Chemical Equilibrium Solved Exercise

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Solved Exercise from chapter 8 chemical equilibrium chemistry book 1 for FSC pre engineering and pre medical for Board of Intermediate and Secondary education. Also For Entry Test Preparation for UET, NUST, PIEAS, GIKI, AIR, FAST, WAH University, UHS, other eningeering Universities and Medical Colleges.

                    TEXT BOOK EXERCISE
 Q1.      Multiple Choice questions
(i)         For which system does the equilibrium constant, Kc has units of (concentration)-1  
(a)        N2 + 3H22NH3
(b)        H2 + I2  2HI
(c)        2NO2   N2 O4
(d)       2HF     H2 + F2
(ii)       Which statement about the following equilibrium is correct
2SO2 + O2 2SO3
(a)       The value of Kp falls with a rise in temperature
            (b)       The value of KP falls with increasing Pressure
            (c)       Adding V2 O5 catalyst increases the equilibrium yield of SO3
(d)       The value of Kp is equal to Kc
(iii)     The pH of 10-3 mol/dm3 of an equous solution of H2SO4 is
            (a)       3          (b)       2.7       (c)       2          (d)       1.5
(iv)      The solubility product of AgCI is 2 x 1010 mol2 dm-6 . The maximum concentration of Ag+ ions in the solution is
            (a)       2 x 10-10 mol dm-3
(b)        1.41 x 10-10 mol dm-3
(c)        1 x 10-10 mol dm-3
(d)       4 x 10-20 mol dm-3
(v)        An ecxcess of aqueous silver nitrate is added to aqueous barium chloride and precipitate is removed by filtration. What are the main ions in the filtrate?
(a)        Ag+ and NO3-1 only
(b)        Ag+ and Ba2+ and NO3-1
(c)        Ba2+ and NO3-1 only
(d)       Ba2+ and NO3-1 and CI-1
Ans.    i)c        ii)a       iii)b      iv)b      v)b
Q2.      Fill in the blanks
(i)         Law of mass action states that the rate at which a reaction proceeds is directly proportional to the product of the active masses of the ______.
(ii)        In an exothermic reversible reaction, _________in temperature will shift the equilibrium towards the forward direction.
(iii)       The equilibrium constant for the reaction 2O3 3O2 is 1055 at 25oC, it tells that ozone is _______at room temperature.
(iv)       In a gas phase  reaction, if the number of moles of reactants are equal to the number of moles of the products, KC of the reaction is ______to the Kp.
(i)                 Buffer solution is prepared by by mixing together a weak base and its salt with _______ or a weak acid and its salt with ________.
Ans.    i)reactants                  ii)decrease                  iii)unstable                  iv)equal           v)strong acid, strong base
Q3.      Label the sentences True or False
(i)                 When a reversible reaction attains equilibrium both reactants and products are present in a reaction mixture.
(ii)             The Kc of the reaction
A+B C+D         is given by
            Kc =
Therefore it is assumed that [A] = [B]=[C]=[D]
(iii)             A catalyst is a compound, which increases the speed of the reaction and consequently increases the yield of the product.
(iv)           Ionic product Kw of pure water at 25oC is 10-14 dm-6 and is represented by an expression
Kw =[H+][OH-]=10-14 mol2 dm-6
(v)               AgCI is a sparingly soluble ionic solid in water. Its solution produces excess of Ag+ and CI- ions.

Ans.    i)True                                     ii)False                                    iii)False                       iv)True                       v)False

Q4.      (a)        Explain the terms “reversible reaction” and “state of equilibrium”
                                    See Section 8.1 and 8.1.2
(b)       Define and explain the law of mass action and drive the expression for the equilibrium constant KC.
                                    See Section 8.1.3
(c)                Write KC for the following reactions
(i)                 Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)
(ii)               Ag+(aq) +Fe2+(aq) Ag(s) + Fe3+(aq)
(iii)             N2(g)+ O2(g) 2NO(g)
(iv)             4NH3(g) 5O2(g) 4NO(g) +6H2O(g)
(v)               PCI5(g) PCI3(g) +CI2(g)
(i) Kc=
(ii) Kc=
(iii) Kc=
(iv) Kc=
(v) Kc=

Q5.      (a) Reversible reactions attain the position of equilibrium, which is dynamic in nature and not static. Explain it.
                                    See Section 8.1.1
(b)       Why do the rates of forward reactions slow down when a reversible in nature and not static. Explain it.
                                    See Section 8.1.1
Q6.      When a graph is plotted between time on X-axis and the concentration of reactants and products on Y-axis for a reversible reaction, the curve becomes parallel to time axis at a certain stage.
            (a)        At what stage the curves become parallel?
            (b)       Before the curves become parallel, the steepness of curves falls? Give reasons.
(c)        The rate of decrease of concentrations of the reactants and rate of increase of concentrations of any of products may or may not be equal for various types of reactions before the equilibrium time. Explain it.
                                                See section 8.1.1
Q7.      (a)        Write down the relationship of different types of equilibrium constants. i.e. Kcand Kp for the following general reactions.
                                    aA + bB cC +dD
                                                            See section 8.1.2
            (b)       Decide the comparative magnitudes of Kc, Kp for the following reactions.
Synthesis of NH3
                        N2(g) +3H2(g)  2NH3(g)
            KC is given by
                        Kc =
            Kp is given by
                        Kp =
            For this reaction, change in number of moles is given by
            n =number of mole of product-number of moles of reactants
                        =2 – (1+3)=-2
            Hence
                        Kp=Kc x (RT)-2
            Or        Kp = Kc x 
            Thus    if T is such that RT > 1, then Kp < Kc
                        If T is such that RT< 1, then Kp > Kc
Dissociation of PCI5
                                    PCI5(g) PCI3(g)+ CI2(g)
                                    Kc is given by according to law of Mass of Action
                                    Kc =
            Kp is given by
                                    Kp =
            For this reaction, change in number of moles is given by        
            n =number of mole of product-number of moles of reactants
                        =2 – 1 =1
                        Kp =Kc x (RT)
            Thus    If T is such that RT > 1, then Kp > K­c
                        If T is such that RT< 1, then Kp <Kc
Q8.      (a)        Write down KC for the following reversible reactions. Suppose that the volume of reaction mixture in all the cases is ‘V’ dm3 at equilibrium stage.
            (i)         CH3 COOH+ CH3 CH2 OH  CH3COOC2H5+ H2O
            (ii)        H2 +12 2HI
            (iii)       2HIH2 +I2
            (iv)       PCI5 PCI3 + CI2
            (v)        N2 + 3H2 2NH3
                                    See section 8.1.3
(b)       How do you explain that some of the reactions mentioned above are affected by change of volume at equilibrium stage?
Q9.      Explain the following two applications fo equilibrium constant. Give examples.
            (i)         Direction of reaction
            (ii)        Extent of reaction
                                    See section 8.1.5
Q10.    Explain the following with reasons.
(a)        The change of volume disturbs the equilibrium position for some of the gaseous phase reactions, but not the equilibrium constant.
(b)       The change of temperature disturbs both the equilibrium position and the equilibrium constant of a reaction.
(c)        The solubility of glucose in water is increased by increasing the temperature.
Q11.    (a)        What is ionic product of water? How does this value vary with the change in T? Is it true that this value is 75 times when the T of water increased form 0oC to 100oC.
                        Ionic Product of water is given by the equation
                        Kw [H+][OH-]
Value of Kw increases with increase in temperature. It is because increase in temperature increase the ionization of H2O. Thus, more H+ or OH- ions are produced. Hence value of Kw increase.
            e.g.
                                    At 25oC           Kw =1 x 10-14
                                and         At 100oC         Kw =7.5 x 10-14
            Further
                        At 0oC             (Kw)o     =0.1 x 10-14_____(1)
                        At 100oC         (Kw)100 =7.5 x 10-14_____(2)
                        Divide eq (2) by eq (1)
                        ==75
                        or         (KW) 100 =75 x (Kw)0
                        Hence, KW at 100oC is 75 times more than at 0oC
(b)       What is the Justification for the increase of ionic product with temperature?
Value of Kw increase with increase in temperature. It is because increase in temperature increase the ionization of H2O. Thus, more H+ or OH- ions are produced. Hence value Kw increases.
(c)        How do you prove that at 25OC in 1 dm3 of water, there are 10-7 moles of H3O+?
            At 25OC
            Kw =[H3O+][OH-14 ____(1)
Since ionization of water gives equal no. of H3O+ and OH- ions
therefore
            [H3O+]=[OH-]
            Hence, eq (1) can be written as
            Kw =[H3O+][H3O+]=10-14
            Or                    [H3O+]2=10-14
            Taking square root on both sides
                        [H3O+]2=10-7 mol/dm3
Hence, at 25oC, water has 10-7 mole/dm3 of H3O+ ions.
Q12.    (a)        Define pH and pOH. How are they related with pKw.
(b)       What happens to the acidic and basic properties of aqueous solutions when pH varies form 0 to 14.
(c)        Is it true that the sum of pHa and pKb is always equal to 14 to all temperatures for any acid? If not why?
Q13.    (a)        What is Lowry bronsted idea of acids and bases? Explain conjugate acids and bases.
(b)       Acetic acid dissolves in water and gives proton to water. But when dissolves in H2SO4, it accepts proton. Discuss the role of acetic acid in both cases.
                        CH3 COOH + H2O  CH3 COO- + H3O+
However, H2SO4 is a stronger acid than acetic acid, therefore, H2SO4 donates proton and acts as an acid while acetic acid accepts proton and acts as a base.
                        H2SO4+ CH3 COOH HSO4- + CH3 COOH2+
Q14.    In the equilibrium PCI5(g)  PCI3(g) + CI2(g)       H=90 kJ/ mol
            (a)        The position of equilibrium
            (b)       Equilibrium constant?
            If         (i)         Temperature is increased
                        (ii)        Volume of the container is decrease.
                        (iii)       Catalyst is added
                        (iv)       CI2 is added
                                    See section 8.2
            Explain your answer.
Q15.    Synthesis of NH3 by Haber’s process is an exothermic reaction.
                        N2  +  3H2 2NH3                       H=92.46 kJ/ mol
(a)        What should be the possible effect of change of temperature at equilibrium stage?
(b)       How does the change of pressure or volume shifts the equilibrium position of this reaction?
            (c)        What is the role of the catalyst in this reaction?
            (d)       What happens to equilibrium position of this reaction if NH3 is removed form the reaction vessel form time to time.
                                    See Section 8.2.1
Q16.    Sulphuric acid is the king of chemicals. It is produced  by the burning of SO2 to SO3through an exothermic reversible process.
            (a)        Write the balanced reversible reaction
            (b)       What is the effect of pressure change on this reaction?
            (c)        Reaction is exothermic but still the temperature of 400-500oC is required to increase the yield of SO3. Give reasons.
                                    See Section 8.2.2
Q17.    (a)        What are buffer solutions? Why do we need them in daily life?
                                    See Section 8.2
(b)       How does the mixture of sodium acetate and acetic acid give us the acidic buffer?
                                    See Section 8.7
(c)        Explain that a mixture of NH4OH and NH4CI gives us the basic buffer?
                                    See Section 8.7
(d)               How do you justify that the greater quantity of CH3COONa in acetic acid decrease the dissociating power of acetic acid and so the pH increases.
CH3COOH is a weak acid and ionizes very small, while CH3COONa is a strong electrolyte and it ionizes in water to greater extent and provides acetate ions.
 CH3COOH +H2O        CH3COO- + H3O+
CH3COONa                  CH3COO- + Na+
            Thus CH3COONa decreases the ionization of CH3COOH due to common CH3COO- ion and pH of solution increases.
(d)       Explain the term buffer capacity.
                        See Section 8.7.1
Q18.    (a)        What is the     solubility product? Derive the solubility product expression for sparingly soluble compounds, AgCI, Ag2CrO4 and PbCI2.
                                    See Section 8.8
(b)       How do you determine the solubility product of a substance when its solubility is provided in grams/100 g of water?
                        See Section 8.8
(c)        How do you  calculate the solubility of a substance from the value of solubility product.
                                    See Section 8.8
Q19.    Kc value for volume for the following reaction is 0.076 at 520oC
                                    2HI   H2            +          I2
            Equilibrium mixture constants [HI] =0.08 M, [H2] =0.01 M. To this mixture more HI is added so that its new concentration is 0.096 M. what will be the concentration of [HI], [H2] and [I2] when equilibrium is re-established.
            Solution
                                    2HI   H2            +          I2
            Equilibrium conc.        0.08               0.01                 0.01
            (mol/dm3)
            Initial conc.            0.096               0.01                 0.01
            After adding more HI
            (mol/dm3)
            Equilibrium conc.      0.096 – 2x       0.01                 0.01
            When equilibrium is re-established
            (mol/dm3)
            According to law of mass action
                        Kc =
                        Kc = =0.016
                              ==0.016
            Taking square root on both sides
                          
                        ==0.126
0.01  + x=0.126 (0.096 – 2x)
0.01  + x =0.0121 – 0.252 x
x + 0.252 x=0.0121 – 0.01
1.252 x=0.0021
x ==
x=0.00168  mol/dm-3
Thus
            Concentrations when equilibrium is re-established are
            [H2] =0.01 + x =0.01 +0.00168          =0.01168 mol dm-3
                [I2] =0.01 + x =0.01 +0.00168            =0.01168 mol dm-3
                [HI] =0.096 - 2  x =0.096 -2x0.00168 =0.0926 mol dm-3
Q20.    The equilibrium constant for the reaction between acetic and ethyl alcohol is 4. Amixture of 3 moles of acetic acid and one mole of C2H5OH is allowed to come to equilibrium. Calculate the amount of ethyl acetate at equilibrium stage in number of moles and grams. Also, calculate the masses of reactants left behind.
Solution
                        CH3COOH + C2H5OH CH3 COOC2H5 +H2O
Initial conc.            3                      1                                  0               0
(mol/dm3)
Equilibrium conc.        3- x                  1 – x                            x                 x
(mol/dm3)
            According to law of mass action
            Kc=
            Kc= 
            X2 =4(3 – x) (1 – x)
            X2 =4(3 – 3x) (1 – x)
            X2 =4(3 –4x +  – x2)
            X2 =12 –16x  +  4 x2)  
Or        12  – 16x + 4x2 – x2 =0
            3x2  – 16x + 12=0
            It is quadratic equation and can be solved by using quadric formula
 Here               a = 3    ,           b = 12              ,           c = 12
            Thus
                        x=
                        x=
                        x=
                        x=
                        x=
            Either  x=               or         x=
            x= 4.43 mol dm-3         or         x= 0.9 mol dm-3
                        x= 4.43 is not possible as it is greater than the initial concentrations of reactants, therefore , x= 0.9 mol dm-3
Therefore
            Moles of ethyl acetate =x=0.9 moles
            Mass of ethyl acetate = 0.9x88=79.46g.
                        Moles of water =x=0.9 moles
                        Mass of water =0.9 x 18=16.2 g.
            Moles of acetic acid     = 3 – x =3 – 0.9 =2.1 moles
            (left behind)
            Moles of ethyl alcohol = 1 – x =1 – 0.9= 0.1 moles
            (Left behind)
            Mass of ethyl alcohol =0.1 x 46=4.6g.
            (left behind)
Q21.    study the equilibrium
                                                H2O + CO  H2 + CO2
            (a)        Write an expression of Kp
                                    Kp= 
(b)       When 1 mole of steam and 1 mole of CO are allowed to reach equilibrium, 33.3% of the equilibrium mixture is hydrogen. Calculate the value of Kp. state the units of Kp.
Solution      
                                                H2O +       CO  H2 + CO2
                        Initial conc.            1                 1         0         0
                        (mol/dm3)
                        Equilibrium conc.      1 – x        1 – x      x         x
                        (mol/dm3)
            Total no. of moles at equilibrium =1 – x + 1 – x + x + x=2
            Hence
                                    % of H2 =
                                    33.3=
                        or        no of moles of H2 =0.67 moles
Hence
            At equilibrium
                        Moles of H2 =x=0.67 moles
                        Moles of CO2 =x=0.67 moles
                        Moles of H2O =1 – x = 1 - 0.67= 0.33 moles
                        Moles of CO =1 – x = 1 - 0.67= 0.33 moles
Hence, Kc is given as
                        Kc =  
                        Kc =  
            Since Kp =Kc (RT) andn =n products  -n rectants =0, therefore
                        Kp=Kc=4

Q22.    Calculate the pH of
            (a)        10-4­ moles/ dm3 of HCI
            HCI ionizes as
                                    HCI       H+ + CI-
            Since HCI is a strong acid, and it is 100% dissociated. Hence 10-4 mol/dm3 of HCI produces 10-4­­­ mol/dm3 of H+ ions.
            Thus
                        [H+] = 10-4­­ mol/dm3
            So
                        pH = -  log [H+]
                        pH = -  log [10-4]
                        pH = 4
            (b)       10-4 moles/dm3 of Ba(OH)2
                        Ba(OH)2 ionizes as
                                                Ba(OH)2       Ba+2   + 2 OH-
            Since Ba(OH)2 is a strong base it is 100% dissociated.
Hence 10-4 mol/dm3 of Ba(OH)2 produces 2 x 10-4 mol/dm3 of OH- ions.
            Thus
                        [OH-]= 2 x 10-4 mol/dm3
            So
                        pOH= - log[OH-]
                        pOH= - log[2 x 10-4]
                        pOH= 3.699
Since
                        pH + pOH=14
            Therefore
                        pH= 14 – pOH
                        =14 – 3.699 =10.301
            (c)        1 mol/dm3 of H2X, which is 50% dissociated
                        H2X ionizes as
                                                H2X       2H+   +  X-
            1mole of H2X produces 2 moles of H+ ions if 100% dissociated
            However, since H2X is 50% dissociated therefore 1 mole of H2X produces 1 mole of H+ ion
            Thus
                        [H+]=1 mol/dm3
            So
                        pH= -log [H+]
                        pH= -log [1] 
            pH= 0
            (d)       1 mol/dm3 of NH4 OH which is 1% dissociated
                        NH4 OH ionizes as
                                                NH4OH       NH4+   + OH-
                        It shows that 1 mole of NH4OH produces 1 mole of OH- ions.
                        NH4OH is only 1% dissociated
            Hence
                        % dissociation =x 100
                        1 = x 100
                        mol of OH- x1=0.01 mol/dm3
            Thus
                        [OH-] =0.01 mol/dm3
            So
                        pOH = - log [OH-]
                        pOH = - log [0.01]
                        pOH = 2
            Since
                         pH + pOH =14
            Therefore
                        pH=14 – pOH
                        =14 – 2=12
Q23.    (a)        Benzoic aicd C6H5COOH is weak mono-basic acid (Ka=6.4 x10-5­­­­mol/dm3). What is the pH of a solution   containing 7.2 g of sodium  benzoate in one dm3 of o.02 mol/dm3 benzoic acid.
            Mass of sodium benzoate                   =7.2 g /dm3
Formula of sodium  benzoate is C6H5COONa
Mol. Mass of sodium benzoate           =144 g/mol
Moles of sodium benzoate                  ==0.05 mol/dm3
            Moles of benzoic acid             =0.02 mol/dm3
                        Ka  of benzoic acid      6.4 x 10-5 mol/dm3
Thus                pKa = - log Ka= - log (6.4 x10-5­­ ) =4.2
Hence, according to Henderson’s eq.
            pH =pKa  + log
            pH =4.2 + log 
            pH = 4.2 + 0.39
            pH =4.59
(b)       A buffer solution has been prepared by mixing 0.2 M CH3COONa and 0.5 M CH3COOH in 1 dm3 of solution. calculate the pH of solution. pKa of acid is 4.74 at 25oC. How the value of pH will change by adding 0.1 mole 0.1 mole of NaOH and 0.1  mole mol HCI respectively.
Solution
                        [CH3 COOH]              =0.5 M
                        [CH3 COONa]                        =0.2 M
                        pKa of CH3 COOH                 =4.74
                                                            pH       =?
            Since pH =pKa + log 
or         pH =pKa + log 
                        pH =4.74  + log 
                        pH =4.74  - 0.4
                        pH =4.34
When 0.1 mole of NaOH is added
NaOH is strong base. It dissociates completely. Therefore, it produces 0.1 moles ofOH- ions. Thus, 0.1 moles of OH- ions reacts with 0.1 moles of CH3 COOH. Hence, out of 0.5 moles of CH3 COOH, 0.4 moles of CH3 COOH are behind.
On the other hand, due to slat formed by the neutralization reaction, conc. of salt (CH3COONa) is increased from 0.2 moles to 0.3 moles.
            Hence, new conc. will be
                        [CH3 COOH]              =0.4 M                        [CH3 COONa]                        =0.3M
            Thus    pH =pKa + log 
                        pH =4.74  + log 
                        pH =4.74  - 0.12
                        pH =4.62
            Addition of 0.1 mole of HCI
HCI is a strong acid. It dissociates completely. Therefore, it produces 0.1 moles of H+ions. Thus, 0.1 moles of H+ ions react with 0.1 moles of CH3COO- ions. Hence, out of 0.2 moles of salt (CH3COO-Na+), 0.1 moles of salt are left behind.
On the other hand, conc. of acid (CH3COOH) is increased form 0.5 moles to 0.6 moles.
Hence, new conc. will be
                        [CH3 COOH]              =0.6 M                        [CH3 COONa]                        =0.1M
            Thus    pH =pKa + log 
                        pH =4.74  + log 
                        pH =4.74  - 0.78
                        pH =3.96
            (See Section 8.7.1  for complete understanding of this numerical)
Q24.    Solubility of CaF2 in water at 25OC is found to be 2.05 x 10-4 mol dm-3. What is the value of Ksp at this T.
                        Solubility of CaF2 = 2.05 x 10-4 mol dm-3
                        According to balanced chemical eq.
                                    CaF2(aq)  Ca2+(aq)  + 2F-(aq)
            At initial stage 2.05 x 10-4       0              0
            (mol/dm3)
            After solubility 0         2.05 x 10-4         2x2.05 x 10-4
            Hence Ksp=[Ca+2][F-]
                        Ksp=[0.05 x 10-4][2 x 2.05 x 10-4­­]2
                        Ksp=3.446 x 10-11 mol3 dm-9­­
Q25.    The solubility product of Ag2CrO4 is 2.6 x 120-2 at 25oC. Calculate the solubility of the compound.
            Ksp of Ag2 CrO4 =2.6 x 10-2
            We know
                                    Ag2CrO4(aq) 2Ag+             CrO42-(aq)
Initial stage (mol/dm3)             Ag2CrO4                             0             0
At equilibrium (mol/dm3)          Ag2CrO4                           2S            S

Hence
            Ksp =[Ag+]2[CrO42-]
            Ksp=[2S]2[S] =2.6 x 10-2
                        4[S]3 =2.6 x 10-2
                        [S]=  
            or         [S]=0.1866 mol/dm3
Hence at equilibrium
                        [Ag+]   =2 x 0.1866 mol/dm3 =0.3732 mol/dm3
and      [CrO42- ]           =0.1866 mol/dm3
Since
            1 mole of Ag2CrO4 gives 1 mole of CrO42-ions, hence
            Solubility of Ag2CrO4=[CrO42-]=0.1866 mol/dm3

I am Mechatronics and Control Engineer from UET, Lahore. Helping Engineering students gives me pleasure.

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