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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What is The Difference Between Reaction Quotient and Equilibrium Constant?

In the previous post we have learnt about equilibrium constant. Let’s revise it quickly.

• Equilibrium constant is a ratio of product and reactant’s molar concentration at equilibrium.
• It is independent of initial concentration of reactants.
• It is specific for a particular reaction.
• It is a temperature dependent constant.
• It is unit less constant because it is the ratio of concentrations.

Equilibrium is a state when reaction seems to be seized because both forward and backward reactions go on at the same rate and no change in the concentration of reactant and product is seen . So what information can we draw by equilibrium constant?

Equilibrium constant is the ratio of molar concentration of product to the molar concentration of reactant and its value gives general information about the extent of reaction. Larger value of equilibrium constant shows that reaction is nearer to the completion when equilibrium is established, as it means product concentration is larger than reactant. Similarly smaller value shows that equilibrium is established at the beginning of the reaction, because reactant concentration is much larger than product concentration. And if its value is equal to 1, it means reaction has finished half way through because product and reactant concentrations are equal at this point.

From above discussion one more concept about equilibrium has been clarified that equilibrium can be established at any point of reaction. It is the situation when rate of forward and backward reaction becomes equal. These are the general applications of equilibrium constant but how do we get information about a particular reaction? For this we need another parameter known as “Reaction Quotient” (Q_c). We can predict the direction of the particular reaction by comparing it with the equilibrium constant.

Reaction Quotient is similar to the equilibrium constant, the only difference is that in reaction quotient, concentrations of product and reactants are at a given time not necessarily at the time of equilibrium. Let’s take an example:

a A + b B ↔ c C + d D
Q_c  = [C]^c [D]^d / [A]^a[B]^b

Here concentration of products and reactants are taken in mole/L at the time t. At the time of equilibrium Q_c becomes equal to the K_c. Let’s see how Qc and Kc help us to predict the direction of the reaction. We will take an example to explain it:

H_2(g) + I_2(g) ↔ 2HI_(g)
K_c = 57.0 at 700K.

At time t molar concentration of H₂ = 0.10M, I₂ = 0.20M and HI = 0.40M so reaction quotient will be:

Q_c = (0.40)2 / (0.10) (0.20)
Q_c = 8.0

At the time t, Q_c is less than K_c and Q_c has to upgrade itself to reach to  K_c. In order to increase the value of Q_c reaction has to produce more product which means reaction will be shifted in forward direction.

If at a certain time t₂ you find Q_cis greater than K_c, then Q_c has to downgrade itself by increasing the concentration of reactants, that means reaction will be shifted in backward direction.

In the above discussion, we have seen that reversible reactions have a tendency to achieve equilibrium state, they adjust themselves either in backward or forward direction to do so. If we want to produce more HI in the above reaction, how can we take advantage of equilibrium. Can we fool it and get more HI? In the next post we will learn it.

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Law of Chemical Equilibrium or Law of mass action

Now you are quite familiar with the term Equilibrium. We know that at equilibrium the rate of forward reaction and the rate of backward reaction are equal to each other. So, the composition of reaction mixture also has a fixed value which means the reactant and product will be present in a fixed ratio. This ratio is the characteristics of that particular reaction and it is governed by the law of equilibrium.  In this post we will learn about it. I want to tell you one thing that Equilibrium is possible only in reversible reactions and not all reactions are reversible. In the coming posts we will study reversible reactions.
 
Let's take an example of a reversible reaction in which reactant ‘A’ reacts with ‘B’ to form products ‘C’ and ‘D’. Now write the balanced equation of this reaction.  

aA+ bB ↔ cC + dD

To get the equilibrium constant K_cwe have to determine the molar concentration of all species at equilibrium. And this is shown by enclosing the specie in square bracket.

K_c = [C]^c [D]^d/ [A]^a [B]^b

Equilibrium Constant Kc 

K_c is known as equilibrium constant of the reaction. And subscript c indicates that the concentrations are expressed in moles per litre (it is also termed as Molarity and shown by symbol M). 

To get the equilibrium constant K_cwe have to determine the molar concentration of all species at equilibrium. Molar concentration is shown by enclosing the species in square bracket. Let's take an example of a real reaction.

H_2(g) + I_2(g)  ↔ HI_(g)

Now write the balanced equation in which number of moles of each element on both sides of arrow becomes equal. In the above equation if we multiply 2 on right side, then both sides will have 2 moles of ‘H’ and 2 moles of ‘I’. So the balanced equation will be:

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

and K_c will be:

K_c = [HI]²/ [H] [I]

Let's find the unit of Equilibrium constant:

K_c = [moles/L]²/ [moles/L] [moles/L]
K_c = [moles/L]

Now try a different reaction. 

HI_(g)  ↔ H_2(g) + I_2(g) 

Now write the balance equation:

2HI_(g)  ↔ H_2(g) + I_2(g)  -----------(2)

And if we name K_c for this reaction is K'_c , then:

K'_c = [H] [I]/ [HI]²

Now find out the unit of K'_c

K'_c = [moles/L] [moles/L] / [moles/L]²
K'_c = 1/ [moles/L]

So, you see the Equilibrium constant has different units for different reactions. 
One more thing, did you notice that 2^nd reaction is the reverse reaction to the 1^stone? And did you find any relation between equilibrium constants of both reactions? 

K_c = 1/ K'_c

If we reverse a reaction, equilibrium constant also gets reversed for that reaction. 
And what will happen if we multiply the reaction 1? Like:

n H_2(g) + n I_2(g)  ↔ n 2HI_(g)

K"_c = [HI]²ⁿ/ [H]ⁿ [I]ⁿ
K"_c = (K_c)ⁿ

Let's try it with real data for reaction 1. Calculate K_c if at  equilibrium, concentration of H₂ is 1.4×10^-2 mol /L , I₂ is 0.12×10^-2 mol/L and HI is 2.52×10^-2 mol/L.

K_c = (2.52×10^-2)²/ ( 1.4×10^-2) ( 0.12×10^-2)
K_c = 46.4 mol/L

Let's take another data set. Calculate K_cif take initially 2.4×10^-2 M of H₂ and 1.38×10 ^-2M of I₂ and at  equilibrium concentration of H₂ becomes 1.4×10^-2 M , I₂ becomes 0.12×10^-2 M and HI is 2.52×10^-2 M.

K_c = (2.52×10^-2)²/ ( 1.4×10^-2) ( 0.12×10^-2)

K_c = 46.4mol/L

Now you have seen that initial concentration doesn't affect the equilibrium constant. Let's take another example. 

Calculate K_c if at  equilibrium concentration of H₂ is 0.77×10^-2 mol /L , I₂is 0.31×10^-2 mol/L and HI is 3.34×10^-2 mol/L.

K_c = (3.34×10^-2)²/( 0.77×10^-2)( 0.31×10^-2)
K_c = 46.4 mol/L

You might get surprised, even after changing the equilibrium concentration, equilibrium constant doesn't change. Equilibrium constant is the characteristic of a given reaction at a particular temperature. If you change the reaction temperature of the same reaction you will get a different value of K_c.

What are the other factors which affect equilibrium and equilibrium constant? In the next post we will discuss it in detail. 

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Chemical Equilibrium

Did you ever try to balance a half filled cold drink bottle horizontally on your finger? Its a difficult task because when you try to balance it, the liquid shifts to one side and the bottle gets imbalanced. But at a point when the amount of liquid on both sides becomes equal, the bottle gets balanced on your finger. This situation is known as equilibrium.  

Reaction is a very common term which we use frequently in our daily life. We will learn about it in chemistry also. When something named A reacts to B and produce a completely different thing C, this all process is known as a chemical reaction.

A + B → C

Here A and B are reactants and C is the product. In this reaction a number of molecules react with each other and a number of product molecules are formed. If you visualize it, you will see there is a pool of molecules which are moving from reactant to product side just like the liquid moves in the bottle. Bottle is the vessel in which this reaction is carried out.    

Equilibrium is that balance point at which the number of reactant molecules are equal to the number of product molecules. But it doesn't mean that equilibrium is a static condition at which reaction gets seized. Actually at equilibrium the rate of formation of product molecules is equal to the rate of break down of product molecules changing them back to reactant molecules again. That's why number of reactant molecules remain equal to the number of product molecules and it seems that reaction has been seized. 

What is Chemical Equilibrium?

When two people of equal strength arm wrestle, there comes a stage when no one could defeat the other, this is the equilibrium stage. At arm wrestling equilibrium the forces applied by both contestants are equal and opposite. Similarly, when in a chemical reaction the rate of forward reaction (forward arrow→) becomes equal to the rate of backward reaction (backward arrow ←), it attains the equilibrium stage which is depicted by two half arrows pointing in the opposite directions.  

In the last post of liquification of gases, we have observed a plateau region in the graph, where liquid CO₂ and gas CO₂ both coexist. At the middle of the plateau we get the equilibrium point. That was an example of gas-liquid equilibrium. 

Similarly, you have seen solid-water equilibrium when you place ice in water and keep them at constant temperature above zero degree. At this condition you will see there is no change in the amount of ice and water because freezing and melting both are going on at equal rate.

You may wonder why I took the example of bottle to explain equilibrium. Because it is one of the characteristics of equilibrium.  Bottle is the vessel in which reaction is carried out, in the language of chemistry bottle is the system. Equilibrium is possible only in the closed system and as the bottle is capped it becomes the closed system. When I talk about the bottle of cold drink, can you guess which equilibrium I am talking about? Here the equilibrium is set between gas dissolved in the liquid and undissolved gas. Closed system is that system which doesn't allow molecules or heat to escape. Another characteristic of equilibrium is fixed parameters like temperature and pressure. 

Let's take an example of real chemical reaction. When we take N₂ and H₂ we get NH₃. Initially N₂and H₂ get combined to form NH₃ in forward reaction. After some time the NH₃ formed by this reaction starts to decompose to N₂ and H₂ in backward reaction.  And after some time the system reaches the equilibrium, when both reactions occur in the same rate and concentrations of product and reactants become equal. If we plot a graph between concentration and time, we can visualize it happening. In one side we take reactant concentration and on the other side we take the product concentration. Reactant vs time curve represents forward reaction, while product vs time curve represents backward reaction.  You can see in the graph that we can reach at equilibrium stage by any of the curve. That means either we took N₂ and H₂ or NH₃ alone, we would reach the equilibrium by same time. 

What happens if we take out some of the product from the system or we change the temperature or pressure of the system? How will it affect the equilibrium? You can answer all these questions but for that you'll have to learn some laws of equilibrium. In the next post we will discuss these laws and learn how to use them to find answers of these questions.

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Liquification of gases

We all are familiar with the different phases of matter viz. gas, liquid and solid. You have learned that the basic difference between these phases is the strength of inter-molecular attraction between their molecules. By changing the strength of inter-molecular attraction between molecules of any phase we can transform it to another phase.  In this post we will learn how we can transform gas into liquid phase and see how the knowledge of gases and their laws helps us to make this transition.  

When one phase transforms into another phase, one intermediate phase occurs during this transition which is present in between these two phases or you can say that this third phase is a bridge between two different phases. In the transition of gas into liquid, vapour is the intermediate phase. 

What are the key changes we have to make in the molecules of gas phase to convert it into liquid phase? We have to bring them closer so they are held together by intermolecular attraction, in order to do it we have to reduce the volume. If we apply pressure we can reduce the volume of gas. Furthermore, we need to reduce temperature to slow down the speed of molecules. That means we need to find that particular temperature, pressure and volume in which we can liquefy real gas, and here Boyle’s lawhelps us.     

In the last post you have learned that, real gases follow ideal behaviour and obey Boyle’s law at higher temperature. You have seen in the graph of pV vs p where a straight line was obtained for ideal gas, which means that its volume cannot be reduced even on applying high pressure or in other words ideal gas cannot be liquefied. On the other hand, when we reduce the temperature of real gases, they deviated from the Boyle’s law. The highest temperature at which a real gas shows deviation from ideal behaviour for the first time is the temperature at which we can liquefy a real gas. This is known as critical temperature (T_c). And corresponding pressure and volume are known as critical pressure (p_c) and critical volume (V_c).

Let’s take an example of CO₂gas. At higher temperature range from 50 ͦ C to 31 ͦ C when pressure is applied it shows perfectly Ideal behaviour as expected by Boyle’s law (to learn more visit DoReal Gases Behave Ideally?). When we reduce the temperature just a bit more to 30.98 ͦ C it shows deviation from Boyle’s law on applying pressure. In the graph this deviation is clearly recognised by a sudden change in curve. At this point we get liquid CO₂ for the first time. This temperature 30.98 ͦ C is the critical temperature (T_c) of CO₂gas. At this temperature on applying pressure CO₂ gas gets compressed and transforms into liquid CO₂.

Liquification of Gas

What happens if we further reduce the temperature? CO₂ gas shows different behaviour on applying pressure at temperature below 30.98 ͦ C.

• On compressing, initially CO₂ gas remains gas till point ‘B’
• On applying still more pressure it shows deviation from Boyle's law and a little liquid CO₂ appears
• On further compression the pressure remains constant for a period (point ‘B’ to ‘C’) and we get a plateau for this phase. In this region we get vapour CO₂that means a state in which liquid and gas coexist. 
• If we further compress it, a steep rise in pressure is observed (point ‘C’ to ‘D’). As the plateau ends we start getting liquid CO₂.

All real gases show similar behaviour as CO₂. CO₂ gas represents all real gases. But every gas has a particular set of critical constants.

So what you have learnt in this post? At critical temperature (T_c) you can liquefy a gas directly into liquid phase. It means that you can skip transition phase. But if you carry out liquification at a temperature lower than critical temperature, you will get the transition phase region or two phase portion in which gas and liquid coexist.

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Do Real Gases Behave Ideally?

In the last post we have discussed the reasons behind the non-ideal behaviour of real gases. As we behave ideally in certain conditions, real gases also do. What are those conditions in which real gases behave like an Ideal gas?

As you know intermolecular forces are responsible for the non-ideal behaviour of real gases. So think about those situations in which these forces have no significance. Inter molecular forces are effective within a small range, if we increase the distance between two molecules, Inter molecular forces will diminish.    

In Boyle’s Law you have learnt that at lower pressure the volume of gas increases, which in turn increases the distance between molecules and in the larger space, volume of molecules can be neglected.  So by decreasing pressure we can beat attractive force by increasing the distance.

But still molecules are not completely free from attractive force; there are chances that they can be affected by the attractive force if any of them passes a nearby molecule. To cancel this possibility of getting caught by attractive force of nearby molecule, we have to increase the speed (or kinetic energy) of molecules and to increase their kinetic energy we have to increase temperature. At higher temperature molecules travel with higher speed and wouldn't get caught by other molecule.

That means at lower pressure and higher temperature real gases behave like an ideal gas. The temperature at which a real gas obeys Boyle’s law is called the Boyle’s temperature or Boyle’s point. Boyle’s point of gas depends on it’s nature.

Now we know how we can modify conditions to force real gases to behave ideally. Can we measure their deviation from ideal behaviour?

We know in Ideal gas equation:

 pV = nRT

pV/ nRT = 1

For an ideal gas the ratio of pV/ nRT is equal to 1. But in case of real gases this ratio deviates from unity. This ratio is defined as ‘Compressibility Factor’ and denoted as Z.

pV/ nRT = Z

When a graph is plotted between Z and pressure, we get a straight line for ideal gas. And for real gases we get different curves showing positive or negative deviation from straight line.

If we derive another equation from it, you will be able to understand the Compressibility Factor better.

pV/ nRT = Z

if we write V as V_real

p V_real / nRT = Z

From ideal gas equation we know:

pV_ideal = nRT

V_ideal = nRT/p

Now place the value of nRT/p in the above equation:

V_real / V_ideal= Z

In a graph of Z vs Pressure for ideal gas, we get a straight line at Z=1 which is parallel to the x axis. It doesn’t mean that with increase in pressure no changes occur to ideal gas, changes occur but these changes occur in such a manner that the ratio of pV/nRT remains unity.

While in case of real gases these changes occur in undisciplined way and the ratio of pV/nRT deviates from unity. Most of the real gases show a three staged curve where they have Z ≈1 at lower pressure, Z < 1 at high pressure and Z > 1 at a still higher pressure.

Stage I:  At lower pressure where Z ≈ 1 all gases show ideal behaviour. As pressure increases most of the real gases show negative deviation where Z < 1, which means V_real is less than V_ideal which signifies that the gas gets compressed more than the ideal gas at increased pressure. Here first time real gases start disobeying the Boyle’s law (p ∝ V^-1).

Stage II: As the pressure further increases, all real gases touch the straight line for an instance when they have Z= 1 and where they behave ideally since V_real is equal to V_ideal. 

Stage III: As the pressure reaches to still higher range, all real gases again deviate from ideal behaviour and show positive deviation where Z > 1. At this stage V_real is more than V_ideal that means the gases no more follow the trend of Boyle’s law of decrease in volume on increasing pressure. At this stage real gases again start disobeying the Boyle’s law (p ∝ V^-1) and it becomes impossible to compress them. 

Now you understand that why real gases obey Ideal gas equation at lower pressure and higher temperature. Does this information benefit to us? In the next post we will see how we can use it in practical life?​

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