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SAMPLE QUESTION PAPER
CHEMISTRY (043)
CLASS – XII – (2014-2015)
QUESTION PAPER DESIGN
CLASS XII CBSE
Types of Questions
S.No
Type of Question
1.
2.
3.
4.
5.
Long Answers (LA)
Value based question
Short Answers-II(SA II)
Short Answers-I(SA I)
Very Short Answer (VSA)
Total
Type of questions
Knowledge
Understanding
Application
HOTS (Higher order thinking skills)
Evaluation and Multi-disciplinary
Marks
for No.
each
Questions
Question
5
03
4
01
3
12
2
05
1
05
26
VSA(1)
2
2
1
SA1(2)
1
2
2
-
SA2(3)
1
4
4
1
2
of Total Marks
15
04
36
10
05
70
VB(4) LA(5)
1
1
1
1
-
Total(%)
07 (10%)
21 (30%)
21 (30%)
10(14%)
11(16%)
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SAMPLE QUESTION PAPER
CHEMISTRY
CLASS – XII (2014 – 2015)
Time Allowed: 3 hr
Maximum marks: 70
General Instructions:
(a)
All questions are compulsory.
(b)
Q.no. 1 to 5 are very short answer questions and carry 1 mark each.
(c)
Q.no. 6 to 10 are short answer questions and carry 2 marks each.
(d)
Q.no. 11 to 22 are also short answer questions and carry 3 marks each
(e)
Q.no. 23 is a value based question and carry 4 marks.
(f)
Q.no. 24 to 26 are long answer questions and carry 5 marks each
(g)
Use log tables if necessary, use of calculators is not allowed.
1.
The following figure shows the variation of adsorption of N2 on charcoal with pressure at
different constant temperatures:
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Arrange the temperatures T1, T2 and T3 in the increasing order.
2. Give the formula of a noble gas species which is isostructural with IBr2-.
3. What is the effect of synergic bonding interactions in a metal carbonyl complex?
4. PCl5 acts as an oxidizing agent. Justify.
5. Write the name of the product formed when benzenediazonium chloride solution is treated with
potassium iodide.
6. Name the crystal defect which reduces the density of an ionic solid? What type of ionic
substances show this defect?
7. The molar conductivity ( λ ) of KCl solutions at different concentrations at 298 K is plotted as
m
shown in the figure given below:
Determine the value of λ0m and A for KCl.
8. Aluminum crystallizes in anfcc structure. Atomic radius of the metal is 125 pm. What is the
length of the side of the unit cell of the metal?
9. Draw the structure of the following compounds:
(i)
H2S2O7
(ii)
XeOF4
OR
Write balanced chemical equations for the following:
(i)
Reaction of chlorine with hot and concentrated NaOH.
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(ii)
Sulphur dioxide is passed through an aqueous solution of Fe (III) salt.
10. 0.5 g of KCl was dissolved in 100 g of water and the solution originally at 200C, froze at
-0.240C. Calculate the percentage dissociation of the salt.
(Given :Kf for water = 1.86 K kg /mol, Atomic mass: K = 39 u, Cl= 35.5 u)
11. State briefly the principles involved in the following operations in metallurgy. Give an
example.
(i)
Hydraulic washing.
(ii)
Zone refining.
i)
What type of deviation from Raoult’s law is observed, when two volatile liquids A and
12.
B on mixing produce a warm solution? Explain with the help of a well labeled vapour
pressure graph.
Consider separate solutions of 0.5 M CH3OH, 0.250 M KCl (aq) and 0.125 M Na3PO4
ii)
(aq). Arrange the above solutions in the increasing order of their Van’t Hoff factor.
13. Write the Nernst equationand calculate the emffor the following cell at 298 K:
Mg(s) / Mg2+ (0.001 M) // Cu2+ (0.0001 M) / Cu(s)
How does Ecellvary with the concentration of both Mg2+ and Cu2+ ions?
(GivenEocell= 2.71 V)
14. Explain the following observations giving appropriate reasons:
(i)
Ozone is thermodynamically unstable with respect to oxygen. .
(ii)
The HEH bond angle of the hydrides of group 15 elements decrease as we move down
the group.
(iii) Bleaching effect of chlorine is permanent.
15.
(i)
Predict the number of unpaired electrons in the tetrahedral [MnBr4]2- ion.
(ii)
Draw structures of geometrical isomers of [Co(NH3)4Cl2]+.
(iii)
Write the formula for the following coordinate compound:
Amminebromidochloridonitrito-N-platinate(II)
16. Explain what is observed when
(i)
Silver nitrate solution is added to potassium iodide solution.
(ii)
The size of the finest gold sol particles increases in the gold sol.
(iii)
Two oppositely charged sols are mixed in almost equal proportions.
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17.
(i) In the following pairs of halogen compounds, which would undergo SN1 reaction
faster? Explain.
Cl
Cl
and
(ii) Amongst the isomeric dihalobenzenes which isomer has the highest melting point and
why?
(iii) Arrange the following haloalkanes in the increasing order of density. Justify your
answer.
CCl4, CH2Cl2 and CHCl3.
18. An organic compound ( A ) has characteristic odour. On treatment with NaOH, it forms
compounds ( B ) and ( C ). Compound ( B ) has molecular formula C7H 8Owhich on oxidation
gives back ( A ). The compound ( C ) is a sodium salt of an acid. When ( C ) is treated with
soda-lime, it yields an aromatic compound ( D ). Deduce the structures of ( A ), ( B ), ( C ) and
( D ). Write the sequence of reactions involved.
19. (a)
Give one chemical test to distinguish between the following pairs of compounds:
(i)
Methylamine and dimethylamine.
(ii)
Aniline and benzylamine
(b) Write the structures of different isomers corresponding to the molecular formula C3H9N,
which will liberate nitrogen gas on treatment with nitrous acid.
20. (a)
Exemplify the following reactions:
(b)
(i)
Rosenmund reduction reaction.
(ii)
Kolbe electrolysis reaction.
Arrange the following compounds in increasing order of their reactivity towards HCN:
Acetaldehyde, Acetone, Di-tert-butyl ketone.
OR
(a)
Predict the products of the following reactions:
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(i)
CH3-CH2-COOH
(i) Cl2 / Red phosphorous
(ii) H2O
(ii)
CH3
(i) CS2
+ CrO2Cl2
(b)
(ii) H3O+
Arrange the following compounds in increasing order of acid strength:
Benzoic acid, 4-Nitrobenzoic acid, 4-Methoxybenzoic acid.
21.
(i)
Identify the monomer in the following polymeric structure:
CN
[ CH2-CH=CH-CH2-CH2-CH ]n
(ii)
On the basis of forces between their molecules in a polymer to which class does
neoprene belong?
(iii)
Can both addition and condensation polymerization result in the formation of a copolymer?
22.
(i)
Which of the following biomolecule is insoluble in water? Justify.
Insulin, Haemoglobin, Keratin.
23.
(ii)
Draw the Haworth structure for α-D-Glucopyranose.
(iii)
Write chemical reaction to show that glucose contains aldehyde as carbonyl group.
John had gone with his mother to the doctor as he was down with fever. He then went to the
chemist shop with his mother to purchase medicines prescribed by the doctor. There he
observed a young man pleading with the chemist to give him medicines as he had nasal
congestion. The chemist gave him cimetidine. John advised and also explained to the young
man that he should only take the medicines prescribed by the doctor.
Answer the following questions:
a) Did the chemist give an appropriate medicine? Justify your answer.
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b) John’s action was appreciated by his mother. List any two reasons.
24.
(a) Write the mechanism of hydration of ethene to form ethanol.
(b)
(c)
How are the following conversions carried out?
(i)
Propanol to propan-2-ol.
(ii)
Propanol to 1-propoxypropane.
Give the structure and the IUPAC name of the major product obtained in the following
reaction:
OH
conc. HNO3
OR
(a)
Write the mechanism of the reaction of HI with methoxymethane.
(b) Identify A and B in the following reactions:
(i)
OH
(i) CO2
NaOH
A
(ii) H+
B
(ii)
Cu, 573 K
C2H5OH
CH3MgBr
A
H2O / H+
B
(c) Give the structure and the IUPAC name of the major product obtained in the following
reaction:
OC2H5
conc. HNO3
conc. H2SO4
25. (a)
A blackish brown coloured solid (A) which is an oxide of manganese, when fused with
alkali metal hydroxide and an oxidizing agent like KNO3, produces a dark green coloured
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compound (B).
Compound (B) on disproportionation in neutral and acidic solution gives a
purple coloured compound (C). Identify A, B and C and write the reaction involved when
compound (C) is heated to 513 K.
(b)
(i)
E0 M3+ / M2+values for the first series of transitionelements are given below.
Answer the question that follows:
E0 (V)
M3+ / M2+
Ti
-0.37
V
-0.26
Cr
-0.41
Mn
+1.57
Fe
+0.77
Co
+1.97
Identify the two strongest oxidizing agents in the aqueous solution from the
above data.
ii)
Copper (I) ion is not known in aqueous solution
iii)
The highest oxidation state of a metal is exhibited in its oxide.
OR
(a)
Write balanced equations to represent what happens when
(i) Cu2+ is treated with KI.
(ii) Acidified potassium dichromate solution is reacted with iron (II) solution.
(ionic equation)
(b)
i)
The figure given below illustrates the first ionization enthalpies of first, second
and third series of transition elements. Answer the question that follows
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Which series amongst the first, second and third series of transition elements have
the highest first ionization enthalpy and why?
ii)
Separation of lanthanide elements is difficult. Explain.
iii)
Sm2+, Eu2+ and Yb2+ ions in solutions are good reducing agents but an
aqueous solution of Ce4+ is a good oxidizing agent. Why?
26.
i)
Graphically explain the effect of temperature on the rate constant of reaction? How can
this temperature effect on rate constant be represented quantitatively?
ii)
The decomposition of a hydrocarbon follows the equation
k = (4.5 x 10 11 s −1 ) e
−28000K
T
Calculate Ea
OR
i)
In the reaction
Q + R → Products
The time taken for 99% reaction of Q is twice the time taken for 90% reaction of Q.
The concentration of R varies with time as shown in the figure below:
What is the overall order of the reaction? Give the units of the rate constant for the
same. Write the rate expression for the above reaction.
ii)
Rate constant for a first order reaction has been found to be 2.54 x 10-3s-1.
Calculate its three-fourth life.
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MARKING SCHEME- CHEMISTRY (043)
SAMPLE PAPER (CLASS - XII)
Q.N
Value Points
o.
1.
T1 < T2 < T3
2.
XeF2
3.
It strengthens the bond between CO and the metal.
Mar
ks
1
1
4.
The oxidation state of P in PCl5 is +5 it cannot increase its oxidation state beyond
+5 but it can decrease from +5 to +3.
1
5.
6.
Iodobenzene
Schottky defect
It is shown by ionic substances in which the cation and anion are of almost similar
sizes. / ionic substances having high coordination number.
The plot is nearly a straight line and can be extrapolated to zero concentration(i.e.
from the intercept) to find the value of 0m = 150.0 S cm2 mol-1
1
1
1
7.
1
1
[Λ mc = Λ m0 - A c ]
½
y
A = - slope =
x
150.0  147.0
=
0.034
½
= 88.23 S cm2 mol-1
8.
r = 125 pm, a = ?
for fcc structure r 
a
2 2
a  125 x 2 x 1.414
= 353.5 pm
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½
½
½+
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½
9.
(i)
1
(ii)
1
OR
(i)
3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O
1
(ii)
2 Fe3+ + SO2 + 2 H2O → 2 Fe2+ + SO42- + 4 H+
( ½ mark to be deducted for an unbalanced chemical equation)
1
10.
KCl is an electrolyte, it undergoes dissociation
 Tf  i K f m
i
½
 Tf
Kf m
Tf = 0 –(-0.24) = 0.24 0
Molar mass of KCl = 39 + 35.5 = 74.5 u
substituting the values
i 
0.24 x 74.5 x100
1.86 x 0.5 x1000
i = 1.92
α is the degree of dissociation
i=1+α
½
½
α =1.92 – 1 = 0.92
Percentage dissociation = 92 %
½
(or any other suitable method)
11.
(i)
(ii)
Hydraulic washing:
Principle involved: differences in gravities of the ore and the gangue particles
e.g.oxide ores ( haematite), native ores Au, Ag (any one example)
Zone refining:
Principle involved: the impurities are more soluble in the melt than in the solid
state of the metal.
e.g. germanium, silicon, boron, gallium and indium (any one example)
12.
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1
½
1
½
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(i)
A warm solution is obtained on mixing the two liquids A and B indicate that the
process of mixing is exothermic. ( ∆H mixing = - ve). So the solution shows a
negative deviation from the Raoult’s law.
½
1
The forces of interaction between A and B molecules are more than in the A-A and B-B
molecules. So the partial vapour pressure of each component will be less and the
partial vapour pressure of the solution will also be less than that from the Raoult’s law.
(ii)
13.
For CH3OH ( non electrolyte)
KCl → K + + ClNa3PO4 → 3 Na+ + PO43CH3OH < KCl < Na3PO4
i=0
i=2
i =4
Ecell  2.71 
0.0591
[ Mg 2  ]
log
2
[Cu 2 ]
½
0.0591
[10 3 ]
log
2
[10  4 ]
½
Ecell = 2.68 V
Ecell increases with the increase in the concentration of Cu2+ ions and decrease in the
concentration of Mg 2+ ions.
(ii)
(iii)
1
Nernst equation:
0
E cell  E cell

14.
(i)
½
Decomposition of ozone into oxygen is an exothermic process (∆H = -ve) and results in an
increase in entropy(∆S = +ve), resulting in large negative Gibbs energy change(∆G = ve). Decomposition of ozone to oxygen is a spontaneous process.
As we move down the group, the size of the central atom increases and the
electronegativity decreases. therefore the bond pair of electrons lie away from the central
atom, the force of repulsion between the adjacent bond pairs decreases. So the bond
angle decreases.
½+
½
1
1
1
The bleaching action of Cl2 is due to the oxidation of coloured substance to
colourless by nascent oxygen.
½
Cl2 + H2O → 2 HCl + [O]
Nascent oxygen
½
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15.
(i)
Mn has the configuration 3d54s 2. Hence the configuration of Mn2+ in [MnBr4]2- is 3d5.
Mn2+
↑
↑
↑
↑ ↑
½
3d
4s
Since it is tetrahedral in shape, the hybridization is sp3.
There are five unpaired electrons.
½
4p
(ii)
(½
+½)
[Pt(NH3)BrCl(NO2)](iii)
16.
(i)
(ii)
(iii)
17.
(i)
When silver nitrate solution is added to potassium iodide solution, a precipitate of silver
iodide is formed which adsorbs iodide ions from the dispersion medium and a
negatively charged colloidal solution is formed.
As the size of the gold sol particles increases, the colour of the solution changes from red to
purple, then blue and finally golden because the colour of colloidal solution depends on
the wavelength of the light scattered by the dispersed particles and wavelength
further depends on the size of the particles.
When oppositely charged sols are mixed in almost equal proportions, neutralization of
their charges occur and precipitation occurs.
1
1
1
Cl
Undergoes SN1 faster.
It is a secondary halide, which forms a secondary carbocation which is more
stable than the primary carbocation so greater will be its ease of formation
from the corresponding alkyl halide and faster will be the rate of reaction.
(ii)
1
The para-isomer
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½
½
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½
Cl
Cl
has the highest melting point as compared to their ortho- and metaisomers.
The para-isomer is more symmetrical and fits into the crystal lattice better, as a result
intermolecular forces are stronger, higher temperature required to melt the paraisomer.
(iii)
18.
CH2Cl2 < CHCl3 < CCl4 ( increasing order of density)
The density increases with the increase in the number of the halogen atoms.
A (C7H6O)
NaOH

 B (C7H8O)
[O]
+
½
½
½
C (sodium salt of an acid)
NaOH + CaO
A
D
(A) undergoes disproportionation in presence of an alkali (Cannizzaro reaction) so there is
no α hydrogen.
(C) undergoes decarboxylation.
A
½
CHO
B
½
CH2OH
C
½
COONa
D
½
CH2OH
CHO
NaOH
COONa
+
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½
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A
B
C
COONa
NaOH + CaO
½
C
19.
(a)
(i)
D
Chemical test
Methylamine
(10 aliphatic amine)
Dimethylamine
(20 aliphatic amine)
Gives offensive smell.
No odour obtained (no
reaction)
Carbylamine test:
To 1 ml of the organic
compound add an alcoholic
solution of KOH and CHCl3
(ii)
Chemical test
Azo dye test:
To 1mL of the organic
compound add HNO2
(NaNO2 + dil. HCl) at 273278 K. then add an alkaline
solution of β naphathol to
the solution.
(b)
Aniline
(10 aromatic amine)
Benzylamine
(10 aliphatic amine)
A brilliant red dye is
obtained
No dye obtained
1
1
( any other suitable test)
Four structural isomers are possible for C3H9N
Only primary amines react with HNO2 to liberate nitrogen gas.
10 amines: CH3CH2CH2NH2
CH3-CH-CH3
½
½
NH2
No reaction for 20 and 30 amines
20 amine: CH3-NH-C2H5
30 amine:
CH3-N-CH3
CH3
20.
(a)
(i)
When acid chloride is hydrogenated over catalyst, palladium on barium sulphate they are
reduced to the corresponding aldehydes. This reaction is called Rosenmund reduction.
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½
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CHO
COCl
H2
½
Pd - BaSO4
Benzoyl chloride
(ii)
(b)
Benzaldehyde
Alkali metal salts of carboxylic acids undergo decarboxylation on electrolysis of their
aqueous solutions and form hydrocarbons having twice the number of carbon atoms present
in the alkyl group of the acid. This reaction is called Kolbe electrolysis reaction.
½
2CH3COOK + 2H2O → CH3-CH3 + 2CO2 + 2KOH + H2
½
Acetaldehyde
Acetone
Di-tert-butyl ketone
: CH3CHO
: CH3COCH3
: (CH3)3C-CO-C(CH3)3
1
Di-tert-butyl ketone < Acetone < Acetaldehyde
OR
(a)
(i)
CH3-CH2-COOH
(i) Cl2 / Red phosphorous
CH3-CH-COOH
(ii) H2O
CH3
1
Cl
CHO
(i) CS2
(ii)
+ CrO2Cl2
(ii) H3O+
Benzoic acid
(b)
COOH
4-Nitrobenzoic acid
O2 N
COOH
4-Methoxybenzoic acid
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1
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COOH
CH3O
4-Methoxybenzoic acid < Benzoic acid <
21.
(i)
1
4-Nitrobenzoic acid
Buta-1,3diene : CH2=CH-CH=CH2
and Acrylonitrile : CH2=CH(CN)
(either name or structure)
½
½
(ii)
Neoprene is classified as an Elastomer.
(the polymeric chain are held together by weakest intermolecular forces)
1
(iii)
22.
(i)
Yes a co-polymer can be formed in addition and condensation polymerization.
1
Keratin is insoluble in water.
It is a fibrous protein in which the polypeptide chains are held together by strong
intermolecular forces, hence insoluble in water.
½
½
(ii)
α-D-Glucopyranose
6
CH2OH
H
5
H
OH
4
HO
3
23.
(a)
(i)
1
H
OH
OH
2
H
(iii)
1
O H
The sequence in the complimentary strand is
ATGCTTGA
1
No the chemist did not give the appropriate medicine.
Cimetidine is an antihistamine, but it is an antacid and not an antiallergic drug.Antacid and
antiallergic drugs work on different receptors. Therefore cimetidine cannot be used to treat
nasal congestion.
½
Critical thinking
Social responsibility
(or any other two reasons)
1
1
1½
(ii)
24.
(a)
Mechanism of hydration of ethene to ethanol by acid catalysed hydration:
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1
½
HBr
H SO
2
4


 CH3CH=CH2 
CH3CH2CH2-OH 
 CH3-CH(Br)-CH3
½
Heat
(b)
(i)
 aq. KOH
CH3-CH(OH)-CH3
Propan-2-ol
Propanol
1
SOCl
2
CH3CH2CH2-OH 
 CH3CH2CH2-Cl
↓NaOH
CH3CH2CH2-ONa
(ii)
½
CH3CH2CH2-Cl + CH3CH2CH2-ONa → CH3CH2CH2-O-CH2CH2CH3
OH
OH
conc. HNO3
½
NO2
O 2N
(c)
½
NO2
2,4,6-Trinitrophenol
OR
The mechanism of the reaction of HI with methoxymethane involves the following steps:
Step I : protonation of ether molecule
(a)
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½
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Step II : nucleophilic attack by I- by SN2 mechanism
½
½
Step III : when HI is in excess and the reaction is carried out at high temperature, methanol
formed in the second step reacts with another molecule of HI and is converted to methyl
iodide.
1
Methyl iodide
A
B
ONa
OH
(b)
COOH
(i)
½+
½
A : CH3CHO
B : CH3-CH(OH)-CH3
½
½
(ii)
OC2H5
OC2H5
conc. HNO3
conc. H2SO4
½
O 2N
1- Ethoxy-4-nitrobenzene
(c)
½
25.
(a)
A : MnO2
B : K2MnO4
C : KMnO4
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½
½
½
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½

2 KMnO4  K2MnO4 + MnO2 +O2
(b)
(i)
Mn3+ and Co3+ are the strongest oxidizing agents from the data given.
(ii)
Copper (I) compounds are unstable in aqueous solution and undergo disproportionation
2 Cu+ → Cu2+ + Cu
Cu2+ (aq) is more stable than Cu+ because it has high negative ∆hydrH0as compared to Cu+.
(iii)
The highest oxidation state of a metal is exhibited in its oxide as oxygen has the ability to
form multiple bonds to metal atoms.
½
+½
½
½
1
OR
(a)
(i)
(ii)
(b)
(i)
(ii)
(iii)
26.
(i)
2 Cu2+ + 4I- → Cu2I2 (s) + I2
Cr2O72-
+
2+
+ 14 H + 6 Fe
→ 2 Cr
1
3+
3+
+ 6 Fe
7 H2O
The third series (5d elements) of the transition elements have the highest first ionization
enthalpy ( Hf to Au).
This is because of the poor shielding of the nucleus by the 4f electrons in the 5d elements
which results in greater effective nuclear charge on the valence electrons.
Due to the lanthanoid contraction the change in the ionic radii in the lanthanoids is very
small, their chemical properties are similar, so the separation is difficult.
1
½
½
1
+3 is the most stable oxidation state of lanthanides. Ions in +2 tend to change to +3 by
losing electrons so act as reducing agents, whereas ios in +4 tend to change to +3 by
gaining electrons so act as oxidizing agents.
1
For a chemical reaction with rise in temperature by 100, the rate constant is nearly doubled.
½
1
(T2 = T 1 + 10 0)
Increasing the temperature of the substance increases the fraction of molecules which
collide with energies greater than Ea (activation energy)
As the temperature increases (T2) , the fraction of molecules having energy equal to or
greater than activation energy gets doubled leading to doubling the rate of reaction.
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1
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k = A e-Ea/ RT (Arrhenius equation)
(ii)
½
k = A e-Ea/ RT

1
Ea
28000 K
 
RT
T
½
Ea= 28000 K x 8.314 J / K / mol
½
Ea = 232.79 kJ / mol
OR
(i)
t 99 % = 2 t 90 % Given for reactant Q.
for first order reaction
t 99 % 
2.303
a
2.303
2.303
log

log 10 2  2 x
k
a  0.99a
k
k
½
t 90 % 
2.303
a
2.303
2.303
log

log 10 
k
a  0.90a
k
k
½
Therefore t 99 % = 2 t 90 % , therefore it is a first order reaction with respect to Q
½
From the graph it is evident that the concentration of R decreases linearly with time,
therefore the order with respect to R is zero.
½
The overall order of the reaction is 1.
Units for rate constant = s-1
Rate = k [Q]1 [R]0
½
½
For a first order reaction
(ii)
t 
[R ]0
2.303
log
k
[R ]
Time required for the 3 /4 th of the reaction
[R]0 = a, [R] = a 
t3 
4
t3 
4
3
1
a a
4
4
½
½
2.303
a
log
1
k
a
4
2.303
log 4
2.54 x 10  3
= 545 s
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½
½
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