Practice Paper
TERM II (2021 – 2022)
Class – XI
Chemistry (043)
Time: 2 hours Maximum Marks: 35
General Instructions:
1. There are 12 questions in this question paper with internal choice.
2. SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
3. SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
4. SECTION C- Q. No. 12 is case based question carrying 5 marks.
5. All questions are compulsory.
6. Use of log tables and calculators is not allowed
SECTION – A
1. Consider the following acid base equilibrium:
H2SO4 (aq) + CH3COO– (aq) ________ CH3COOH (aq) + HSO4–(aq)
Write the two conjugate acid-base pair in the given equilibrium
Acid – H2SO4 & Conjugate Base – HSO4–
Base -CH3COO–& Conjugate Acid – CH3COOH
2. An alkali metal nitrate (A) when heated strongly produces a brown gas (B). Identify the compound A and B and write the complete reaction.
A – LiNO3 , B – NO2 ½+1/2
4LiNO3 ——–> 2Li2O +4 NO2 + O2
3. The solubility product of silver carbonate at 298 K is Ag2CO3 is 8 x 10-12. Calculate the solubility of the salt.
Ag2CO3 2Ag+ +CO32-
Ksp = [Ag+]2 [CO32-] = 8 x 10-12 1/2
If the solubility of salt be S
[Ag+] =2S; [CO32-] = S
(2S)2 (S) = 8×10-12 1/2
4S3 = 8 x10-12
S3 = 2 x 10-12 1
S = 1.26 x 10-4
SECTION – B
4. An alkene ‘A’ on ozonolysis gives two moles an aldehyde ‘B’ having molecular formula C2H4O.
(a) Write the IUPAC name of A and B
A- But-2-ene , B – Ethanal (molecular mass : 44)
(b) Draw the geometrical isomers of alkene A. Which isomer is non polar and why?
Trans but-2-ene is nonpolar. ½
.In trans-but-2-ene, the two methyl groups are in opposite directions,
Therefore, dipole moments of C-CH3 bonds cancel each other
OR
Rotation around C-C single bonds in an alkane leads to different spatial arrangements
Of atoms in space. Each structure thus obtained due to C-C single bond rotation in alkane is called conformer. Newman’s projection of structure of two extreme conformations of ethane is shown below
(a) Predict the total number of conformation of ethane molecule?
Infinite number of conformations are possible for ethane.
(b) Which conformation of ethane has lowest torsional strain?
Staggered conformation of ethane has lowest torsional strain.
(c) Is it possible to separate the extreme conformations of ethane? Give reason.
It is not possible to separate the extreme conformations of ethane. The energy difference between the two extreme conformation is so small that at even at ordinary temperatures, the thermal or kinetic energy gained by ethane molecule is sufficient enough to overcome this energy.
5. (a) Write van-der-waal Equation of state for n moles of gas.
(P + an2/V2) (V – nb) = n x R x T
(b) If the pressure correction factor in the van-der-waal equation for one mole is neglected then what will be value of compressibility factor? Mention the nature of deviation shown by the gas.
If pressure correction term is neglected, then van-der-waal gas equation for 1 mole of gas is reduced to,
P.(V-b) = RT PV – Pb = RT
PV/RT + Pb/RT = 1
Z +Pb/RT = 1
Z = 1-Pb/RT
Z < 1
The gas is exhibiting negative deviation from ideal behaviour
OR
(a) 3.1 g of a gas at 101oC occupies same volume as 0.187 g of Hydrogen gas at 37oC at same pressure. Calculate the molar mass of the gas.
Mass of gas (mg) = 3.1 g
T = (273+101) = 374 K
Molar Mass = Mg gram.mole-1
Volume of gas = Vg
PVg =ngRT
Vg = ( mgRT) /(PxMg )
Mass of hydrogen (m) = 0.187
T = (273+37) = 310 K
Molar Mass of hydrogen = 2 gram.mole-1
Volume of gas of H2 = V
V = ( mRT) / (PxM)
Both the gas occupies same volume V = Vg 1/2
(mgRT) / (P x Mg) = (mRT) / (MxP)
( 3.1 x 374 ) / Mg = ( 0.187 x 310) / 2 1/2
Mg = 40 gram.mole-1
(b) A plot of Pressure (P) vs Volume (V) for a gas at different values of temperature are shown in the figure below.
Which law is shown by the graph? Arrange the temperatures in increasing order.
6. Arrange the following in the increasing order of their property indicated:
(a) Propene, Benzene, Propene (pka values).
Propyne< Propene < Benzene
(b) Toluene, Chlorobenzene, Nitrobenzene, Phenol (reactivity towards an electrophile)
Nitrobenzene <Chlorobenzene<Toluene <Phenol
(c) n-Pentane, ethane, 2-methybutane, 2,2-dimethylpropane (boiling point)
ethane<2,2-dimethylpropane <2-methybutane <n-Pentane
7. At 197oC, equilibrium constant Kc for decomposition of PCl5 is 8 x10-3. If decomposition is depicted as, PCl5 (g) ——— PCl3 (g) + Cl2 (g)
(a) Calculate the value of Kc for the reverse reaction at same temperature.
PCl5 (g) ——– PCl3 (g) + Cl2 (g)
Kc = 8 x 10-3
For reverse reaction, equilibrium constant = Kc’
Kc’ = 1/Kc 1/2
Kc’ = 1/(8×10-3 )
Kc’ = 1251/2
(b) Calculate the value of Kp for the reaction at same temperature
Kp = Kc x (RT)∆n
∆n = (2-1) =1
T = (273 + 197)K = 470 K
Kp = 8 x 10-3 x (0.082 x 470)
Kp = 3.08 x 10-1
OR
(a) State La Chatelier’s Principle.
Change in any of the factor that determine the equilibrium conditions of a system to change in such a manner so as to reduce or counteract the effect of the change.
(b) At 500 K, the equilibrium constant Kc for the reaction
N2 (g) +3H2 (g) ——— 2NH3(g) is 0.061
The concentration of N2, H2 and NH3 at certain stage of reaction is 0.4 M, 0.3M and 0.06M respectively. In which direction the reaction will proceed.
Qc = [NH3]2 / [N2] x [H2]3
Qc = [0.06]2 / [0.4] x [0.3]3
Qc = 0.331/2
Since, Qc >Kc, The reaction will proceed in backward direction, i.e NH3 will decompose to produce N2 and H2.
8.(a) What is the structural difference between chlorides of boron and aluminium?
Chloride of boron (BCl3) is monomeric and Chloride of Aluminium (AlCl3) is dimeric.
(b) Common oxidation state of Group 14 element is +2 & +4. Which member of Group 14 elements forms stable compound in +2 oxidation state compared to +4 oxidation state. Give reason
Lead (Pb) forms stable compound in +2 oxidation state due to inert pair effect
OR
(a) Compare the B-F bond length in BF3 and BF4— ion.
B-F bond has partial double bond character in BF3 due to pπ-pπ back bonding between Boron & Fluorine. So B-F bond length in BF3 is less compared to BF4–
(b) Which member of Group 14 element has unique tendency to form p-pie (π) bond? Explain.
Carbon has unique tendency to form pie (π) bond because of its small size.
(c) Which allotrope of carbon is soluble in benzene?
Fullerene is soluble in benzene.
9. Account for the following:
(a) Halides of boron on hydrolysis produces [B(OH)4]– but halides of Al produces [Al(H2O)6]3+ ion
Due to unavailability of d orbital in boron, it cannot extend its covalency beyond 4
(b) CO is highly poisonous gas
CO form stable complex with Fe present in haemoglobin.
(c) Gallium has higher ionisation enthalpy than aluminium
Electron present in poor shielding 3d orbitals increase the effective nuclear charge in Galium.
8. Carry out the following conversion
(a) Ethyne to Acetophenone
(b) Benzene to parachlorotoluene
(c) Phenol to nitrobenzene
OR
(a) Complete the following sequences of following reaction
(i) P = Br-CH2-CH2-CH2-CH3
Q = CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
(ii)
(b) Distinguish between propene and propane by a chemical reaction.
Propene can decolourise red colour of bromine but propane cannot decolourise red colour of Bromine water
11. (a) Alkali metal dissolve in liquid ammonia and form deep blue solution.
What is the change in magnetic nature of the solution when sodium is dissolved in ammonia and allowed to stand for some time?
When sodium is dissolved in liquid ammonia the solution become paramagnetic and when allowed to stand for some time it become diamagnetic.
(b) Arrange the following hydroxides in decreasing order of alkalinity
NaOH, Be(OH)2, KOH, LiOH, CsOH
Be(OH)2 < LiOH < NaOH < KOH < CsOH
(c) Explain why Ba imparts colour in flame but Magnesium does not impart any colour.
As magnesium has high ionisation enthalpy, the energy of the flame is not sufficient to excite the electron.
SECTION – C
12. Read the passage given below and answer the questions that follow
Thermodynamics deals with energy changes in chemical or physical processes an enables to study these changes quantitatively to make useful predictions. The energy stored within a substance is called the internal energy of the system. The change in internal energy (∆U) of a system is the sum of all the energy inputs and outputs to and from the system. Work (W) and heat (Q) are the two modes of internal energyof the system. According to first law of thermodynamics the energy of isolated system is constant and mathematically it is expressed as ∆U = Q + W. Heat exchange at constant pressure for any process gives the measure of enthalpy change (∆H) of the system. Every chemical reaction is associated with change of material as well as energy. The enthalpy change associated with a chemical reaction at 1 bar pressure and 298.15 K is called standard reaction enthalpy (∆rHo). According to Hess’s law, for any chemical reaction the net heat exchange will be same whether the process occurs in one stage or in many stages. According to second law of thermodynamics, in a spontaneous chemical reaction the net entropy of both the system as well as surrounding will increase, i.e., ∆ Stotal> 0. Gibbs free energy is another thermodynamic quantity that helps in predicting the direction of spontaneity of a process. Gibbse free energy is defined as G = H – TS where H = U+PV. For any chemical process at constant pressure and temperature Gibbs free energy of the system decreases, i.e, ∆G < 0. U, H, S and G all are state function and extensive property where as q and w are path function.
(a) Consider the following sublimation process: I2 (solid) à I2 (vapour)
At which condition of temperature the above reaction will be spontaneous?
For the sublimation process: I2 (solid) —-> I2 (vapour)
∆H > 0 and ∆S system >0,
The process will be spontaneous when, ∆H> T∆S
T > (∆H / ∆S)
(b) Although work done by the system is a path function, but when work is carried out Under specific process then it is independent of the path. What is the process when work behaves as a state function?
According to first law of thermodynamics, ∆U = Q + W
For adiabatic process, Q = 0 1/2
∆U =W, thus work behaves as a state function
(c) State the relation between ∆U and ∆H for the given reaction at constant temperature and pressure
C2H6 (g) + 7/2 O2 (g) ————–> 2CO2 (g) + 3H2O (g).
∆H = ∆U + ∆nRT1/2
∆n = (3+2) – (1+3.5) = 0.5
∆H = ∆U + 0.5 RT
(d) 8 gram of oxygen is compressed isothermally and reversibly at a temperature of 27oC from 10L to 5L. Calculate q and w for the process. Oxygen gas is behaving like an ideal gas. Given log 2=0.3, R =2 cal K-1mole-1
Mass of oxygen = 8 g
No. of moles (n) = 8/32 = 0.25
T= (273+27)K = 300 K
V1 = 10 L & V2 = 5 L
W = – 2.303nRTlog(V2/V1)1/2
W = -2.303 x 0.25 x 2 x 300 log (5/10) 1/2
W = 103.635 cal1/2
∆U = Q + W
For isothermal process, ∆U =0
Q = -W = -103.635 cal
OR
(d) Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH (l) + 3 /2 O2 (g) → CO2 (g) + 2H2O(l); ΔrH 0 = –726 kJ mol–1
C (graphite) + O2 (g) → CO2 (g); ΔcH 0 = –393 kJ mol–1
H2 (g) + 1/ 2 O2 (g) → H2O(l); Δf H 0 = –286 kJ mol–1
The formation of CH3OH is represented by equation
C (graphite) + 2H2(g) +1/2 O2(g) ————> CH3OH(l)
CH3OH (l) + 3 /2 O2 (g) → CO2 (g) + 2H2O(l); ΔrH0 = –726 kJ mol–1
CO2 (g) + 2H2O(l) ———> CH3OH (l) + 3 /2 O2 (g) ΔrH0 = 726 kJ mol–1
H2 (g) + 1/ 2 O2 (g) → H2O(l); ΔfH0 = –286 kJ mol–1
2H2(g) + O2 (g) → 2H2O(l); ΔfH0= –572 kJ mol–1
The formation of CH3OH can be considered to be the sum of following reactions
CO2 (g) + 2H2O(l) ———> CH3OH (l) + 3 /2 O2 (g) ΔrH0= 726 kJ mol–1
C (graphite) + O2 (g) → CO2 (g); ΔcH0 = –393 kJ mol–1
2H2(g) + O2 (g) → 2H2O(l); Δf H0= –572 kJ mol–1
ΔfH0 = (726 -393 -572) kJ mol-1 = -239 kJmole-1