Chemical equilibrium Reversible and irreversible chemical reactions

In this lesson we will study equilibrium in in the physical systems , equilibrium in chemical systems ,types of chemical reaction reversible or irreversible ,rate of chemical reactions and factors affecting of chemical reactions such as nature of chemical reactants , concentration of reactants , temperature , catalysts ,light and finally pressure

The equilibrium system:

It is apparently s stationary system but in reality dynamic .

[1]Equilibrium in the physical; systems :

-If we heated a water in a closed vessel :

1-there are two opposite processes (evaporation and condensation)

2-At the beginning of heating rate of evaporation is predominates which is followed by increase in the vapor pressure

Vapor pressure :the pressure due to water vapor in air at a certain temperature

3-the evaporation process continues until the vapor pressure equals saturated water vapor pressure  

Saturated water vapor pressure:

The maximum water vapor pressure in air at a certain temperature  

State of equilibrium :

The rate of evaporation =the rate of condensation

The number of water molecules which evaporate= the number of  water vapor molecules condensate

Types of chemical reactions: according to equilibrium:

1)complete (irreversible) reaction           2) Reversible reaction

1)complete reaction:

– it goes in one direction (approximately forward) GR?

-because the products includes gas or precipitate that cant react to produce reactants

Examples:

1)adding sodium chloride solution to silver nitrate solution where a white precipitate of silver chloride is formed

2)adding a strip of magnesium metal to Hydrochloric acid  solution  hydrogen gas  evolved

2)reversible reactions:

When adding one mole of acetic acid with one mole of ethyl alcohol , one mole of ester (ethyl acetate ) and one mole of water should be formed

If we put blue litmus paper it turns into red GR?

• Due to acetic acid where the reaction is reversible reaction which moves in both direction :forward and backward

• 

Chemical equilibrium in reversible reactions:

It is a dynamic system takes place when:

-the rate of forward reaction = the rate of backward reaction

-The concentration of the reactants and products are not changed  

-The equilibrium remains unchanged since all reactants and products are still found in the medium of reaction (no gas evolve , no precipitate is formed and temperature and pressure not changed

Rate of chemical reaction:

-The change in concentration of reactants per unit time

       -The concentration unit is expressed as : mole/liter

       -The time : second or minute

– During the complete chemical reaction:

The concentration of  the reactants decreases till consumed and the concentration of products increases

– During the reversible chemical reaction:

-The decrease in concentration of reactants and increasers the concentration of product occurs  till the equilibrium sate  

Types of chemical reaction according their speeds:

1)instantaneous  reaction: take very short time

Reaction of silver nitrate with sodium chloride to form sparingly soluble precipitate of silver chloride  

2)comparatively slow

Reaction between plants oil and caustic soda to form soap  and glycerol

3)very slow: iron rusting : needs several months  

Factors affecting the rate (speed)of chemical reactions:

1)Nature of reactants

2)concentration of reactants

3)reaction temperature

4)pressure

5)catalyst

1)Nature of reactants

           a)Type of bonds in reactants                                b)surface area of reactants

Iconic bond       covalent bond

A)Type of bonding in reactants:

– Iconic bond: reactants takes place very fast(instantaneous) GR? Because it occurs between IONS

Example: reaction between sodium chloride and silver nitrate

-Covalent bond: the reaction is slow GR? Because it occurs between molecules

Example: organic reactions

B)surface area exposed to reaction:

-Mass of block Zinc + volume of dilute hydrochloric acid : reaction is slow

-The same mass of zinc powder+ the same volume of hydrochloric acid:reaction is fast  GR:

-because the exposed area to the reaction is case of block zinc is small then that in case of zinc powder

2)concentration of reactants

Increases of concentration of reactants cause increase of number of molecules of reactants that increases the collision between reactants molecules that increases the speed of the reaction

Waage and Guldberg(Norwegian scientists):

-they established the law of mass action that explain the  the relationship between the  concentration of reactants and the speed of chemical reaction

Law of mass action:

At constant temperature the rate of chemical reaction is directly proportional to the result of multiplication of the reactant concentration each raise to the power of the number of molecules or ions in the balanced equation

Excrement:

By gradual addition of the following reactants:

FeCl₂(aq)       +     3NH₄SCN(aq)  →  Fe(SCN)₃ (aq) +3NH₄Cl(aq)

Iron III chloride ammonium thiocyanate       iron III thiocynate

(pale yellow)         (colorless)                            (blood red)

The color of the reaction mixture become blood red GR?

-due to the formation of iron III thiocynate

– If an excess of iron(III) chloride is added the red color of the solution increases indication the formation of more iron III thiocynate

-The equilibrium state occurs:

when: the rate of backward reaction (r2) = the rate of forward reaction (r1)

Note:

• [ ]       :the concentration in unit( moles/liter)
• K₁ and K₂ :rate constants for the forward reaction and backward reaction
• At equilibrium: r₁=r₂

The product   K1/K2   is a constant  called Equilibrium constant of the reaction (K_c)

At equilibrium, the rates of the forward and reverse chemical reactions are equal:

aA + bB ⇌ cC + dD

The ratio between the concentrations of the products and reactants is a constant, known as the equilibrium constant, K_c:

K_c = [C]^c[D]^d/[A]^a[B]^b

In this equation, the square brackets indication concentration of the chemical species.

The exponents are the coefficients from the chemical equation.

The equilibrium constant for the reverse reaction, K’_c, is given by the following:

K’_c = 1/K_c = [A]^a[B]^b/[C]^c[D]^d

Example of law of mass action:

According to this law, “the rate of reaction of a substance is proportional to the product of molar concentration of reactants at a constant temperature at any given time”. Molar concentration is called active mass.

Active mass is the number of moles dissolved in one litre of solution.

eg. CH3COOH + C2H5OH ➝ CH3COOC2 H5 + H2O

According to law of mass action, Rate of forward reaction

a [CH3COOH] [C2H5OH] = K1 [CH3COOH] [C2H5OH] When K2 = rate constant for backward reaction At equilibrium,

Rate of backward reaction a [CH3COOC2H5] [H2O] = K2 [CH3COOC2H5] [H2O]

Where K2 = rate constant for backward reaction

At equilibrium, Rate of forward reaction = Rate of backward reaction K2 [CH3COOH] [C2H6OH] – K2 [CH3COOC2H5] [H2O]

When to Use the Law of Mass Action

Remember, the law of mass action only applies in cases of dynamic equilibrium. Regardless of the arrows in a chemical equation, make sure the following statements are true:

• The chemical equation represents the reaction of a closed system. That is, there is no heat or mass entering or leaving the system.
• Temperature remains a constant. At equilibrium, temperature does not change. Similarly, the equilibrium constant for a reaction depends on temperature. It’s value at one temperature may differ from K_c at another temperature.

Law of Mass Action Examples

For example, write the equilibrium constant expression for the dissociation of sulfuric acid into hydrogen and sulfate ions:

H₂SO₄ ⇌ 2H⁺ + SO₄^2-

Answer: Kc = [H⁺]²[SO₄^2-]/[H₂SO₄]

For example, if you know K_c is 5×10⁵ for the reaction:

HCOOH + CN⁻ ⇌ HCN + HCOO⁻

Calculate the equilibrium constant for the reaction:

HCN + HCOO⁻ ⇌ HCOOH + CN⁻

Answer: The second equation is the reverse of the first equation.

K’_c = 1/K_c = 1/(5 x 10⁵) = 2 x 10^-6

Example: calculate the equilibrium constant of the reaction:

 H${}_2$(g) + I${}_2$(g) $\to$ 2HI(g)

If the concentration of I2 ,H2 and HI are 0.221,0.221 and 1.563 mole/liter respectively

solution: The equilibrium constant for the reaction 

H2(g)​+I2(g)​⇌2HI(g)​ is $32$ at a given temperature.
The equilibrium concentration of I2​ and $HI$ are 0.5×10⁻³M and 8×10⁻³M respectively.

The equilibrium concentration of H2​ is:

solution:

• So increases the temperature causes increase the activated molecules that cause increase the rate of reaction
• The rate of chemical reaction is doubled by increasing the temperature by 10 C

Excrement: the effect of temperature in the rate of chemical reaction:

-by placing of nitrogen dioxide gas flask (reddish brown) in cold water the reddish brown color decreases until become colorless

-when putting it at room temperature (25 c)  the reddish brown color appears again

 So the color degree increases as temperature increases as the following:

        2NO2​(g)$\rightleftharpoons$N2​O4​(g)+heat

     reddish brown                    colorless

So if an exothermic reaction has reached the equilibrium state  decrease in temperature force the reaction to go in the forward direction to liberate heat

4)Effect of pressure in the rate of chemical reaction:

If the reactants and products are in gaseous state the concentration is expressed by using partial pressure

for the reaction    : N_2​(g)+3H_2​(g)⇌2NH_3​(g)        ΔH=-92Kj

4 moles reacts to form 2 moles from ammonia molecule

So the formation of ammonia gas is followed by a decrease in the number of moles and consequently a reduction in the volume

-by increasing the pressure or cooling on a gaseous reaction under equilibrium  a shift in the direction of reducing the volume (or the direction in which volume is less) takes place 

The equilibrium constant(partial pressure :Kp

So the formation of ammonia gas is followed by a decrease in the number of moles and consequently a reduction in the volume

-by increasing the pressure or cooling on a gaseous reaction under equilibrium  a shift in the direction of reducing the volume (or the direction in which volume is less) takes place 

The equilibrium constant(partial pressure :Kp

Example: write the Kp for the following equation

5)effect of catalyst:

Catalyst are substances that little amount of it cause a change in the rate of chemical reaction without itself being change or change equilibrium position

Why we use a catalyst ?to increase the rate of chemical to

Do decrease the consumed energy in industries and decrease the cost of products

-it decreases the activation energy of reaction so it used in 90% of industrial process (food ,petrochemicals industries)  

– it is used in catalytic convertors in the car GR?

To convert the harmful gases of burning fuel to decrease air pollution

-the catalyst are: metals , metal oxides or compounds

Enzymes: proteins molecules produced in living cells act as catalyst in biological and industrial process

6)Effect of light:

In photosynthesis process chlorophyll in plants absorbs light and form carbohydrates in presence of carbon dioxide and water

-Photographic films contain silver bromide in a gelatinous layer

When light falls on it positive Ag ions accept electrons from negative bromide ions and converted to  silver metal

-The increase the light intensity increases the amount of silver formed

Ag⁺ + e  >>>  Ag

Examples:

The so-called water-gas shift reaction:

$H_{2}O\left(g\right)+CO\left(g\right)\rightleftharpoons H_2+C{O}_2$, is useful in the production of hydrogen gas from steam and carbon monoxide. An equilibrium is reached at 700 K when $p_{H^{}}=p_{C{O}^{}}=0.755$ and $p_{H^{}O}=p_{CO}=0.245$. W

hat is $K_p$ for the reaction at that temperature?

Step 1: Read through the given information and take note of the partial pressure of each reactant and product.

$p_{H^{}}=p_{C{O}^{}}=0.755$and $p_{H^{}O}=p_{CO}=0.245$

Step 2: Write the expression for the reaction’s $K_p$.

Kp=(pH2)(pCO2) / (pH2O)(pCO)

Calculating an Equilibrium Constant Kp Using Partial Pressures

Dark brown nitrogen dioxide gas fades to colorless dinitrogen tetroxide gas by the reaction:

$$2N{O}_2\left(g\right)\rightleftharpoons N_2{O}_4\left(g\right)$$

When equilibrium is reached, $p_{N{O}^{}}=2.052$ and $p_{N^{}{O}^{}}=0.474$. What is $K_p$ for the reaction?

Step 1: Read through the given information and take note of the partial pressure of each reactant and product.

$p_{N{O}^{}}=2.052$ and $p_{N^{}{O}^{}}=0.474$

Step 2: Write the expression for the reaction’s $K_p$.

$K_p=\frac{{\left({}_{{}^{}{}^{}}\right)}^{\mathrm{1\; /}}}{{\left({}_{N{}^{}}\right)}^2}$

Step 3: Plug in the numerical values of the partial pressures, and calculate $K_p$.

$K_p=\frac{(0.474)\; /}{(2.052{)}^2}=0.113$

Calculating an Equilibrium Constant Kp Using Partial Pressures

Partial pressures at equilibrium for the reaction :

$H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$ are given in the table. What is $K_p$ for the reaction?

Step 1: Read through the given information and take note of the partial pressure of each reactant and product.

 These are conveniently found in the table.

Step 2: Write the expression for the reaction’s $K_p$.

$K_p=\frac{p{(}_{HI}{)}^{\mathrm{2\; /}}}{\left(p_{H^{}}\right)\left(p_{I^{}}\right)}$

Step 3: Plug in the numerical values of the partial pressures, and calculate

$K_p=\frac{(4.368{)}^{\mathrm{2\; /}}}{\left(0.316\right)\left(1.316\right)}=45.9$

 References:

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