Lewis Acid Base Theory

You have studied the Lewis dot structure in earlier posts. The same G.N. Lewis who has given the Lewis dot structure has defined acid and base in more generalized term. Every species has electron pairs, so he gave his definition in the terms of electron pairs.

Lewis acids are those which accept electron pairs and Lewis bases are those which donate electron pairs. Let’s check that BF₃ fits in which category?

Lewis Acid and Base

Draw the Lewis dot structure of BF₃molecule. You will find that B has 6 electrons it has incomplete octet. That means it is ready to accept an electron pair to fulfill its octet. Thus it falls under the category of Lewis acids. It means that the species which have incomplete octet (electron deficit) can act as Lewis acid like AlCl₃, BCl₃, Mg²⁺ and so on.

Lewis Vs Bronsted theory

Lewis theory Vs Bronsted theory

Let’s check whether the Brönsted base NH₃ is also a Lewis base or not? When you draw the Lewis dot structure of NH₃ you will find that is has 3 bonded pairs and 1 lone pair of electrons. It can donate one lone pair of electron and acts as Lewis base. Species those are electron rich can act as Lewis base like H₂O, OH^-, Cl^-, O^2- and so on.

If you observe closely, the Brönsted concept and Lewis theory apparently do not have much difference. Brönsted base accepts H⁺(proton) and Lewis base donate electron pair which is the same thing, only difference is in the language.

NH_3(aq) ⟶ NH₄⁺_(aq) 

Brönsted base NH_3(aq) accepts H⁺and becomes  NH₄⁺_(aq). When you draw the structure you will find that NH₃ provides electrons to make bond with H⁺. It means that in either case (Brönsted and Lewis) base provides electrons. Similarly acid accepts electrons in both cases.

How do we know which acid or base is stronger? Is there any method to measure the strength?  In the next post we will explore it.

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Brönsted -Lowry Acid Base Theory

Johannes Brönsted and Thomas M. Lowry gave a generalized definition of acids and bases. They defined them by a common term “proton (H+)”. Acids are those which give proton and bases are those which accept proton.

Brönsted Acid and Base

Let’s see if this theory can explain the basic nature of NH₃.

NH_3(aq)+ H₂O_(l) ⇌ NH₄⁺_(aq) + OH^- _(aq)

Here NH₃accepts proton from H₂O, hence it is called as base.

Conjugate Acid - Base Pair

This theory not only defines acid and base but also clarifies their relation with each other. Acid and base are like the two sides of a coin. As either side of a coin cannot stay alone, acid and base also can’t stay alone. Each of them has its counterpart which is named as conjugate.

In the above equation NH₃ accepts proton and becomes NH₄⁺, here it acts as base.

NH_3(aq) ⟶  NH₄⁺_(aq) 

When you see the reverse reaction, you will see NH₄⁺donates proton and becomes NH₃, thus it acts as acid. NH₃is a base and NH₄⁺is its conjugate acid or vice versa.

NH_3(aq) ⟶ NH₄⁺_(aq) 

Similarly H₂O gives proton to NH₃ and becomes OH^- , so H₂O acts as an acid and OH^- is its conjugate base, which accepts proton from NH₄⁺in reverse reaction. Let’s see one more equation.

HCl_(aq)+ H₂O_(l) ⇌ H₃O⁺_(aq) + Cl^- _(aq)

Here HCl gives proton H⁺ and acts as acid, while H₂O accepts proton and acts as base. I hope now you can find their respective conjugate. HCl has its conjugate base Cl^- and H₂O has its conjugate acid OH^-.

Conjugate Acid-Base Pair

In the above two equations you have seen that H₂O acts as acid when it comes with NH₃ and acts as base when it comes with HCl. That means acid and base are comparative terms.

For example when 2 comes with 1, 2 looks bigger than 1 but if it comes with 3, it looks smaller. Similarly H₂O acts as acid when it comes with NH₃ and acts as base when it comes with stronger acid HCl.

Which factor decides the strength of an acid or base? Readiness to give off the proton decides the strength of any acid. If we compare two acids, the one which readily gives off the proton is the stronger acid. And similarly the one which accepts proton readily is the stronger base.

It is very easy to figure out the corresponding conjugates for acids and bases. If you want to find conjugate acid of any species just add proton (H⁺) to it and if you want to find conjugate base, subtract proton (H⁺) from it. Let’s practice few examples of conjugate acid- base pair:

Species	Conjugate Acid	Conjugate Base
NH3	NH4+	NH2-
H2O	H3O+	HO-
HSO4-	H2SO4	SO42-

Now you must be able to guess the nature of species and also to find the conjugate acid and base of any species. But what happens to those species which lack a Hydrogen? For example, how can we find out whether BF₃ is an acid or base? Arrhenius concept and Brönsted -Lowry acid base theory both are not able to help us in this case. So how do we find the right answer? In the next post we will try to find out its answer.

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What are Acid Base and Salt?

The name acid has been derived from the Latin word “acidus” which means sour. You must have experienced a number of foods which taste sour, like lemons, oranges, grapes and many others. Lemons have citric acid and ascorbic acid is found in oranges. You will be thrilled to know that your body also produces hydrochloric acid in stomach which helps in digestion of food. You can check yourself if something has acid in it or not by simply putting it on a litmus paper. If the paper turns red that means it has acid. Acid changes the colour of litmus paper to red. 

As everything has its counterpart, acid also has its counterpart which is named as Base. It tastes bitter. Baking soda, washing soda are common examples of bases. If you try a litmus paper test, a base will turn it into blue. And when acid and base come together they cancel the effect of each other and thus salt comes into existence. Table salt is the most abundant salt in the nature.

What are acid base and salt?

How can we define an acid and a base? Initially scientist Svante Arrhenius gave a theory which is named after him the Arrhenius theory. According to it acids are those substances which give hydrogen ion when dissolved in water.

HX_(aq) ⟶  H⁺ + X^-_(aq)

H⁺ is very reactive and can’t live alone so it combines with O of H₂O and forms H₃O⁺ ion, it is called hydronium ion. I have used X with H which represents halogens group 17 elements, but why? Think about it.

HX_(aq) + H₂O_(aq) ⟶  H₃O⁺ _(aq) + X^-_(aq)

And bases are the substances which give OH^- ion when dissolved in water.

MOH_(aq) ⟶ M⁺ + OH^-_(aq)

Here I have used M with OH and it represents metals of group 1, think about it too. Arrhenius theory fits well in this case, but it is limited to the aqueous solutions only. It doesn’t explain acidic or basic behaviour of substances lacking H⁺ or OH^-ions. Like ammonia NH₃ which is a base but doesn’t have OH^-ion.

In the next post we will see what solutions have been given by the scientists for such problems.

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Chemical Equilibrium at a Glance

Chemical Equilibrium at a Glance

Equilibrium Constant for a general reaction and its multiples

Le Chatelier’s principles

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Le Chatelier’s principle: Temperature change

In the last post we have seen how the system dealt with the concentration change and pressure change, today we will see what happens if we change the temperature or add some foreign substances like catalyst or noble gases to the system.

You know that equilibrium constant depends on temperature, if we change the temperature, system will no longer be in equilibrium. How can a system control its temperature itself? Energy is released when new bonds are formed and this provides heat to the system. And energy is needed when a bond is broken which is supplied by the system in the form of heat. Now you can guess how the system can deal with it.

When reactants combine to form products, some old bonds are broken and some new bonds are formed. And when we subtract the energies involved, we get to know how much energy is used or released in that particular reaction. If the energy of the reactants is more than that of the products, then energy will be released in the reaction, such reactions are called exothermic reactions. And its opposite is called endothermic reactions, here energy is required.

N_2(g) + 3H_2(g) ⇌ 2NH_3(g)  ......... ∆E = 92.38 kJ mol^-1

The above reaction is an example of exothermic reaction, which means some amount of energy is released. If we increase the temperature of the system, then the system will shift the reaction in backward direction so that it can consume some of the heat. And if we lower the temperature then system will make the forward reaction faster to produce more heat. That means if we want to produce more ammonia we have to keep the temperature low.  

Effect of Temperature change on Equilibrium

Effect of catalyst addition

Catalysts are those substances which speed up the reaction without being involved it in. Suppose you are participating in a race and suddenly you find that a furious dog is chasing you, then what will happen? Naturally you will run like hell. Here the dog is neither participating in the race nor is it involved the race, but its presence speeds up your running. So dog acts as a catalyst.

Catalysts can help a system to achieve equilibrium sooner but their presence don’t create any disturbance because they don’t participate in the reaction.

Similarly addition of noble gases don’t alter the equilibrium because they are noble in nature and do not participate in reaction.

So you have learnt how Le Chatelier’s principle help us to predict the direction of the reaction and help us to understand how a system deals with the changes. Now you will be able to understand what Le Chatelier’s principle states, it states that “a change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner so as to reduce or to counteract the effect of the change”.

In the next post we will try to sum up all the findings of equilibrium.​

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What Is the Relation between Kc and Kp?

In the last post we learned about equilibrium constant Kc, for which we expressed concentration of reactants and products in terms of molarity (mols/L). If all species in a reaction mixture are gases, it becomes difficult to measure their concentration in molarity. For such a reaction mixture, it is convenient to measure concentration of their participants in terms of Partial Pressure.

You are quite familiar with the gases and you know how they exert pressure. If two or more types of gas molecules are present in a container, how will you decide which gas exerts maximum pressure? And how much pressure is exerted by each gas?

Let's try a different example, imagine that 4 members of yellow team, 6 members of green team and 10 members of orange team are jumping on the stage. Their combined efforts exert pressure on the floor of the stage. So what do you think which team contributes more? Obviously orange team contributes more because 10 out of 20 members are from Orange team. It means the team with larger fraction (team members/ total members) contributes more. Or, we can say that fraction of team members is proportional to the pressure exerted by team. Pressure exerted by an individual team is called the partial pressure of that particular team. Total pressure exerted on the stage is the sum of partial pressure of all teams.

P_total = P₁+ P₂+ P_3........

Similarly, when all participants in a reaction vessel are in gaseous state, their concentration is determined by their partial pressure. Let’s find out how we can relate partial pressure to the concentration.

From Ideal gas equation we know that:

PV= nRT

P= nRT/V

n/V is concentration in moles per litre, so

P= cRT

So we can say that:

P = [concentration of gas] RT

At constant temperature we can say that pressure of gas is proportional to its concentration:

P is proportional to c

Let’s take a reaction as example:

H_2(g) + I_2(g) ↔ 2HI_(g)           ....................(1)

For this reaction equilibrium constant will be:

K_c= [HI]² / [H] [I]

Or, if we write in terms of partial pressure, then K_c will become K_p

K_p= (P_HI)² / (P_H) (P_I)

Since P = cRT we can write:

K_p= (P_HI)² / (P_H) (P_I) = [HI]² (RT)²/ [H]RT [I]RT

K_p= K_c

Here you have seen that K_p = K_c but, it doesn't happen always. If it is not true then what is the relation between them. Let’s try to find out their relation:

a A + b B ↔ c C + d D

K_c  = [C]^c [D]^d / [A]^a [B]^b

K_p= (P_C) (P_D) / (P_A) (P_B)

K_p= (P_C) (P_D) / (P_A) (P_B) = [C]^c(RT)^c [D]^d(RT)^d / [A]^a (RT)^a[B]^b (RT)^b

K_p= K_c (RT)^(c+d)-(a+b)

K_p= K_c (RT)^Δn

Relation between Kp and Kc

Where Δn = (number of moles of gaseous products - number of moles of gaseous reactants) in a balanced chemical equation.

In equation (1) number of moles of reactants 2 and number of moles of gaseous product is 2, that’s why for this reaction K_p= K_c.

Let’s check this relation for another reaction:

N_2(g) + H_2(g) ↔ 2NH_3(g)

It is not a balanced equation since number of H isn’t equal on both sides of arrow. First we write the balanced equation:

N_2(g) + 3H_2(g) ↔ 2NH_3(g)             .................(2)

This reaction has total 4 moles of reactants and 2 moles of product, thus we get

Δn = 2-4 = -2

If the above relation is correct, we would get:

K_p = K_c(RT)^-2

Let’s try to find out:

K_c= [NH₃]² / [N] [H]³

And

K_p= (P_NH3)² / (P_N) (P_H)³

K_p= (P_NH3)² / (P_N) (P_H)³ = [NH₃]² (RT)² / [N]RT [H]³ (RT)³

K_p = K_c(RT)^-2

Yes, we have successfully proved it.

Like K_c, K_p is also a unit-less constant and since it is the ratio of pressures, its unit depends on it. For equation 1, it is unit-less quantity but for equation 2 its unit is bar^-2.

I hope you have understand the concept of K_p and its relation with K_c. Let’s try to solve a problem:

For reaction 2NOCl_(g) ↔ 2NO_(g) + Cl_2(g) value of K_c is 3.75×10^-6 at 1069K. Calculate the K_p for the reaction at the same temperature.

You know that:  K_p = K_c (RT)^Δn 

This reaction has 2 moles of reactants and total 3 moles of product, thus we get

Δn = 3-2 = 1

So,

K_p= K_c (RT)

K_p= 3.75×10^-6 (0.0831)(1069)

K_p= 0.033

Now we have learnt how to calculate equilibrium constants, but we don't know its significance. What information can we draw from it? In the next post we will explore its significance.

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Le Chatelier’s Principle: Concentration and Pressure Change

Now you are familiar with the term ‘equilibrium’. In this post we will try to understand its nature. You know that equilibrium is established under particular conditions of temperature, pressure and concentration. If you change any of these conditions it gets disturbed. But it has a peculiar quality that when anything disturbs it, it tries to overcome that disturbance and regain its peace. How does it overcome disturbances? Let’s try to learn it.

Equilibrium happens in reversible reactions or systems. When equilibrium gets disturbed due to any change, the system works to nullify those changes and regain its equilibrium. It is known as Le Chatelier’s principle (I am not giving its proper definition). In this post we will see how a system deals with the change in concentration and pressure?

Effect of Concentration Change

H_2(g) + I_2(g) ⇌ 2HI_(g)

If we add some H₂ or I₂ in above reaction mixture, system will no longer be in equilibrium. To regain its equilibrium, the system will work to reduce the concentration of H₂ or I₂. So it works in forward direction to consume excess H₂ or I₂and regain its equilibrium.

If we add some HI, then the system will start working in backward direction and regain its equilibrium. If we remove some HI then what will it do? It will work in forward direction and produce more HI to cancel out the changes. And if we frequently remove some HI from the system it will continually produce HI.

Effect of concentration change on equilibrium

Effect of Pressure Change

In previous post (Ideal Gas Equation) we have seen that pressure is inversely proportional to the volume and directly proportional to the number of moles. Pressure change affects only those systems or reactions which involve gaseous reactants and products but has no effect on solid and liquid reactants or products because pressure change does not cause much effect on them.

Pressure change affects those reactions in which total number of moles of reactants and total number of moles of product are different. Let’s take an example:

CO_(g) + 3H_2(g) ⇌ CH_4(g) + H₂O_(g)

As you can see, in this reaction 4 moles of reactants are being converted into 2 moles of products.
If we reduce the volume of the reaction vessel by half then pressure of the system will be doubled (P∝V^-1). By doing this, we have disturbed the equilibrium of the system. So it shifts the reaction in forward direction (pressure ∝ number of moles), thus it can reduce the number of moles and the pressure of the system.

Le Chatelier’s Principle: effect of pressure change

Similarly if we reduce the pressure of the system, it will shift the reaction in backward direction and by increasing the number of moles the system will cancel the effect of pressure and resume its equilibrium.  

Now you have seen how smartly a system reacts and regains its equilibrium. In the next post we will see how it deals with the other changes.

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