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Study exam IIT JEE 2004 chemistry questions
Typology: Exams
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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.
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
Sol. ∆H =0, ∆q (^) p = ∆U-W
W = PdV = 100×1 atmmL = 10 -2^ KJ = ∆U
Sol.
F
F F
F
Xe (square planar) (sp^3 d^2 )
F
O
F
S
F
F
(Trigonal bipyramidal) (sp^3 d)
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 +→
Sol. a) CH 3
H H. H
.
CH 3 .
H
b) Less stability is due to Vander Waal’s strain
Sol. Cl O 2 7 < SO 3 < CO 2 < B O 2 3 <BaO
Sol. 3KF + AlF 3 → K 3 AlF (^6)
K (^) 3AlF 6 + 3BF 3 → AlF 3 + 3KBF (^4)
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
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
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
π π π
− ^ π ^ = σ σ σ σ σ (^)
bond order =
2 2 2 2 2 2 y x (^) 2Pz^2
2 6
O , , , , 2p
in [O (AsF )]
π
bond order
Bond length order is O 2 O 2 O 2
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
10 atm P 10
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
and α << 1
Sol. a) In first case
i) ∆Tb = Kb×m
0.17 = 1.7× 3
ii) In second case ∆Tb = Kb×m
0.13 = (^3)
Benzoic acid dimerises in benzene
b) Since at end point molarity of salt =
∵ pH of salt of weak acid and strong base
pH =
[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 ∆ →
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.
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 (^3) 0 = - 0.84 F
Cu
Cu
∆G 0 = - 0.59 F
Ecell = E° - (^) c
log K n
0.59 =
log Kc 1 K (^) c = 10 10
Sol.
C H 5 10 O is CH 3
→H+ H 3 C
O
Highly stable carbocation
H O^2 → H 3 C O
+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.