculating Equilibrium Concentrations species

- Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined.
E.g.2 [H2]o = [I2]o = 24 mM, were mixed and heated to 490oC in a container. Calculate equil. composition. Given that Kc = 46.

Solution: set up an equilibrium table and solve for unknown after substitution into equilibrium expression.

H2(g) +I2(g)2HI(g)

- Equilibrium concentrations of each component are in last row. Substitute into equilibrium expression and solve for x.
- It may be necessary to rearrange so that quadratic equation can be used.
- .
- Rearrange to ax2 + bx + c = 0; determine a, b, c and substitute into quadratic equation:

Download Presentation

Connecting to Server..