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Equilibrium is a relatively less important topic in JEE Main Chemistry. However, it forms an easy chunk of the Physical Chemistry syllabus. A total of 2 questions are usually asked carrying 8 marks from this topic. Therefore, the weightage of Equilibrium in the JEE Main is around 2-3%.
Although a low weightage compared to other topics like Atomic Structure and Electrochemistry, the questions are quite scoring. Some of the topics tested under Equilibrium are Le Chatelier's Principle, Relation between Kp and Kc, Solubility Product, and Salt Analysis. More than one question was asked from Equilibrium in the JEE Main Session on January 8th, 2020. Check JEE Main Chemistry Syllabus
Before you proceed towards studying the notes for Equilibrium, take a look at the following question -
What is the order of the basic character?
1) I > II > III > IV
2) IV > III > I > II
3) II > I > III > IV
4) IV > I > II > III
To ace Equilibrium in JEE Main Chemistry, refer to the study notes given below.
Equilibrium is the state of a process in which the system's properties such as temperature, pressure, and concentration display no change over time. Two opposing processes are involved in all processes which attain equilibrium. It reaches equilibrium when the concentrations of the two competing processes are equal. The equilibrium is considered Physical Equilibrium because the competing mechanisms require only physical shifts. If chemical reactions are the opposite processes, the equilibrium is called Chemical Equilibrium
Physical equilibrium can be described as equilibrium in physical processes.
|Solid ⇔ Liquid H2O(s) ⇔ H2O(l)||At a constant pressure, melting point is fixed.|
|Liquid ⇔ Gas H2O(l) ⇔ H2O(g)||PH2) constant at a specific temperature.|
|Solute (s) ⇔ Solute (solution) Salt(Solid) ⇔ Salt(in solution)||At a specific temperature, the conc. of solute in a solution is constant.|
|Gas (g) ⇔ GAs (aq.) CO2(g) ⇔ CO2(in solution)||At a specific temperature, gas (g) or gas(aq.) is constant.|
An example of solid - liquid equilibrium is the relationship of ice and water. In a close network, ice and water achieve equilibrium at 0oC. Around this point the rate of ice melting is equal to that of water freezing. Equilibrium is seen here as -
H2O(s) ⇔ H2O(l)
Water evaporation in a closed tank is an example of liquid-gas equilibrium, where evaporation rate equals condensation rate. Equilibrium is seen here as -
H2O(l) ⇔ H2O(g)
As you add more and more salt in water that is taken in a glass jar and mixed with a glass rod, after you dissolve it a bit, you will figure out that there is no more salt going to the water and it sets down at the bottom. The solution is now reported to be saturated and in a state of equilibrium. At this point, several salt molecules from the undissolved salt enter the solution (dissolution) and the same quantity of dissolved salt is deposited back (precipitation). Therefore, the rate of dissolution at equilibrium is equal to that of precipitation.
Salt(Solid) ⇔ Salt(in solution)
Within a closed vessel, the dissolution of a gas in a liquid under pressure forms a gas-liquid equilibrium. Cold drink bottles are the perfect example of this type of equilibrium. The equilibrium within the bottle is-
CO2(g) ⇔ CO2(in solution)
A reaction is considered a reversible reaction when not only the reactants react to form the products under specific conditions but also the products react to form reactants under the same conditions.
If a reaction can not take place in the reverse direction , i.e. the products formed do not react under the same condition to give back the reactants, it is called an irreversible reaction.
Typically, a chemical equilibrium is represented as -
where A and B are reactants, and C and D are products.
The double arrow between the left part and the right part indicates that shifts exist in both directions.
Based on the degree of reaction, chemical reactions can be divided into three groups before equilibrium is reached.
A) Reactions which are almost complete.
B) Certain reactions which only continue to a very limited degree.
C) Those reactions which continue to such an extent that the equilibrium concentrations of reactants and products are comparable.
In fact, the state of equilibrium is fluid and not static. When the rate of forward reaction is equal to that of backward reaction, a reaction is assumed to have reached equilibrium.
If all the reactants and products of any equilibrium reaction are in the same physical state, this is known as homogeneous equilibrium.
N2(g) + 3H2(g) 2NH3 (g)
Here, all the reactants and products are in the same physical state.
Heterogeneous equilibrium is the type of equilibrium in which the state of one or more reacting species varies, i.e. both reactants and products are not in the same physical condition.
2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(l)
Here, the reactant is in the solid state while the product is in the solid, liquid, and gaseous matter states.
Chemical reactions occur mostly in solutions. Solution chemistry plays a very critical role in the process. All chemical substances consist of either polar or nonpolar units (called ions). The operation of these substances is more apparent and noticeable in solutions. The action of these substances is based on the nature and conditions of the medium in which they are applied. Consequently , it is important to understand the principles which regulate their actions in solution.
This kind of equilibrium is found in quickly ionizing substances, or in polar compounds where ionization can be caused. Because of the ease of ionization in solvent medium, ionic and polar substances are more easily soluble in polar solvents. These solutions become rich in mobile charge carriers (ions) with the dissolution of ionic and polar substances in the solvent, and thus can conduct electricity. Substances capable of conducting electricity are called electrolytes, while non-conducting substances are called non-electrolytes.
Check Gujarat Board Released Question Banks for JEE Main
i) When the reversible reaction is conducted in a closed system, equilibrium can be achieved.
ii) This can be done from either side of the reaction.
iii) A catalyst can speed up the equilibrium approach but does not change the equilibrium state.
iv) It is of a dynamic nature, i.e. the reaction does not end but reactions both forward and backward take place at the same time.
v) Changes in pressure, concentration, or temperature favor one of the reactions (forward or backward) resulting in a shift in one direction of equilibrium point.
Guldberg and Waage developed a relationship between the rate of chemical reaction and the concentration of the reactants or, in the context of the law of mass action, their partial pressure. According to this law, “the rate at which a compound reacts is directly proportional to its active mass, and the rate of a chemical reaction is directly proportional to the sum of the active masses of reactants each elevated to a value equal to the related stoichiometric coefficient occurring in the balanced chemical equation”.
rate of reaction ∝ [A]a.[B]b
rate of reaction = K[A]a[B]b
where K is the rate constant or velocity constant of the reaction at that temperature.
Unit of rate constant (K) = [moles/lit]1–n time–1 (where n is order of reaction).
The rate of forward and backward reactions is equal in equilibrium and the concentration of reactants and products remains constant as shown in the figure after achieving equilibrium.
Active mass = number of moles/volume in litres
Active mass of solid is taken to be unity.
Also, Active mass of reactant (a) = Conc. × activity coefficient
i.e. a = Molarity × f for dilute solution f = 1
Applying Law of mass action for general reversible reaction,
aA + bB cC + dD
Rate of forward reaction [A]a[B]b
or Rf = Kf [A]b [B]b
Similarly for backward reaction
Rb = Kb[C]c [D]d
At equilibrium Kf[A]a[B]b = Kb[C]c[D]d
The above equation is called equilibrium equation and Kc is known as equilibrium constant.
The subscript ‘c’ shows that Kc is displayed in concentrations of mol L–1.
At a specific temperature, the product of concentrations of the reaction products raised in the balanced chemical equation to the respective stoichiometric coefficient divided by the product of concentrations of the reactants raised to their corresponding stoichiometric coefficients has a constant value. This is known as the Law of Chemical Equilibrium or Equilibrium Law.
In the reverse reaction, the equilibrium constant is the inverse of the equilibrium constant for the forward path reaction.
Equilibrium Constant in terms of Partial Pressures (Kp)
For reactions that involve gases,
it is more feasible to display the equilibrium constant in terms of partial pressure.
The ideal gas equation is represented as,
where c or n/v is the concentration expressed in mol/m3 (mol/L3) & R= 0.0831 bar litre/mol K.
This is an indication that at constant temperature, the pressure of any gas is directly proportional to its concentration.
For reaction in equilibrium,
We can write, With the usage of Ideal Gas Equation we get,
PA = CART = [A] RT
PB = CBRT= [B] RT
Pc = CCRT = [C] RT
Pd = CDRT= [D] RT
Substituting the above values,
where n = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. The equation given above represents the relationship between kc and kp.
Equilibrium Constant of a Reaction that involves Condensed Phase
The representation for the equilibrium constant of a reaction between a condensed substance (solid or liquid) and a gaseous state is determined by considering only the gaseous species concentrations or partial pressures. Steady integration of condensed phases is combined with constant equilibrium.
a) A huge value of KC or KP signals that the forward reaction reaches completion or very nearly so.
b) A low numerical value of KC or KP signals that the forward reaction does not reach to any significant extent.
c) A reaction will most probably reach a state of equilibrium in which both, reactants and products are present if the numerical value of KC or Kp is not very huge or low.
d) The equilibrium constant of a forward reaction and its backward reaction are reciprocal to each other.
e) If a chemical equation is multiplied by a specific factor, its equilibrium constant should be increased to a power equal to that factor so as to avail the equilibrium constant for the new reaction.
The Reaction Quotient “Q”
Look at the following equilibrium.
PCl5 (g) PCl3(g) + Cl2 (g)
When the reaction is not in equilibrium, this ratio is known as ‘QC’. QC is the general term used for the ratio given above at any point of time. At equilibrium, QC becomes KC.
In a similar manner, PCl2 PPCl2 / PPCl2 is known as QP and at equilibrium, it turns into KP.
Suppose the reaction is at equilibrium, Q = Kc
A net reaction progresses from left to right (forward direction) if Q < KC.
A net reaction progresses from right to left (the reverse direction) if Q >Kc
"When an equilibrium is exposed to a change of concentration, temperature or external pressure, the equilibrium will move in that direction where the effects of these changes are nullified."
It can be understood by example given below. Overall, we can also foresee the direction of equilibrium with the theoretical assumption in mind.
PCl5 ——> PCl3 + Cl2
Assume that this reaction is at equilibrium and the moles of Cl2, PCl3, and PCl5 at equilibrium are a, b, and c respectively, and the total pressure is PT.
PT = (a+b+c) RT / V
KP = abRT / cV
Suppose d moles of PCl3 is included in the system, the value of Q would be, a(b+d)RT / cV. This is evdiently more than KP. Therefore, the system will move reverse to reach equilibrium.
If the volume of the system is increased, the Q becomes abRT / cV' where V' > V. Q becomes lesser, and the system will move forward to obtain equilibrium.
If a noble gas is added at constant pressure, it leads to an increase of the system’s volume and hence, the reaction moves forward.
If the noble gas is added at constant volume, the expression of Q remains as Q = abRT / cV and the system remains in equilibrium.
Thus, for using the Le Chatlier’s principle, change the expression of KP and KC into basic terms and later notice the effect of different changes.
A solution is said to be saturated when it stays in contact with undissolved solute.
In a saturated solution of an electrolyte, two equilibria exist. This can be represented as -
Solid unionized ions
On application of the law of action to the ionic equilibrium,
As the solution is saturated, the concentration of unionised molecules of the electrolyte is stable at a specific temperature, i.e., [AB] = K'= constant.
[A+] [B-] = K[AB] = KK’ = Ks (constant)
Ks is called the solubility product. It is described as the product of the concentration of ions in a saturated solution of an electrolyte at a particular temperature.
|NaCl (Strong acid + Strong Base) 2. Ch3COONa (Weak acid + Strong base) 3. NH4Cl (Strong acid + Weak base) 4. CH3COONH4 (Weak acid + Weak base)||Neutral Base Acidic *||No Hydrolysis h = √kw/Cka h = √kw/Ckb h = √kw/(ka + kb)||- Kh = kw/ka Kh = kw/Ckb Kh = kw/(ka + kb)||- pH=1/2[pkw + pka + logC] pH=1/2[pkw- pkb - logC] pH=1/2[pkw + pka - pkb]|
In the scenario of salt of weak acid and weak base, the nature of medium after hydrolysis is determined in the following manner:
(i) If Ka = Kb, the medium should be neutral.
(ii) If Ka > Kb, the medium should be acidic.
(iii) If Ka < Kb, the medium should be basic.
The degree of hydrolysis of salts of weak acids and weak bases is not affected by dilution as there is no concentration term in the expression of degree of hydrolysis.
Q1. Find out the pH of the solution when 0.1 M CH3 COOH (50 ml) and 01. M NaOH (50 ml) are mixed, [Ka (CH3COOH)=10-5]
CH3 COOH CH3 COO_ + H+ …(I)
NaOH → Na+ + OH-
H+ + OH_ H2O …(II)
(I) + (II)
CH3COOH + OH- CH3COO- + H2O . (III)
0.05-X 0.05-x x
Keq of eq. (III) = Ka/Kw
Conc. of H2O remains constant
109 = x/(0.05-x)2
because the value of eq. Const. is very high
Here for x» 0.05
a = 7.0710-6
pOH= 6-log 7.07
pOH= 6 – 0.85
pH= 14-6+0.85 = 8.85
Q2. The solubility product of Pb3 (PO4)2 is 1.5 x 10-32. Calculate the solubility in gms/litre.
Solubility product of Pb3 (PO4)2 = 1.5 10–32
Pb3 (PO4)2 3Pb2+ + 2PO43-
Suppose x is the solubility of Pb3 (PO4)2
Then, Ksp = (3x)3 (2x)2 = 108 x5
x = 1.692 10–7 moles/lit
Molecular mass of Pb3(PO4)2 = 811
x = 1.692 ´ 10–7 ´ 811 g/lit = 1.37 10–4 g/lit
Solubility product is
Ksp(SrC2O4) = [Sr2+] [C2O42–] = (5.4 10–4)2- = 2.92 10–7
Q3. Calculate the concentration of H+, HCO3- and CO32-, in a0.01M solution of carbonic acid if the pH is 4.18.
Ka1(H2CO3) = 4.45 10–7 and Ka2 = 4.69 10–11
pH = – log[H+]
4.18 = – log [H+]
[H+] = 6.61 10–5
H2CO3 H+ + HCO3-
HCO3- H+ + CO32-
[CO32-] = 4.8 10–11
Q4. Find out the molar solubility of Mg(OH)2 in 1MNH4Cl
KspMg(OH)2 = 1.8 10–11
Kb(NH3) = 1.8 10–5
Mg(OH)2(s) Mg++ + 2OH– K1 = Ksp
2NH4+ + 2OH- 2NH4OH K2 = 1/K2b
Q5. How much of AgBr can dissolve in 1.0 L of 0.4 M NH3? Assume that [Ag (NH3)2]+ is the sole complex formed if Kf [Ag(NH3)2+]=1.0108, Ksp (AgBr)= 5.010-13
AgBr Ag+ + Br-
Ag+ + 2NH3 Ag (NH2)2+
Let x= solubility ,
Then, x= [Br-]=[Ag+]+[Ag(NH3)2+]
Q1. Trick to solve the question: Consider the solubility to be x. Balance the equation by splitting the compounds - Ag2Cr04 and AgNO3. Enter the values found in the final equation for Ksp and find the answer to x.
Q2. Trick to solve the question: Split HA into its elements. Correlate the rates of HA andHX to frame an equation. Equate HX to 1M. Enter the values found and derive the answer.
Q3. Trick to solve this question: Write down the formula for pH. Substitute the values of 1 and 2 given in the formula. Use M1V1 = M2V2 to find the value of V2.
|Week 1 (June 7 to 14)||Physical Chemistry has the most weightage in JEE Main Chemistry. Study important chapters from this section such as Atomic Structure, Electrochemistry, and Equilibrium first. Then, move on to study the chapters with lesser weightage like Surface Chemistry and Solid State. Choose two topics everyday and go through them. Revise the basics well. Use NCERT books.|
|Week 2 (June 14 to 20, 2020)||Study Organic Chemistry. Pay more attention to important chapters such as Aromatic Compounds, Alkyl Halides, Alcohol, and Ether. Then, move on to study less important chapters such as Biomolecules and Carbonyl Compounds. Use NCERT books.|
|Week 3 (June 21 to 27, 2020)||Study Inorganic Chemistry. Start by studying the most important chapters like Chemical Bonding, p Block, and s Block. Then study less important chapters like Metallurgy and Qualitative Analysis. Use NCERT books.|
|Week 4 (June 28 to -July 4, 2020)||Revise all the formulas and important concepts. Solve as many previous year’s papers as you can. Pick up questions from the Chemistry section and attempt them as fast as you can (within 45 minutes-1 hour).|
|Week 5 (July 5 to 11, 2020)||Attempt mock tests everyday.|
|Week 6 (July 12 to 17, 2020)||Do one final revision.|
|JEE Main 2020 (July 18 to 23, 2020)||Research the analysis of the ongoing papers.|