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Time: 2 hours
Note: Question number 1 to 10 carries 2 marks each and 11 to 20 carries 4 marks each.
- For the given reaction A + B → Products Following data were given Initial conc. (m/L). Initial conc. (m/L) Initial rate [mL-1^ s -1^ ] [A] [B] 0.1 0.1 0. 0.2 0.1 0. 0.1 0.2 0. a) Write the rate equation. b) Calculate the rate constant.
Sol. a) Let the order w.r.t A & B are x any y respectively r = K[A]x^ [B] y 0.05 = K[0.1]x^ [0.1] y 0.1 = K[0.2]x^ [0.1] y or 2 = [2] x
x = 0.05 = K[0.1] x [0.1] y
0.05 = K[0.1]x^ [0.2] y 1 = [2] y
y= b) rate equation = r = K[A] [B] 0
0.1 = K[0.2] K = 0.5 Sec
- 100 ml of a liquid contained in an isolated container at a pressure of 1 bar. The pressure is steeply increased to 100 bar. The volume of the liquid is decreased by 1 ml at this constant pressure. Find the ∆H & ∆U.
Sol. ∆H =0, ∆q (^) p = ∆U-W
W = PdV = 100×1 atmmL = 10 -2^ KJ = ∆U
- Draw the shape of XeF 4 and OSF (^) 4according to VSEPR theory. Show the lone pair of electrons on the central atom
Sol.
F
F F
F
Xe (square planar) (sp^3 d^2 )
F
O
F
S
F
F
(Trigonal bipyramidal) (sp^3 d)
- The structure of D-Glucose is as follows CHO
H OH
HO H
H OH
H OH
HO
a) Draw the structure of L – Glucose. b) Give the reaction of L – Glucose with Tollens reagent.
Sol.
CHO
H OH
HO H
H OH
HO H
HO
(L −glu cos e)
H OH
HO H
H OH
HO H
HO
O
O
Ag( NH )^3 2 +→
- a) Draw New mann`s projection for the less stable staggered form of butane. b) Relatively less stability of the staggered form is due to i) Torsional strain. ii) Vander Waal’s strain. iii) Combination of the above two.
Sol. a) CH 3
H H. H
.
CH 3 .
H
b) Less stability is due to Vander Waal’s strain
- Arrange the following oxides in the increasing order of Bronsted basicity. Cl O , BaO, SO , CO , B O 2 7 3 2 2 3
Sol. Cl O 2 7 < SO 3 < CO 2 < B O 2 3 <BaO
- AlF 3 is insoluble in anhydrous HF but when little KF is added to the compound it becomes soluble. On addition of BF 3 ,AlF 3 is precipitated. Write the balanced chemical equations.
Sol. 3KF + AlF 3 → K 3 AlF (^6)
K (^) 3AlF 6 + 3BF 3 → AlF 3 + 3KBF (^4)
- The crystal AB (rock salt structure) has molecular weight 6.023 y amu. where y is an arbitrary number in
amu.. If the minimum distance between cation & anion is 1/ 3 y nm and the observed density is 3 20 Kg / m. Find the a) density in 3 Kg / m and b) type of defect
∴ Ψ^2 = 0 = 0
3 2 r / a 0 0
1 1 1 r 2 e (^4 2) a a
^ −
^ −^ ×
π (^)
0
r 2 0 a
−^ =
0
r a
⇒ 2a 0 = r
b) λ =
34
3
h 6.626 10 mv (^100 10 )
−
−
×
× ×
λ = 6.626× 10 -35^ m = 6.626× 10 -25^ A° c) Yis (^) 84Po 206
- On the basis of ground state electronic configuration arrange the following molecules in increasing O-O bond length order. KO2, O2, O 2 [AsF6].
Sol.
2 1 2 2 2 2 2 y^ y 2 x (^2 ) z z
2p 2p O 1s , 1s , 2s , 2s , 2p 2p 2p
∗ ∗ ∗ ∗
π π = σ σ σ σ σ (^) π^ π
bond order =
2 *^ 2Py^2 2 2 2 2 y (^2) 2Pz 2 * (^) 2Pz 1 x
- 2P O 2 1s , (^) 1s , (^) 2s , (^) 2s, (^) 2P in [KO ] 2
π π π
− ^ π ^ = σ σ σ σ σ (^)
bond order =
2 2 2 2 2 2 y x (^) 2Pz^2
- 2P * 1 2 1s^ 1s^ 2s^ 2s^ 2P y
2 6
O , , , , 2p
in [O (AsF )]
π
- ^ π = σ σ σ σ σ (^) π
bond order
Bond length order is O 2 O 2 O 2
- a) In the following equilibrium
N (^) 2O (^) 4(g) 2NO 2 (g) when 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that 0 ∆G (N O ) (^) f 2 4 =100KJ 0 ∆G (NO ) (^) f 2 =50KJ i) Find ∆G of the reaction ii) The direction of the reaction in which the equilibrium shifts b) A graph is plotted for a real gas which follows Vander Waal’s equation with PVm taken on Y – axis & P on X – axis. Find the intercept of the line where Vm is molar volume
Sol. a) i) N (^) 2O (^) 4(g) 2NO 2 (g)
Reaction quotient = 2 2 4
2 NO
N O
P 100
10 atm P 10
∆G° reaction = 2 ∆G f ( NO 2 ) − ∆Gf ( N O 2 4 )
D D
0 = 100 – 100 ∆G = ∆G° + RT lnk ∴∆G =RT ln Q = 2.303 × .082 × 298 × log 9.9= 56.0304 Lit atm. = Positive
ii) Therefore reaction will shift towards backward direction.
b) (^2) m m
a P (v b) RT v
∴ (^) + (^) − =
2 2
aP PV P b RT PV P
(^) + − (^) = [PV) 2 P+aP 2 ][(PV) –b)] = P(PV) 2 RT ⇒ P[(PV) 2 + aP] (PV-bP)=P(PV) 2 RT Put P = 0 ⇒ (PV) 3 = (PV) 2 RT Intercept = RT
- a) 1.22 g C6H 5 COOH is added into two solvent and data of ∆Tb and Kb are given as:- i) In 100 g CH 3 COCH 3 ∆Tb= 0. K (^) b = 1.7 Kg Kelvin /mol ii) In 100 g benzene, ∆Tb = 0.13 and Kb = 2.6 Kg Kelvin/mol Find out the molecular weight of C6H (^) 5COOH in both the cases and interpret the result. b) 0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH at end point. Given Ka(HA)=5× 10
and α << 1
Sol. a) In first case
i) ∆Tb = Kb×m
0.17 = 1.7× 3
M × 100 × 10 −
⇒ M = 122
ii) In second case ∆Tb = Kb×m
0.13 = (^3)
M 100 10
× ′× −
×
M′ = 244
Benzoic acid dimerises in benzene
b) Since at end point molarity of salt =
M
∵ pH of salt of weak acid and strong base
pH =
( pKw pK a logc)
[14 + 5.3010 + [-1.3010] ⇒ pH = 9.
Convert
NO (^2)
to
NO (^2)
OH
in not more than four steps. Also mention the temp and reaction condition.
Sol. NO^2
Conc.H SO^2 4 →
NO (^2)
NO (^2)
→NH HS^4
NO (^2)
NH 2
NaNO 2 / HCl 0 C −5 C D D →
NO (^2)
N
NCl
NO (^2)
OH
H O 2 ∆ →
- Cl
2 5 2 5 3 2 6 5 3 2
KCN C H ONa / C H OH H O SOCl DMF (A)^ C H CHO / (B)^ (C)^ CH NH (D)
→ ∆ → ∆→ →
Identify A to D.
- Find the equilibrium constant for the reaction
Cu
Cu
Given that ECu (^) + 2 /Cu (^) + =0.15V D
EIn (^) + 2 / (^) In+ = −0.4V D
EIn (^) + 3 / (^) In+ = −0.42 V D
Sol. Cu +2^ +e-^ → Cu +^ ∆G 10 = - 0.15 F
In
- → In + + ∆G (^2) 0 = +0.4 F In
→ In
∆G (^3) 0 = - 0.84 F
Cu
Cu
∆G 0 = - 0.59 F
- nFE° = - 0.59F 0 − Ecell F = −0.59F 0 ECell =0.
Ecell = E° - (^) c
log K n
0.59 =
log Kc 1 K (^) c = 10 10
- An organic compound ‘P’ having the molecular formula C5H (^) 10O treated with dil H (^) 2SO 4 gives two compounds, Q & R both gives positive iodoform test. The reaction of C 5 H (^) 10O with dil H (^) 2SO 4 gives reaction 10^15 times faster then ethylene. Identify organic compound of Q & R. Give the reason for the extra stability of P.
Sol.
H 2 C O
CH 3
C H 5 10 O is CH 3
H 2 C O
CH 3
CH 3
→H+ H 3 C
C
O
CH 3
CH 3
Highly stable carbocation
H O^2 → H 3 C O
CH 3
+C H OH 2 5
P is stabilized by resonance
Note: FIITJEE solutions to IIT − JEE, 2004 Mains Papers created using memory retention of select FIITJEE students appeared in this test and hence may not exactly be the same as the original paper. However, every effort has been made to reproduce the original paper in the interest of the aspiring students.
