NCERT Solutions for Class 11 Chemistry: Chapter 7 (Equilibrium)

NCERT Solutions for Class 11 Chemistry: Chapter 7 (Equilibrium) are provided on this page for the perusal of CBSE class 11 chemistry students. Detailed, step-by-step solutions for each and every intext and exercise question listed in chapter 7 of the NCERT class 11 chemistry textbook can be accessed here. Furthermore, the NCERT solutions provided on this page can also be downloaded as a PDF for free by clicking the download button provided above.

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NCERT Solutions for Chemistry – Class 11, Chapter 7: Equilibrium

“Equilibrium” is the seventh chapter in the NCERT class 11 chemistry textbook. Several important concepts such as equilibrium constants, buffer solutions, and the common-ion effect are explained in this chapter. Along with structured questions, the NCERT solutions for class 11 chemistry (chapter 7) provided on this page come with detailed explanations to help students learn and understand the key concepts related to chemical equilibrium in a seamless manner. The important subtopics associated with chapter 7  – “Equilibrium” are listed below.

Subtopics included in NCERT Class 11 Chemistry Chapter 7 – “Equilibrium”

  1. Solid-liquid Equilibrium
    • Liquid-vapour Equilibrium
    • Solid – Vapour Equilibrium
    • Equilibrium Involving Dissolution Of Solid Or Gases In Liquids
    • General Characteristics Of Equilibria Involving Physical Processes
  2. Equilibrium In Chemical Processes – Dynamic Equilibrium
  3. Law Of Chemical Equilibrium And Equilibrium Constant
  4. Homogeneous Equilibria
    • Equilibrium Constant In Gaseous Systems
  5. Heterogeneous Equilibria
  6. Applications Of Equilibrium Constants
    • Predicting The Extent Of A Reaction
    • Predicting The Direction Of The Reaction
    • Calculating Equilibrium Concentrations
  7. Relationship Between Equilibrium Constant K, Reaction Quotient Q And Gibbs Energy G
  8. Factors Affecting Equilibria
    • Effect Of Concentration Change
    • Effect Of Pressure Change
    • Effect Of Inert Gas Addition
    • Effect Of Temperature Change
    • Effect Of A Catalyst
  9. Ionic Equilibrium In Solution
  10. Acids, Bases And Salts
    • Arrhenius Concept Of Acids And Bases
    • The Bronsted-lowry Acids And Bases
    • Lewis Acids And Bases
  11. Ionization Of Acids And Bases
    • The Ionization Constant Of Water And It’s Ionic Product
    • The Ph Scale
    • Ionization Constants Of Weak Acids
    • Ionization Of Weak Bases
    • The Relation Between Ka And Kb
    • Di- And Polybasic Acids And Di- And Polyacidic Bases
    • Factors Affecting Acid Strength
    • Common Ion Effect In The Ionization Of Acids And Bases
    • Hydrolysis Of Salts And The Ph Of Their Solutions
  12. Buffer Solutions
  13. Solubility Equilibria Of Sparingly Soluble Salts
    • Solubility Product Constant
    • Common Ion Effect On Solubility Of Ionic Salts.

NCERT Solutions for Class 11 Chemistry Chapter 7

Q.1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

I)What will be the final vapour pressure and  what will happen when equilibrium is restored finally?

II)Write down, how initially the rates of evaporation and condensation got changed? 

III)Write down the effect observed when there was a change in vapour pressure.

 Ans.

(I)Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

(II)On increasing the volume of the container, the rates of evaporation will increase initially because now more space is available. Since the amount of the vapours per unit volume decrease on increasing the volume, therefore, the rate of condensation will decrease initially.

(III)On increasing the volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a large space.

 

Q.2.What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?

2SO2(g)+O2(g)2SO3(g)2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g)

Ans.

As per the question,

2SO2(g)+O2(g)2SO3(g)2SO_{2}(g)+O_{2}(g)\leftrightarrow 2SO_{3}(g) (Given)

 

Kc=[SO3]2[SO2]2[O2]=(1.9)2M2(0.6)2(0.82)M3=12.229M1K_{c}=\frac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\ =\frac{(1.9)^{2}M^{2}}{(0.6)^{2}(0.82)M^{3}}\\ \\ =12.229M^{-1}(approximately)

Hence, K for the equilibrium is 12.229 M–1.

 

Q.3. At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms

I2(g)↔ 2I (g)

Calculate Kp for the equilibrium

Ans.

Partial pressure of Iodine atoms (I)

pI=40100×ptotal=40100×105=4×104Pap_{I}=\frac{40}{100}\times p_{total}\\ \\ =\frac{40}{100}\times 10^{5}\\ \\ =4\times 10^{4}Pa

Partial pressure of I2 molecules,

pI=60100×ptotal=60100×105=6×104Pap_{I}=\frac{60}{100}\times p_{total}\\ \\ =\frac{60}{100}\times 10^{5}\\ \\ =6\times 10^{4}Pa

Now, for the given reaction,

Kp=(pI)2pI2=(4×104)2Pa26×104Pa=2.67×104PaK_{p}=\frac{(p_{I})^{2}}{p_{I_{2}}} =\frac{(4\times 10^{4})^{2}Pa^{2}}{6\times 10^{4}Pa}\\ \\ =2.67\times 10^{4}Pa

 

Q.4.Write the expression for the equilibrium constant, Kc for each of the following reactions:

(i)2NOCl(g)2NO(g)+Cl2(g)2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g)

(ii)2Cu(NO3)2(s)2CuO(s)+4NO2(g)+O2(g)2Cu(NO_{3})_{2}(s)\leftrightarrow 2CuO(s)+4NO_{2}(g)+O_{2}(g)

(iii)CH3COOC2H5(aq)+H2O(1)CH3COOH(aq)+C2H5OH(aq)CH_{3}COOC_{2}H_{5}(aq)+H_{2}O(1)\leftrightarrow CH_{3}COOH(aq)+C_{2}H_{5}OH(aq)

(iv)Fe3+(aq)+3OH(aq)Fe(OH)3(s)Fe^{3+}(aq)+3OH^{-}(aq)\leftrightarrow Fe(OH)_{3}(s)

(v)I2(s)+5F22IF5I_{2}(s)+5F_{2}\leftrightarrow 2IF_{5}

Ans.

KC=[NOg]2[Cl2(g)][NOCl(g)]2K_{C}=\frac{[NO_{g}]^{2}[Cl_{2_{(g)}}]}{[NOCl_{(g)}]^{2}}

(ii)KC=[CuO(s)]2[NO2(g)]4[O2(g)][Cu(NO3)2(g)]2=[NO2(g)]4[O2(g)]K_{C}=\frac{[CuO_{(s)}]^{2}[NO_{2_{(g)}}]^{4}[O_{2_{(g)}}]}{[Cu(NO_{3})_{2(g)}]^{2}}\\ \\ =[NO_{2(g)}]^{4}[O_{2(g)}]

(iii)KC=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)][H2O(l)]=CH3COOH(aq)[C2H5OH(aq)][CH3COOC2H5(aq)]K_{C}=\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}][H_{2}O_{(l)}]}\\ \\ =\frac{CH_{3}COOH_{(aq)}[C_{2}H_{5}OH_{(aq)}]}{[CH_{3}COOC_{2}H_{5(aq)}]}

(iv)KC=Fe(OH)3(s)[Fe(aq)3+][OH(aq)]3=1[Fe(aq)3+][OH(aq)]3K_{C}=\frac{Fe(OH)_{3(s)}}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}\\ \\ =\frac{1}{[Fe^{3+}_{(aq)}][OH^{-}_{(aq)}]^{3}}

(v)KC=[IF5]2[I2(s)][F2]5=[IF5]2[F2]5K_{C}=\frac{[IF_{5}]^{2}}{[I_{2(s)}][F_{2}]^{5}}\\ \\ =\frac{[IF_{5}]^{2}}{[F_{2}]^{5}}

 

Q.5.Find out the value of Kc for each of the following equilibria from the value of Kp:

(i)2NOCl(g)2NO(g)+Cl2(g);Kp=1.8×102at500K2NOCl(g)\leftrightarrow 2NO(g)+Cl_{2}(g);\: \: \: K_{p}=1.8\times 10^{-2}\: \: at\: 500K

(ii)CaCO3(s)CaO(s)+CO2(g);Kp=167at1073KCaCO_{3}(s)\leftrightarrow CaO(s)+CO_{2}(g);\: \: \: K_{p}=167\: \: at\: 1073K

Ans.

The relation between Kp and Kc is given as:

Kp=Kc(RT)ΔnK_{p}=K_{c}(RT)^{\Delta n}

(a) Given,

R = 0.0831 barLmol–1K–1

Δn=32=1\Delta n = 3-2 =1

T = 500 K

Kp=1.8×1021.8\times 10^{-2}

Now,

Kp = Kc (RT) ∆n

1.8×102=Kc(0.0831×500)1Kc=1.8×1020.0831×500=4.33×104(approximately)\Rightarrow 1.8\times 10^{-2}=K_{c}(0.0831\times 500)^{1}\\ \\ \Rightarrow K_{c}=\frac{1.8\times 10^{-2}}{0.0831\times 500}\\ \\ =4.33\times 10^{-4}(approximately)

(b) Here,

∆n =2 – 1 = 1

R = 0.0831 barLmol–1K–1

T = 1073 K

Kp= 167

Now,

Kp = Kc (RT) ∆n

167=Kc(0.0831×1073)ΔnKc=1670.0831×1073=1.87(approximately)\Rightarrow 167=K_{c}{(0.0831\times 1073)^{\Delta n}}\\ \\ \Rightarrow K_{c}=\frac{167}{0.0831\times 1073}\\ \\ =1.87(approximately)

 

Q.6. For the following equilibrium, Kc=6.3×1014at1000KNO(g)+O3(g)NO2(g)+O2(g)K_{c}=6.3\times 10^{14}\: \: \: at\: 1000K\\ \\ NO(g)+O_{3(g)}\leftrightarrow NO_{2(g)}+O_{2(g)}

Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc , for the reverse reaction?

Ans.

For the reverse reaction, Kc=1Kc=16.3×1014=1.59×1015K_{c}=\frac{1}{K_{c }}\\ \\ =\frac{1}{6.3\times 10^{14}}\\ \\ =1.59\times 10^{-15}

 

Q.7. Explain why solids and pure liquids can be ignored while writing the equilibrium constant expression?

Ans.

This is because molar concentration of a pure solid or liquid is independent of the amount present.

Mole concentration= NumberofmolesVolumeMass/molecularmassVolume=MassVolume×Molecularmass=DensityMolecularmass\frac{Number\: of\: moles}{Volume}\\ \\ \frac{Mass/molecular\:mass}{Volume}\\ \\ =\frac{Mass}{Volume\,\times\, Molecular \:mass}\\ \\ =\frac{Density}{Molecular\: mass}

Though density of solid and pure liquid is fixed and molar mass is also fixed .

Molar concentration are constatnt.

 

Q.8.

Reaction between N2 and O2– takes place as follows:

2N2(g)+O22N2O(g)2N_{2}(g)+O_{2}\leftrightarrow 2N_{2}O(g)

If a solution of 0.933 mol of oxygen and 0.482 mol of nitrogen is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc =2.0×10372.0\times 10^{-37}, determine the composition of equilibrium solution.

Ans.

Let the concentration of N2O at equilibrium be x.

The given reaction is:

2N2(g)             +               O2(g)                               2N2O(g)

Initial conc.     0.482 mol                     0.933 mol                                  0

At equilibrium(0.482-x)mol            (1.933-x)mol                              x mol

[N2]=0.482x10[O2]=0.933x210,[N2O]=x10[N_{2}]=\frac{0.482-x}{10}\cdot [O_{2}]=\frac{0.933-\frac{x}{2}}{10},[N_{2}O]=\frac{x}{10}

The value of equilibrium constant  is extremely small. This means that only small amounts . Then,

[N2]=0.48210=0.0482molL1and[O2]=0.93310=0.0933molL1[N_{2}]=\frac{0.482}{10}=0.0482molL^{-1} \: \:and\:\: [O_{2}]=\frac{0.933}{10}=0.0933molL^{-1}

Now,

Kc=[N2O(g)]2[N2(g)][O2(g)]2.0×1037=(x10)2(0.0482)2(0.0933)x2100=2.0×1037×(0.0482)2×(0.0933)x2=43.35×1040x=6.6×1020K_{c}=\frac{[N_{2}O_{(g)}]^{2}}{[N_{2(g)}][O_{2(g)}]}\\ \\ \Rightarrow 2.0\times 10^{-37}=\frac{(\frac{x}{10})^{2}}{(0.0482)^{2}(0.0933)}\\ \\ \Rightarrow \frac{x^{2}}{100}=2.0 \times 10^{-37}\times (0.0482)^{2}\times (0.0933)\\ \\ \Rightarrow x^{2}=43.35\times 10^{-40}\\ \\ \Rightarrow x=6.6\times 10^{-20} [N2O]=x10=6.6×102010=6.6×1021[N_{2}O]=\frac{x}{10}=\frac{6.6\times 10^{-20}}{10}\\ \\ =6.6\times 10^{-21}

 

Q.9.

Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:

2NO(g)+Br2(g) ⇒ 2NOBr(g)

When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at a constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2.

Ans.

The given reaction is:

2NO(g)+Br2(g)                              2NOBr(g)

2mol       1mol                                2mol

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from 0.05182\frac{0.0518}{2} mol or Br, or 0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br2] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518 = 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br2] = 0.0437 – 0.0259 = 0.0178 mol

 

Q.10. At 450 K, Kp= 2.0×10102.0 \times 10^{10}/bar for the given reaction at equilibrium.

2SO2(g)+O2(g) ⇒ 2SO3(g)

What is Kc at this temperature?

Ans.)

For the given reaction,

∆n = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K–1 mol–1

Kp=2.0×1010bar12.0 \times 10^{10}bar^{-1}

We know that,

Kp=Kc(RT)Δn2.0×1010bar1=Kc(0.0831LbarK1mol1×450K)1Kc=2.0×1010bar1(0.0831LbarK1mol1×450K)1=(2.0×1010bar1)(0.0831LbarK1mol1×450K)=74.79×1010Lmol1=7.48×1011Lmol1=7.48×1011M1K_{p}=K_{c}(RT)\Delta n\\ \\ \Rightarrow 2.0\times 10^{10}bar^{-1}=K_{c}(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}\\ \\ \Rightarrow K_{c}=\frac{2.0\times 10^{10}bar^{-1}}{(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)^{-1}}\\ \\ =(2.0\times 10^{10}bar^{-1})(0.0831L\, bar\, K^{-1}mol^{-1}\times 450K)\\ \\ =74.79\times 10^{10}L\, mol^{-1}\\ \\ =7.48\times 10^{11}L\, mol^{-1}\\ \\ =7.48\times 10^{11}\, M^{-1}

 

Q.11. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?

2HI(g) ⇒ H2(g)+I2(g)

Ans.

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm.

Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

2HI(g)                                     H2(g)       +      I2(g)

Initial conc.               0.2 atm                                        0                     0

At equilibrium        0.4 atm                                          0.16               2.15

2                      2

=0.08atm         =0.08atm

Therefore,

Kp=pH2×pI2pHI2=0.08×0.08(0.04)2=0.00640.0016=4.0K_{p=}\frac{p_{H_{2}}\times p_{I_{2}}}{p^{2}_{HI}}\\ \\ =\frac{0.08\times 0.08}{(0.04)^{2}}\\ \\ =\frac{0.0064}{0.0016}\\ \\ =4.0

Hence, the value of Kp for the given equilibrium is 4.0.

 

Q.12. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction

N2(g)+3H2(g) ⇒ 2NH3(g) is 1.7×1021.7\times 10^{2}

 Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Ans.

The given reaction is:

N2(g)+3H2(g) 2NH3(g)

The given concentration of various species is

[N2]= 1.5720molL1\frac{1.57}{20}mol\, L^{-1} [H2]= 1.9220molL1\frac{1.92}{20}mol\, L^{-1} [NH3]= 8.3120molL1\frac{8.31}{20}mol\, L^{-1}

Now, reaction quotient Qc is:

Q=[NH3]2[N2][H2]3=((8.13)20)2(1.5720)(1.9220)3=2.4×103Q=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(\frac{(8.13)}{20})^{2}}{(\frac{1.57}{20})(\frac{1.92}{20})^{3}}\\ \\ =2.4\times 10^{3}

Since, ?? ≠ ??, the reaction mixture is not at equilibrium.

Again, ?? > ??. Hence, the reaction will proceed in the reverse direction.

 

Q.13. The equilibrium constant expression for a gas reaction is,

Kc=[NH3]4[O2]5[NO]4[H2O]6K_{c}=\frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}

Write the balanced chemical equation corresponding to this expression.

Ans.

The balanced chemical equation corresponding to the given expression can be written as:

4NO(g)+6H2o(g)4NH3(g)+5O2(g)

 

Q.14. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 60% of water (by mass) reacts with CO according to the equation, 

H2O(g)  +  CO(g) ⇒ H2(g)   +  CO2(g)

Calculate the equilibrium constant for the reaction.

Ans.

The given reaction is:

H2O(g)  +  CO(g) ↔ H2(g)   +  CO2(g)

Compound  H20 CO H2 CO2
Initial Conc.  0.1M 0.1M 0 0
Equilibrium Conc.  0.06M  0.06M 0.04M 0.04M

Therefore, the equilibrium constant for the reaction,

Kc = ([H2][CO2])/([H2O][CO]) = (0.4*0.4)/(0.6*0.6) = 0.444

Q.15. At 700 K, equilibrium constant for the reaction

H2(g)+I2(g) ⇒ 2HI(g)

is 54.8. If 0.5 molL–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Ans.

It is given that equilibrium constant ?? for the reaction

H2(g)+I2(g)                              2HI(g)   is  54.8.

Therefore, at equilibrium, the equilibrium constant Kc for the reaction

H2(g)+I2(g)                              2HI(g)

[HI]=0.5 molL-1 will be 1/54.8.

Let the concentrations of hydrogen and iodine at equilibrium be x molL–1

[H2]=[I2]=x mol L-1

Therefore, [H2][I2][HI]2=Kcx×x(0.5)2=154.8x2=0.2554.8x=0.06754x=0.068molL1(approximately)\frac{[H_{2}][I_{2}]}{[HI]^{2}}=K^{‘}_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.5)^{2}}=\frac{1}{54.8}\\ \\ \Rightarrow x^{2}=\frac{0.25}{54.8}\\ \\ \Rightarrow x=0.06754\\ \\ x=0.068molL^{-1}(approximately)

Hence, at equilibrium, [H2]=[I2]=0.068 mol L-1.

 

Q.16. What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2ICl(g) ⇒ I2(g)+Cl2(g) ;  Kc =0.14

 

Ans.

The given reaction is:

2ICl(g)                                           I2(g)      +   Cl2(g)

Initial conc.                 0.78 M                                           0             0

At equilibrium      (0.78-2x) M                                       x M           x M

Now, we can write, [I2][Cl2][IC]2=Kcx×x(0.782x)2=0.14x2(0.782x)2=0.14x0.782x=0.374x=0.2920.748x1.748x=0.292x=0.167\frac{[I_{2}][Cl_{2}]}{[IC]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x^{2}}{(0.78-2x)^{2}}=0.14\\ \\ \Rightarrow \frac{x}{0.78-2x}=0.374\\ \\ \Rightarrow x=0.292-0.748x\\ \\ \Rightarrow 1.748x=0.292\\ \\ \Rightarrow x=0.167

Hence, at equilibrium,

[H2]=[I2]=0.167 M

[HI]= (0.782×0.167)M=0.446M(0.78-2\times 0.167)M\\ \\ =0.446M

 

Q.17. Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?

C2H6(g) ⇒ C2H4(g) +H2(g)

Ans.

Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,

 

C2H6(g)                                         C2H4(g)    +     H2(g)

Initial conc.                 4.0 M                                           0                      0

At equilibrium          (4.0-p)                                            p                     p

We can write,

pc2H4×pH2pc2H6=KPp×p40p=0.04p2=0.160.04pp2+0.04p0.16=0\frac{p_{c_{2}H_{4}}\times p_{H_{2}}}{p_{c_{2}H_{6}}}=K_{P}\\ \\ \Rightarrow \frac{p\times p}{40-p}=0.04\\ \\ \Rightarrow p^{2}=0.16-0.04p\\ \\ \Rightarrow p^{2}+0.04p-0.16=0

Now,

p=0.04±(0.04)24×1×(0.16)2×1=0.04±0.802=0.762(Takingpositivevalue)=0.38p=\frac{-0.04\pm \sqrt{(0.04)^{2}-4\times 1\times (-0.16)}}{2\times 1}\\ \\ =\frac{-0.04\pm 0.80}{2}\\ \\ =\frac{0.76}{2}\: \: \: \: \: \: \: \: \: (Taking \:positive\: value)\\ \\ =0.38

Hence, at equilibrium,

[C2H6] – 4 – p = 4 – 0.38

= 3.62 atm

 

Q.18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:

CH3COOH(l)+c2H5OH(l) ⇒ CH3COOC2H5(l)+H2O(l)

(i)Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction)

(ii)At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.    

(iii)Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

Ans.

(i)Reaction quotient,

Qc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]Q_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}

(ii) Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess.

The given reaction is:

CH3COOH(l)+c2H5OH(l)                          CH3COOC2H5(l)+H2O(l)

Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, equilibrium constant for the given reaction is:

Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.171V×0.171V0.829V×0.009V=3.919=3.92(approximately)K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.171}{V}\times \frac{0.171}{V}}{\frac{0.829}{V}\times \frac{0.009}{V}}=3.919\\ \\ =3.92(approximately)

(iii)Let the volume of the reaction mixture be V.

CH3COOH(l)+c2H5OH(l)CH3COOC2H5(l)+H2O(l)

Initial conc.                                                                        0                        0

At equilibrium

=            =

Therefore, the reaction quotient is,

Kc=[CH3COOC2H5][H2O][CH3COOH][C2H5OH]=0.214V×0.214V0.786V×0.286V=0.2037=0.204(approximately)K_{c}=\frac{[CH_{3}COOC_{2}H_{5}][H_{2}O]}{[CH_{3}COOH][C_{2}H_{5}OH]}\\ \\ =\frac{\frac{0.214}{V}\times \frac{0.214}{V}}{\frac{0.786}{V}\times \frac{0.286}{V}}=0.2037\\ \\ =0.204(approximately)

Since Qc<Kc , equilibrium has not been reached.

 

Q.19. A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5×1010.5\times 10^{-1}mol L–1. If value of Kc is 8.3×1038.3\times 10^{-3}, what are the concentrations of PCl3 and Cl2 at equilibrium?

PCl5(g) ⇒ PCl3(g)   +   Cl2(g)

Ans.

Consider the conc. Of both PCl3 and Cl2 at equilibrium be x molL–1. The given reaction is:

PCl5(g)PCl3(g      +      Cl2(g)

At equilibrium 0.5×1010molL10.5\times 10^{-10}molL^{-1}         x mol L-1      x mol L-1

It is given that the value of equilibrium constant , Kc is 8.3×1010molL38.3\times 10^{-10}molL^{-3}

Now we can write the expression for equilibrium as:

[PCl2][Cl2][PCl3]=Kcx×x0.5×1010=8.3×103x2=4.15×104x=2.04×102=0.0204=0.02(approximately)\frac{[PCl_{2}][Cl_{2}]}{[PCl_{3}]}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{0.5\times 10^{-10}}=8.3\times 10^{-3}\\ \\ \Rightarrow x^{2}=4.15\times 10^{-4}\\ \\ \Rightarrow x=2.04\times 10^{-2}\\ \\ =0.0204\\ \\ =0.02(approximately)

Therefore, at equilibrium,

[PCl3]=[Cl2]=0.02mol L-1

 

Q.20. One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

 FeO (s) + CO (g) ⇒ Fe (s) + CO2 (g); Kp= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and ???2= 0.80 atm?

Ans.

For the given reaction,

FeO(g)            +       CO(g)                             Fe(s)              +           CO2(g)

Initialy,              1.4 atm                                                      0.80 atm

 

Qp=pCO2pCO=0.801.4=0.571\frac{p_{CO_{2}}}{p_{CO}}\\ \\ =\frac{0.80}{1.4}\\ \\ =0.571

Since Qp > Kp , the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.

Now, let the increase in pressure of CO = decrease in pressure of CO2 be p. Then, we can write,

Kp=pCO2pCO0.265=0.80p1.4+p0.371+0.265p=0.80p1.265p=0.429p=0.339atmK_{p}=\frac{p_{CO_{2}}}{p_{CO}}\\ \\ \Rightarrow 0.265=\frac{0.80-p}{1.4+p}\\ \\ \Rightarrow 0.371+0.265p=0.80-p\\ \\ \Rightarrow 1.265p=0.429\\ \\ \Rightarrow p=0.339 atm

Therefore, equilibrium partial of CO2,pCO=0.80-0.339=0.461 atm

And, equilibrium partial pressure of CO,pCO=1.4+0.339=1.739 atm

 

Q.21. Equilibrium constant, Kc for the reaction

N2(g)+3H2(g) ⇒ 2NH at 500 K is 0.061

At a specific time, from the analysis we can conclude that composition of the reaction mixture is, 2.0 mol L–1 H2 , 3.0 mol L –1 N2 and 0.5 mol L–1 NH3. Find out whether the reaction is at equilibrium or not? Find in which direction the reaction proceeds to reach equilibrium.

Ans.

N2(g)        +        3H2(g)                                  2NH3

At a particular time:  3.0 mol L-1          2.0mol L-1                              0.5 mol L-1

So,

Qc=[NH3]2[N2][H2]3=(0.5)2(3.0)(2.0)3=0.0104\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}}\\ \\ =\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\\ \\ =0.0104

It is given that Kc=0.061

   QcKcQc \neq K_{c}, the reaction is not at equilibrium.

    Qc<KcQc < K_{c}, the reaction preceeds in the forward direction to reach at equilibrium.

 

Q.22.Bromine monochloride(BrCl) decays into bromine and chlorine and reaches the equilibrium:

2BrCl(g) ⇒ Br2(g)   +   Cl2(g)

For which Kc= 42 at 600 K.

If initially pure BrCl is present at a concentration of 5.5×1055.5 \times 10^{-5}molL-1 , what is its molar concentration in the mixture at equilibrium?

Ans.)

Let the amount of bromine and chlorine formed at equilibrium be x. The given reaction is:

2BrCl(g)                                               Br2(g)      +        Cl2(g)

Initial conc. 5.5×1055.5\times 10^{-5}                   0                         0

At equilibrium 5.5×1052x5.5\times 10^{-5}-2x         x                          x

Now, we can write,

[Br2][Cl2][BrCl]2=Kcx×x(5.5×1052x)2=42x5.5×1052x=6.48x=35.64×10512.96x13.96x=35.64×105x=35.6413.96×105=2.55×105\frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}=K_{c}\\ \\ \Rightarrow \frac{x\times x}{(5.5\times 10^{-5}-2x)^{2}}=42\\ \\ \Rightarrow \frac{x}{5.5\times 10^{-5}-2x}=6.48\\ \\ \Rightarrow x=35.64\times 10^{-5}-12.96x\\ \\ \Rightarrow 13.96x=35.64\times 10^{-5}\\ \\ \Rightarrow x=\frac{35.64}{13.96}\times 10^{-5}=2.55\times 10^{-5}

So,,at equilibrium

[BrCl]=5.5×105(2×2.55×105)=5.5×1055.1×105=0.4×105=4.0×106molL1[BrCl]=5.5\times 10^{-5}-(2\times 2.55\times 10^{-5})\\ \\ =5.5\times 10^{-5}-5.1\times 10^{-5}\\ \\ =0.4\times 10^{-5}\\ \\ =4.0\times 10^{-6}molL^{-1}

 

Q.23.At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass

 C(s)+CO2 (g) ⇒ 2CO (g)

Calculate Kc for this reaction at the above temperature

Ans.)

Let us assume that the solution is of 100g in total.

Given, mass of CO = 93.55 g

Now, the mass of CO2 =(100 – 93.55)=6.45 g

Now, number of moles of CO, nCO=93.528=3.34moln_{CO}=\frac{93.5}{28}=3.34\: mol

Number of moles of CO2, nCO2=6.4544=0.146moln_{CO_{2}}=\frac{6.45}{44}=0.146\: mol

Partial pressure of CO,

PCO= nCOnCO+nCO2×ptotal=3.343.34+0.146×1=0.958atm\frac{n_{CO}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{3.34}{3.34+0.146}\times 1=0.958\, atm

Partial pressure of CO2, PCO2=nCO2nCO+nCO2×ptotal=0.1463.34+0.146×1=atmP_{CO_{2}}=\frac{n_{CO_{2}}}{n_{CO}+n_{CO_{2}}}\times p_{total}=\frac{0.146}{3.34+0.146}\times 1=\, atm

Therefore, Kp= [CO]2[CO2]=(0.938)20.062=14.19\frac{[CO]^{2}}{[CO_{2}]}\\ \\ =\frac{(0.938)^{2}}{0.062}\\ \\ =14.19

For the given reaction,

∆n = 2 – 1 = 1

We know that,

Kp=Kc(RT) Δn\Delta n

14.19=Kc(0.082×1127)1Kc=14.190.082×1127=0.154(approximately)\Rightarrow 14.19=K_{c}(0.082\times 1127)^{1}\\ \\ \Rightarrow K_{c}=\frac{14.19}{0.082\times 1127}\\ \\ =0.154(approximately)

 

Q.24.Calculate a) ∆G0 and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K

NO(g)+12O2(g)NO2(g)NO(g)+\frac{1}{2}O_{2}(g)\leftrightarrow NO_{2}(g)  

 Where;

fG° (NO2) = 52.0 kJ/mol

 ∆fG° (NO) = 87.0 kJ/mol

 ∆fG° (O2) = 0 kJ/mol

Ans.)

(I)We know that,

∆G° = RT log Kc

∆G° = 2.303 RT log Kc

Kc=35.0×1032.303×8.314×298=6.134Kc=antilog(6.134)=1.36×106K_{c}=\frac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ =6.134\\ \\ ∴ K_{c}=antilog(6.134)\\ \\ =1.36\times 10^{6}

Therefore, the equilibrium constant for the given reaction Kc is 1.36×1061.36\times 10^{6}

 

Q.25. Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(I) PCl5(g) ⇒ PCl3 +Cl2 (g)

(II) Cao(s) + CO2 (g) ⇒ CaCO3(s)

(III) 3Fe (s) + 4H2O (g) ⇒ Fe3O4 (s) +4H2 (g)

Ans.

(I) The number of moles of reaction products will increase. According to Le Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.

(II) The number of moles of reaction products will decrease.

(III) The number of moles of reaction products remains the same.

 

Q.26. Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction.

(I)COCl2 (g) ⇒ CO (g) +Cl2 (g)

(II)CH4 (g) +2S2 (g) ⇒ CS2 (g) + 2H2S (g)

(III)CO2 (g) +C (s) ⇒ 2CO (g)

(IV)2H2 (g) +CO  (g) ⇒ CH3OH(g)

(V)CaCO3 (s) ⇒ Cao (s) + CO2 (g)

(VI)4NH3 (g) +5O2 (g) ⇒ 4NO (g) + 6H2O (g)

Ans.)

When pressure is increased:

The reactions given in (i), (iii), (iv), (v), and (vi) will get affected.

Since, the number of moles of gaseous reactants is more than that of gaseous products; the reaction given in (iv) will proceed in the forward direction

Since, the number of moles of gaseous reactants is less than that of gaseous products, the reactions given in (i), (iii), (v), and (vi) will shift in the backward direction

 

Q.27. The equilibrium constant for the following reaction is 1.6×1051.6 \times 10^{5}at 1024 K.

H2 (g) + Br2 (g) ⇒ 2HBr (g)

Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

 

Ans.

Given, ?? for the reaction i.e., H2 (g) + Br2 (g)2HBr (g) is 1.6×1051.6 \times 10^{5}.

Therefore, for the reaction   2HBr (g)H2 (g) + Br2 (g)  the equilibrium constant will be,

Kp= 1Kp=11.6×105=6.25×106\frac{1}{K_{p}}\\ \\ =\frac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}

Now, let p be the pressure of both H2 and Br2 at equilibrium.

2HBr (g)                                         H2 (g)     +     Br2 (g)

Initial conc.                   10                                                         0                         0

At equilibrium           10-2p                                                       p                         p

 

Now, we can write,

pHBr×p2pHBr2=Kpp×p(102p)2=6.25×106p102p=2.5×103p=2.5×102(5.0×103)pp+(5.0×103)p=2.5×102(1005×103)=2.5×102p=2.49×102bar=2.5×102bar(approximately)\frac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\ \frac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \frac{p}{10-2p}=2.5\times 10^{-3}\\ \\ p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})=2.5\times 10^{-2}\\ \\ p=2.49\times 10^{-2}bar=2.5\times 10^{-2}bar (approximately)

Therefore, at equilibrium,

[H2]=[Br2]= 2.49×102bar2.49\times 10^{-2}bar [HBr]= 102×(2.49×102)bar=9.95bar=10bar(approximately)10-2\times (2.49\times 10^{-2})bar\\ \\ =9.95 bar=10 bar(approximately)

 

Q.28.Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:

CH4(g)+H2O(g)CO(g)+3H2(g)CH_{4}(g)+H_{2}O(g)\leftrightarrow CO(g)+3H_{2}(g)

(I)Write as expression for Kp for the above reaction.

(II) How will the values of Kp and composition of equilibrium mixture be affected by

 (i)Increasing the pressure

(ii)Increasing the temperature

(iii)Using a catalyst?

Ans.)

(I)For the given reaction,

Kp=pCO×pH23pCH4×pH2O\frac{p_{CO}\times p_{H_{2}}^{3}}{p_{CH_{4}}\times p_{H_{2}O}}

(II) (i) According to Le Chatelier’s principle, the equilibrium will shift in the backward direction.

(ii) According to Le Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.

(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

 

Q.29. Describe the effect of:

I) Removal of CO

II) Addition of H2

III) Removal of CH3OH on the equilibrium of the reaction:

IV) Addition of CH3OH

2H2 (g)+CO (g) ⇒ CH3OH (g)

Ans.)

(I) On removing CO, the equilibrium will shift in the backward direction.

(II) According to Le Chatelier’s principle, on addition of H2, the equilibrium of the given reaction will shift in the forward direction.

(III) On removing CH3OH, the equilibrium will shift in the forward direction.

(IV) On addition of CH3OH, the equilibrium will shift in the backward direction.

 

Q.30. At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3×103 8.3 \times 10^{-3} . If decomposition is depicted as,

PCl5(g)PCl3(g)+Cl2(g)PCl_{5}(g)\leftrightarrow PCl_{3}(g)+Cl_{2}(g)

rH° = 124.0 kJmol–1

a) Write an expression for Kc for the reaction.

b) What is the value of Kc for the reverse reaction at the same temperature?

c) What would be the effect on Kc if

(i) more PCl5 is added

(ii) pressure is increased?

(iii) The temperature is increased?

Ans.)

(a)Kc=[PCl3(g)][Cl2(g)][PCl3(g)]K_{c}=\frac{[PCl_{3}(g)][Cl_{2}(g)]}{[PCl_{3}(g)]}

(b)Value of Kc for the reverse reaction at the same temperature is:

Kc=1Kc=18.3×103=1.2048×102=120.48K_{c}^{‘} = \frac{1}{K_{c}}\\ \\ =\frac{1}{8.3 \times 10^{-3}} = 1.2048 \times 10^{2}\\ \\ =120.48

(c)(i)Kc would remain the same because in this case, the temperature remains the same.

(ii)Kc is constant at constant temperature. Thus, in this case, Kc would not change. (iii)In an endothermic reaction, the value of Kc increases with an increase in temperature. Since the given reaction in an endothermic reaction, the value of Kc will increase if the temperature is increased.

 

Q.31. Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,

CO(g)+H2O(g)CO2(g)+H2(g)CO(g)+H_{2}O(g)\leftrightarrow CO_{2}(g)+H_{2}(g)

If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that Pco=PH2O  = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C

Ans.)

Let the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:

CO(g)       +    H2O CO2(g)      +     H2(g)

Initial conc.           4.0 bar          4.0 bar                            0                     0

At equilibrium      4.0-p             4.0-p                                p                     p

Given Kp = 10.1

PCO2×PH2PCO×PH2O=KPp×p(4.0p)(4.0p)=10.1p4.0p=3.178p=12.7123.178p4.178p=12.712p=12.7124.178p=3.04\frac{P_{CO_{2}}\times P_{H_{2}}}{P_{CO}\times P_{H_{2}O}} = K_{P}\\ \\ \Rightarrow \frac{p\times p}{(4.0-p)(4.0-p)}=10.1\\ \\ \Rightarrow \frac{p}{4.0-p} =3.178\\ \\ \Rightarrow p =12.712-3.178p\\ \\ 4.178p = 12.712\\ \\ p =\frac{12.712}{4.178}\\ \\ p=3.04

So, partial pressure of H2 is 3.04 bar at equilibrium.
Q.32. Predict which of the following reaction will have appreciable concentration of reactants and products:

(a)Cl2(g)2Cl(g);Kc=5×1039Cl_{2}(g)\leftrightarrow 2Cl(g);K_{c}=5 \times 10^{-39}

(b)Cl2(g)+2NO(g)2NOCl(g);Kc=3.7×108Cl_{2}(g)+2NO(g)\leftrightarrow 2NOCl(g);K_{c}=3.7 \times 10^{8}

(c)Cl2(g)+2NO2(g)2NO2Cl(g);Kc=1.8Cl_{2}(g)+2NO_{2}(g)\leftrightarrow 2NO_{2}Cl(g);K_{c}=1.8

Ans.)

If the value of Kc lies between 10–3 and 103 , a reaction has appreciable concentration of reactants and products. Thus, the reaction given in (c) will have appreciable concentration of reactants and products.

 

Q.33. The value of Kc for the reaction 3O2 (g) ⇒ 2O3 (g) is 2.0×10502.0\times 10^{-50} at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6×102 1.6 \times 10^{-2} , what is the concentration of O3?

Ans.)

Given,

3O2 (g)2O3 (g)

Then, Kc = [O3(g)]2[O2(g)]3 \frac{ [O_{3}(g)]^{2}}{[O_{2}(g)]^{3}}

Given that Kc = 2.0×1050 2.0 \times 10^{-50} and [O2(g)] = 1.6×1021.6 \times 10^{-2}

Then,

2.0×1050=[O3(g)]2[1.6×102]3[O3(g)]2=2.0×1050×(1.6×102)3[O3(g)]2=8.192×1056[O3(g)]2=2.86×1028M 2.0 \times 10^{-50} = \frac { [O_{3}(g)]^{2} }{ [1.6 \times 10^{ -2 }]^{3}}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.0 \times 10^{-50} \times (1.6 \times 10^{-2})^{3}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 8.192 \times 10^{-56}\\ \\ \Rightarrow [O_{3} (g)]^{2} = 2.86 \times 10^{-28} M

So, the conc. of O3 is 2.86×1028M2.86 \times 10^{-28}M.

 

Q.34. The reaction, CO(g)  +  3H2(g)    CH4(g)  +  H2O(g)CO_{(g)} \; + \; 3H_{2(g)} \; \rightarrow \; CH_{4(g)} \; + \; H_{2}O_{(g)}