chemical reaction concluding part

CHEMICAL EQUILIBRIUM

Have you ever try adding salt to fixed volume of water continuously? Try that and note what will happen at a point. You’ll observe that it will get to a stage that the salt will not dissolve again! Yes, you’ll have your solid salt back in the water. At this point, its not that the salt is not dissolving, but as the salt is dissolving its also precipitating back to give you the solid sample. i.e. dissolution and precipitation is occurring at same rate. In other word, forward and backward reaction occurs simultaneously. Hence, the system is in equilibrium.

Now what is chemical equilibrium?

This is a state where there is no observable change in the properties of the chemical system. Yes a state where there is no observable change in the properties of the chemical system. How do I mean? I make an illustration of saturated salt solution above, in that case of the saturated solution of sodium chloride, it’s not that the salt is not dissolving, but as the salt is dissolving, the same numbers of mole of the salt particles that are dissolving are also precipitating at the same time. Therefore, making it look like the salt is not dissolving, so we cannot notice any changes in the properties of that system.

NaCl _(s) dissolution  NaCl _(aq)

precipitation

 The properties of the system that remain unchanged in equilibrium include:

1.      Concentration

2.      Colour

3.      Density

4.      Pressure

Note: that a reversible reaction can only reach equilibrium in a closed system. A closed system is a form of close vessel where all the reacting species are constant.

How to derive equilibrium constant for a chemical reaction in equilibrium

Before we do this, let’s take a look at the  law of mass action.

Law of mass action state that “at constant temperature, the rate of chemical reaction is directly proportional the active masses”.  Active mass means the concentration of the reacting substance raised to the power of their numbers of mole. i.e. consider the equation below

aA + bB → cC + dD

Where a, b, c and d are the numbers of mole of the reactant and the products. However, law of mass action can be expressed mathematically as : R₁ α [A]^a [B]^b  ,R2 α [C]^c [D]^d  where R1 and R2 equals to the rate of backward reaction and rate of forward reaction respectively.

Also R1 and R2 equals to K1 and K2 respectively. And K1 and K2 are constants of backward and forward reaction respectively.

So at equilibrium K1 ꞊ K2 .i.e.  [A]^a [B]^b  ꞊ [C]^c [D]^{d  .}

Therefore equilibrium constant (Kc) can be written as [C]^c [D]^d 

                                                                                       [A]^a [B]^b  .

I mean Kc ꞊ [C]^c [D]^d 

                    [A]^a [B]^b 

So if you are asked to write Kc. For any reaction, all you have to do is to write the R2/R1. For example. Write the equilibrium constant Kc for the reaction below.

2HBr + Cl₂ → 2HCl + Br₂

solution

Kc꞊ [HCl]² [Br₂]

       [HBr]² [Cl₂]

That is all you need to write as answer.

Factors Affecting Chemical Equilibrium

There are some factors that affect the position of  chemical equilibrium, these factors was studied by a scientist known as le chatelier, who propounded le chatelier’s principle.

Le chatelier’s principle state that “if there is a change in any factors affecting chemical system in equilibrium, the position of the equilibrium change so as to annul the effect of the changes.  

Now what are the factors that affect chemical equilibrium system and how does it affect it?

1.      Change in temperature: consider the reaction below

N2 + O2 ↔ 2NO  ∆H ꞊ + 90.4 KJ/mol.

The reaction above is endothermic reaction, increasing the temperature of the reaction will cause equilibrium to shift to the right and forward reaction will be favoured. Meaning that more of nitrogen (iv) oxide will be produced. Because the system originally need more heat from the surroundings to proceed. However, if the system is cooled down, backward reaction will be favoured. The reverse is the case if the reaction is exothermic.

2.      Change in Pressure: note that for pressure to affect a system in equilibrium, two things must be considered, which are (i) at least one of the reactants or products must be a gas. (ii) the total number of moles of both side must be different. Having considered those two factors, le chatelier says that high pressure will favour the side with lowest numbers molecule. Taken for example, consider the equation below

3H₂ + N₂ ↔ 2NH₃  

If we look at the reactant side, the total numbers of mole is 4 while that of the product side is 2. Therefore going by the rule, high pressure will favour forward reaction. That is more ammonia will be generated, while low pressure will favour backward reaction and vice versa.

3.      Change in concentration: consider the equation below

3Fe3 + 4H₂O → Fe₃O₄ + 4H₂

The higher the number of particles, the higher the effective collision and the higher the rate of the reaction. Therefore adding more reactant to that reaction above will favour forward reaction, and removing some of the reactant will favour backward reaction.

4.      Presence of catalyst: it should be noted that catalyst have no effect on equilibrium system, it only alter the rate of a chemical reaction.