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CHAPTER 8
CHEMICAL EQUILIBRIUM
TEXT BOOK EXERCISE
Q1. Multiple Choice questions(i) For which system does the equilibrium constant, Kchas units of
(concentration)-1
(a) N2+ 3H2 2NH3
(b) H2+ I2 2HI
(c) 2NO2 N2O4
(d) 2HF H2+ F2(ii) Which statement about the following equilibrium is correct
2SO2+ O2 2SO3
(a) The value of Kpfalls with a rise in temperature
(b) The value of KPfalls with increasing Pressure
(c) Adding V2O5catalyst increases the equilibrium yield of
SO3
(d) The value of Kpis equal to Kc
(iii) The pH of 10-3
mol/dm3
of an equous solution of H2SO4is(a) 3 (b) 2.7 (c) 2 (d) 1.5
(iv) The solubility product of AgCI is 2 x 1010
mol2dm
-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 bariumchloride and precipitate is removed by filtration. What are the
main ions in the filtrate?
(a) Ag+and NO3
-1only
(b) Ag+and Ba
2+and NO3
-1
(c) Ba2+
and NO3-1
only
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(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 3O2is 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, KCof 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 bothreactants and products are present in a reaction mixture.
(ii) The Kcof 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 Kwof pure water at 25oC is 10-14dm-6and is
represented by an expression
Kw=[H+][OH
-]=10
-14mol
2dm
-6
(v) AgCI is a sparingly soluble ionic solid in water. Its solution
produces excess of Ag+
and CI-ions.
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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 KCfor the following reactions
(i) Sn2+(aq) + 2Fe3+(aq) Sn4+(aq)+ 2Fe2+(aq)(ii) Ag+(aq) +Fe
2+(aq) Ag(s)+ Fe
3+(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 8.1.1Q6. 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?
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(b) Before the curves become parallel, the steepness of
curves falls? Give reasons.
(c) The rate of decrease of concentrations of the reactantsand 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 Kpfor the following general
reactions.
aA + bB cC +dD
See section 8.1.2(b) Decide the comparative magnitudes of Kc, Kpfor the
following reactions.Synthesis of NH3
N2(g) +3H2(g) 2NH3(g)
KCis given by
Kc=
Kpis given byKp=
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=Kcx (RT)-2
Or Kp= Kcx
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)
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Kcis given by according to law of Mass of Action
Kc=
Kpis 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
=21 =1
Kp=Kcx (RT)
Thus If T is such that RT > 1, then Kp> Kc
If T is such that RT< 1, then Kp
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(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 100
oC.
Ionic Product of water is given by the equation
Kw[H+][OH
-]
Value of Kwincreases 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 100
oC Kw=7.5 x 10
-14
Further
At 0oC (Kw)o =0.1 x 10
-14_____(1)
At 100o
C (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 0
oC
(b) What is the Justification for the increase of ionic
product with temperature?
Value of Kwincrease with increase in temperature. It isbecause increase in temperature increase the ionization of
H2O. Thus, more H+
or OH-ions are produced. Hence value
Kwincreases.
(c) How do you prove that at 25O
C in 1 dm3of water, there
are 10-7
moles of H3O+?
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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
-7mol/dm
3
Hence, at 25oC, water has 10-7mole/dm3of 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 pHaand pKbis 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.
CH3COOH + H2O CH3COO-+ H3O
+
However, H2SO4is a stronger acid than acetic acid, therefore,
H2SO4donates proton and acts as an acid while acetic acid accepts
proton and acts as a base.
H2SO4+ CH3COOH HSO4-+ CH3COOH2
+
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.
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(iii) Catalyst is added
(iv) CI2is added
See section 8.2Explain your answer.
Q15. Synthesis of NH3by Habers 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 NH3is 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 SO2to 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 acidgive 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
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(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, Ag2CrO4and 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. Kcvalue 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
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Equilibrium conc. 0.08 0.01 0.01
(mol/dm3)
Initial conc.
0.096 0.01 0.01After adding more HI(mol/dm3)
Equilibrium conc.
0.0962x 0.01 0.01When 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 (0.0962x)0.01 + x =0.01210.252 x
x + 0.252 x=0.01210.01
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. A mixture of 3 moles of acetic acid and one
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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 leftbehind.
Solution
CH3COOH + C2H5OH CH3COOC2H5+H2OInitial conc.
3 1 0 0(mol/dm3)
Equilibrium conc.
3- x 1x x x(mol/dm3)
According to law of mass action
Kc=
Kc=
X2
=4(3x) (1x)
X2
=4(33x) (1x)
X2
=4(34x + x2)
X2
=1216x + 4 x2)
Or 1216x + 4x
2
x
2
=03x216x + 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=
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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 = 3x =30.9 =2.1 moles
(left behind)
Moles of ethyl alcohol = 1x =10.9= 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+ CO2Initial conc.
1 1 0 0(mol/dm3)
Equilibrium conc. 1x 1x x x
(mol/dm3)
Total no. of moles at equilibrium =1x + 1x + x + x=2
Hence
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% 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 =1x = 1 - 0.67= 0.33 moles
Moles of CO =1x = 1 - 0.67= 0.33 moles
Hence, Kcis given as
Kc=
Kc=
Since Kp=Kc(RT) and n =n products -nrectants =0, therefore
Kp=Kc=4
Q22. Calculate the pH of
(a) 10-4moles/ dm3of HCI
HCI ionizes as
HCI H+
+ CI-
Since HCI is a strong acid, and it is 100% dissociated. Hence 10-
4mol/dm
3of HCI produces 10
-4mol/dm
3of H
+ions.
Thus
[H+] = 10
-4mol/dm
3
So
pH = - log [H+]pH = - log [10
-4]
pH = 4
(b) 10-4
moles/dm3of Ba(OH)2
Ba(OH)2 ionizes as
Ba(OH)2 Ba+2
+ 2 OH-
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Since Ba(OH)2is a strong base it is 100% dissociated.
Hence 10-4
mol/dm3of Ba(OH)2produces 2 x 10
-4mol/dm
3of OH
-ions.
Thus [OH-]= 2 x 10
-4mol/dm
3
So
pOH= - log[OH-]
pOH= - log[2 x 10-4
]
pOH= 3.699
Since
pH + pOH=14
Therefore
pH= 14pOH=143.699 =10.301
(c) 1 mol/dm3of 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/dm
3
So
pH= -log [H+]
pH= -log [1]
pH= 0
(d) 1 mol/dm3of NH4OH which is 1% dissociated
NH4OH ionizes as
NH4OH NH4+
+ OH
-
It shows that 1 mole of NH4OH produces 1 mole of OH-
ions.
NH4OH is only 1% dissociated
Hence
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% dissociation = x 100
1 = x100
mol of OH- x1=0.01 mol/dm
3
Thus
[OH-] =0.01 mol/dm
3
So
pOH = - log [OH-]
pOH = - log [0.01]
pOH = 2Since
pH + pOH =14
Therefore
pH=14pOH
=142=12
Q23. (a) Benzoic aicd C6H5COOH is weak mono-basic acid
(Ka=6.4 x10-5
mol/dm3). What is the pH of a solution
containing 7.2 g of sodium benzoate in one dm3of o.02
mol/dm3benzoic acid.Mass of sodium benzoate =7.2 g /dm
3
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 Hendersons eq.
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pH =pKa + log
pH =4.2 + logpH = 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 dm3of solution.
calculate the pH of solution. pKaof 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[CH3COOH] =0.5 M
[CH3COONa] =0.2 M
pKa of CH3COOH =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 addedNaOH is strong base. It dissociates completely. Therefore, it
produces 0.1 moles of OH-ions. Thus, 0.1 moles of OH
-ions
reacts with 0.1 moles of CH3COOH. Hence, out of 0.5 moles of
CH3COOH, 0.4 moles of CH3COOH 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
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[CH3COOH] =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 HCIHCI 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
[CH3COOH] =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 CaF2in water at 25O
C is found to be 2.05 x 10-4
mol dm-3
. What is the value of Kspat this T.Solubility of CaF2= 2.05 x 10
-4mol dm
-3
According to balanced chemical eq.CaF2(aq) Ca
2+(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-]
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Ksp=[0.05 x 10-4
][2 x 2.05 x 10-4
]2
Ksp=3.446 x 10-11
mol3dm
-9
Q25. The solubility product of Ag2CrO4is 2.6 x 120
-2
at 25
o
C.Calculate the solubility of the compound.Kspof Ag2CrO4=2.6 x 10
-2
We know
Ag2CrO4(aq) 2Ag+
CrO42-
(aq)Initial stage (mol/dm3)
Ag2CrO4 0 0At 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/dm
3=0.3732 mol/dm
3
and [CrO42-
] =0.1866 mol/dm3
Since
1 mole of Ag2CrO4gives 1 mole of CrO42-
ions, hence
Solubility of Ag2CrO4=[CrO42-
]=0.1866 mol/dm3