Video Lesson

Lesson Objectives

At the end if the lesson , you will be able to :

• Define reaction quotient
• Use the equilibrium quotient to predict the direction of the reaction and
• The position of equilibrium
• Calculate equilibrium concentrations given initial concentrations
• Determine whether the reactants or products are favored in a chemical reaction given the equilibrium constant
• List factors that affect chemical equilibrium
• State Le-Chatliers principle
• Use Le-Chatliers principle to explain the effect of changes in temperature,
• Pressure, concentration and presence of catalyst on a reaction
• State the effect of changes in concentration, pressure/volume an temperature on Keq
• Perform an activity to demonstrate the effect of changes in concentratio on the position of equilibrium
• Perform an activity to demonstrate the effect of changes in temperature on the position of equilibrium
• Perform an activity to determine KC for esterification of an organic acid;
• Define optimum conditions
• Explain how Le-Chatliers principle is applied in the Haber process
• (production of NH3 ) and in the Contact process (production of H2SO4

Brainstorming Question

• The equilibrium constant, Kc for the following reaction is 54.3 at 430 °C. H2 (g) + I2 (g) ⇌ 2HI (g) Suppose that in a certain experiment, 0.243, 0.146, and 1.98 moles of H2, I2 and HI, respectively, are placed in a 1.0 L container at 430 °C. Will there be a net reaction to form the H2, O2 or more HI ?

Key Terms/Concepts

Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction, such as :

• The tendency of the reaction to occur (but not the speed of the reaction)
• Whether or not a given set of concentrations represent an equilibrium condition.
• The Equilibrium position that will be achieved from a given set of initial concentrations.

Position of Equilibrium:- kc can have three types of values.

1. When kc > 1 , the value of numerator is greater than the value of denominator, thus, formation of products is favored at equilibrium.
2. When kc < 1 , the value of numerator is less than the denominator. So the formation of products is not favored at equilibrium.
3. When kc = 1 , the rate of forward reactionequals the rate of the reverse reaction.

Extent of Reaction

 The size of kc tells whether an equilibrium favors reactants, products or none of these.
 A very large numerical value of kc or kp means that at equilibrium the reaction system will consist of mostly products. i.e. the equilibrium lies to the right.
 Reactions with very large equilibrium constants go essentially to completion.
 A very small value of kc or kp means that the system at equilibrium will consist of mostly reactants. i.e. the Equilibrium position is far to the left.

Reaction Quotient (Q)

 Is used to determine the direction of reaction in order to achieve equilibrium.
 Law of mass action is applied, but using initial concentrations instead of equilibrium concentrations.

Consider the general reaction aA + bB ⇌ cC + dD

Qc = [C]^c [D]^d / [A]^a [B]^b…………………..Reaction Quotient interms of molarities
Q_p= (P_C)^c (P_D )^d / (P_A)^a (P_B)^b…………………Reaction Quotient interms of partial pressure.

Applications of Reactions Quotient

 When Qc = kc , the reaction attains equilibrium.
 When Qc < kc , the system must shift to the right, consuming reactants and forming products, to attain equilibrium.

 When Qc > kc, the system must shift to the left,consuming products and forming reactants, until equilibrium is reached.

EXAMPLE

1. At a certain temperature, the reaction: CO(g) + Cl₂ (g) ⇌ COCl₂ (g) has an equilibrium constant Kc = 13.8. Is the following mixture an
   equilibrium mixture? If not, to which direction (forward or backward) will the reaction proceed to reach equilibrium?
   [CO]₀ = 2.5 mol L^–1; [Cl₂]₀ = 1.2 mol L^–1, and [COCl₂]₀ = 5.0 mol L^–1.

Solution:
Given: [CO]₀ = 2.5 mol L^–1 , [COCl₂]₀ = 5.0 mol L^–1 and [Cl₂]₀ = 1.2 molL^–1

KC = [COCl₂] / [CO]x [Cl₂] = 5.0 mol L^–1/( 2.5 mol L^–1 )x(1.2 mol L^–1 )
KC = 13.8
The expression for Q_C is: Q_C= [COCl₂ ]/[Cl₂ ][ CO ] = 5.0mol/L ÷(2.5 x 1.2)

Q_C = 1.6
Hence Qc < Kc thus the reaction mixture is not an equilibrium mixture. So, the reactants should combine to form more products to reach equilibrium, i.e., the reaction proceed further to the right (forwarded reaction).

2. At 700K, the reaction: 2𝑆𝑂₂(𝑔) + 𝑂₂(𝑔) ⇌ 2𝑆𝑂₃(𝑔) has an equilibrium constant kc = 4.3 × 10⁶

• 𝑎. Is a mixture with the following concentrations at equilibrium?
  𝑆𝑂₂ = 0.1𝑀, 𝑆𝑂₃ = 10𝑀 , 𝑂₂ = 0.1𝑀
  𝑏. If not at equilibrium, predict the direction in which a net reaction will occur to reach the new equilibrium.
  Solution: a. 𝑄𝑐 = [𝑆𝑂₃]²/[𝑆𝑂₂]² [𝑂₂] = (10)² / (0.1)²(0.1)
  • Q𝑐 = 1.0 × 10⁵
    • Here 𝑄𝑐 < 𝐾𝑐 , therefore, the rxn is not at equilibrium with given concentrations.

• b. When 𝑄𝑐 < 𝐾𝑐 , the system must shift to the left to reach a new equilibrium.

The equilibrium constant of a reaction enables us to:

• Predict the composition of an equilibrium mixture (or predict extent of reaction);
• Predict the direction of the reaction
• calculate the equilibrium concentrations of reactants and products from initial concentrations.

A. Predict the composition of Chemical equilibrium

KC or KP value is used predict the extent of the reactions at equilibrium. Very large values of KC or KP indicate that at equilibrium the reaction system consists of mainly products and the equilibrium lies to the right, i.e., the reaction goes near to completion.
An equilibrium mixture contains about as much product as can be formed from the given initial amounts of reaction .

However, if KC or KP values are very small, the equilibrium mixture consists of mainly reactants and the reaction lies far to the left.
Note that a reaction goes essentially to completion if KC or KP 10¹⁰ and not at all if KC or KP 10^-10

Exercise

Predict whether the formation of product is favoured for the following reactions:
a. N2 (g) + 3H2 (g) ⇌ 2NH3 (g) KC = 3.6 × 108
b. N2O4 (g) ⇌ 2NO2 (g)
KC = 5.0 × 10-3
c. 2CO2 (g) ⇌ 2CO (g) + O2 (g)
KC = 4.45 × 10-24

2. Arrange the following reactions in order of their increasing tendency
   to proceed towards completion (least extent to greatest extent):
   a. CO + Cl2 ⇌ COCl2 Kc = 13.8
   b. 2CO2 ⇌ 2CO + O2 Kc = 2.0 × 10-6
   c. 2NOCl ⇌ 2NO + Cl2, Kc = 4.7 × 10-4

B. Predict the direction of the reaction .

To predict the direction of a reaction of a reaction, we first calculate the reaction quotient ( QC ) and compare it with the value of the equilibrium constant.

The reaction quotient is the ratio of concentrations of products to the concentrations of reactants each raised to the power equal to its stoichiometric coefficient. QC has the same form as KC, but involves concentrations that are not necessarily equilibrium concentrations.

For the general reaction: aA + bB ⇌mM + nN

Where [A], [B], [M] and [N] are concentrations at any stage during the reaction. At the initial stages of the reaction, the amount of product formed is low, therefore, the value of Q is small. As the reaction progresses, the value of Q also increases due to the increase in concentration of the products.

When the reaction attains equilibrium, Q becomes equal to the equilibrium constant.
To predict the direction of the reaction, Q is compared with Keq:
(1) Q = Keq, the reaction is at equilibrium.
(2) Q < Keq, then the reaction will proceed to the right (the forward direction) to consume the reactants and to form more products untill the equilibrium is reached.

(3) Q > Keq, the reaction will proceed to the left (the reverse direction) to reduce the products until the equilibrium is reached.

Examples

1. At a certain temperature, the reaction: CO(g) + Cl₂ (g) ⇌ COCl₂ (g) has an equilibrium constant Kc = 13.8. Is the following mixture an
   equilibrium mixture? If not, to which direction (forward or backward) will the reaction proceed to reach equilibrium?
   [CO]₀ = 2.5 mol L^–1; [Cl₂]₀ = 1.2 mol L^–1, and [COCl₂]₀ = 5.0 mol L^–1.
   Solution:
   Given: [CO]0 = 2.5 mol L^–1[COCl₂]₀ = 5.0 mol L^–1[Cl₂]₀ = 1.2 molL^–1
   • KC = [COCl₂] / [CO]x [Cl₂] = 5.0 mol L^–1/( 2.5 mol L^–1 )x(1.2 mol L^–1 )
     KC = 13.8
     The expression for Q_C is: Q_C= [COCl₂ ]/[Cl₂ ][ CO ] = 5.0mol/L ÷(2.5 x 1.2)
     • Q_C = 1.6
       Hence Qc < Kc thus the reaction mixture is not an equilibrium mixture. So, the reactants should combine to form more products to reach equilibrium, i.e., the reaction proceed further to the right (forwarded reaction

Exercises

1. At a start of a reaction, there are 0.0218, 0.0145, ad 0.0783 moles ofH2, I2 and HI, respectively, in a 3.5 liter reaction vessel at 430 °C. Decide whether or not the reaction is at equilibrium. If not, predict in which direction the reaction will proceed. The equilibrium reaction is:
   H2 (g) + I2 (g) ⇌ 2HI (g) KC = 54.3 at 430 °C
2. At a very high temperature, the following reaction has KC = 1.0 × 10^-13
   2HF (g) ⇌ H₂ (g) + F₂ (g)
   At a certain time, the concentrations of HF, F2 and H2 were 0.4 M, 0.004 M and 0.001 M, respectively. Is the system at equilibrium? If not,
   what must happen for the equilibrium to be reached?
3. The decomposition of NOBr is represented by the equation:
   2NOBr (g) ⇌2NO (g) + Br₂ (g) KC= 0.0169
   At equilibrium the concentrations of NO and Br₂ are 1.05 × 10^-2 M and 5.24 × 10^–3 M, respectively. What is the concentration of NOBr?

C. Calculate the equilibrium Concentration ,Kc

Once equilibrium constant is determined for a reaction, it can be used to calculate the concentrations or partial pressures in an equilibrium mixture. To calculate equilibrium concentrations, follow the following steps:

• Step1. Write the balanced equation for the reaction.
• Step2. Under the balanced equation, make a table that lists for each substance involved in the reaction:
  • a. the initial concentration,
  • b. the change in concentration on going to equilibrium, and
  • c. the equilibrium concentration.
    • In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.
• Step3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.
• Step4. Calculate the equilibrium concentrations from the calculated value of x.
• Step5. Check your results by substituting them into the equilibrium equation.

• Example1. Use K to find conc. of reactants and products at equilibrium. Given K= 2.7x 10-7 and initial amounts of [A]=1.0 [B]=0 and [C]=0 then find
• concentrations at equilibrium for reaction 2A ⇌ B + 3C
  • a. Make table of initial, change, and final in symbolic form
    • 2A ⇌ B + 3C

	2A ⇌	B +	3C
Initial	1.0	0	0 0
Change	-2x	+x	+3x
Equilibrium	1.00-2x	x	3x

b. Write Equilibrium expression and substitute in symbols
Kc= [B][C]³/[A]² = [x][3x]³/[1-2x]²

c. Substitute in for K and solve math exactly or simplify to approximate
2.7×10^-7 = (27)(x⁴) / [1-2x]²
Since Kc is small then x is small so approximate 1-2x ~ 1

• 2.7×10^-7 = (27)(x⁴) / [1-2x]²

• 1.0 x10^-8= x⁴

• (1.0 x10^-8)^1/4 = (x⁴)^1/4

• 1.0 x 10^-2 = x

………………………………………END………………………………….

1. Write the balanced equation for the reaction and then the KC or KP expressions.
2. List the initial concentrations.
3. Calculate QC or QP
   to determine the direction of the reaction, if initial
   concentrations of the reactants and products are given.
4. Define the change in concentration needed to reach equilibrium and find the
   equilibrium concentrations by using the change of the initial concentrations.
5. Substitute the equilibrium concentrations into the KC or KP expression, solve for the unknown.
6. Check whether the calculated equilibrium concentrations give the correct value of KC or KP

Example .

Calculate the pH of a 0.50 M HF solution at 25°C. The ionization of HF is given by HF(aq)+H2O (l) H3O+(aq)+F–(aq)is 6.8×10-4
Solution:

The species that can affect the pH of the solution are HF, and the conjugate base F–, Let x be the equilibrium concentration ofH3O+and F-ions in molarity(M). Thus,

H2 (g) + I2 (g) ⇌ 2HI (g)
[ ]
[ ][ ]
2
C
2 2
HI
K 69
H

0.5)(6.8×10–4)=3.4×10–4
Rearranging this equation gives

x=∫3.4×10–4=1.8×10 -2M

Thus, we have solved for x without using the quadratic equation. At equilibrium, we have
[HF]=(0.50–0.018)M=0.48M, [H3O+]=0.018M [F–]=0.018 M And the pH
of the solution is pH =–log(0.018) =1.74

The equilibrium constant depends upon the following factors:

Temperature at which the experiment is performed.

The form of equations which describe the equilibrium.

For example, for the reaction N2(g) + 3H2(g) → 2NH3(g) If the reaction is multiplied by2,
2N2(g)+6H2(g)→4NH3(g)the expression for Kc becomes

if the reaction is reversed, 2NH3(g)→N2(g)+3H2(g) The KC becomes

the equilibrium constant does not depend upon the initial concenetrations of the reactants.Kc and K p are independent of pressure.

Exercise