NIIT University, Neemrana
Course Title: Chemistry Course Code: CHM 101
Discipline: BT, ECE, CS Academic Year: 2012-13
Program: B.Tech (1st year & IInd Year) Total Max. Marks: 40
Date: Dec 20th 2012
Time : 10:30-1:30
Answers of comprehensive exam
Q1. Select True or False for each statement.
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a. Half life of first order reaction is t1/2=0.692/K. (False)
b. HCl is diamagnetic molecule. (True)
c. F2 is Paramagnetic molecule.(False)
d. Inversion of sugar in acid is a zero order reaction.(False)
e. Gaseous decomposition of hydrogen iodide is a second order reaction.(False)
f.
The Thermal decomposition of nitrogen penta dioxide in the gaseous phase is a first order reaction.
(True)
g. The free energy of only chemical process can be converted into electrical energy in galvanic cell.
(True)
h. In galvanic cell, the cathode has a higher potential than the anode. (True)
i.
If the two concentrations are same, the system is at equilibrium and no net change can occur.
(True)
j.
Resonance has no role in molecular orbital theory. (True)
Q2. A. Calculate the bond order of NO and NO+ using Molecular orbital Theory and predict their magnetic
behavior.
03
NO+: (σ2s)2(σ2s*)2(π2p)4(σ2p)2, B.O. = 3, 0 unpaired e− (diamagnetic)
NO: (σ2s)2(σ2s*)2(π2p)4(σ2p)2(π2p*)1, B.O. = 2.5, 1 unpaired e− (paramagnetic)
Q2. B. Balance the given chemical equations
02
A. Fe2O3 (s) + H2 (g) –––> Fe (s) + H2O (l)
Fe2O3 (s) + 3 H2 (g) –––> 2 Fe (s) + 3 H2O (l)
B. Al (s) + O2 (g) –––> Al2O3 (s)
4 Al (s) + 3 O2 (g) –––> 2 Al2O3 (s)
C. Zn(NO3)2 (aq) + NaOH (aq) –––> NaNO3 (aq) + Zn(OH)2 (s)
Zn(NO3)2 (aq) + 2 NaOH (aq) –––> 2 NaNO3 (aq) + Zn(OH)2 (s)
D.
CCl4 (g) + O2 (g) –––> COCl2 (g) + Cl2 (g)
2 CCl4 (g) + O2 (g) –––> 2 COCl2 (g) + 2 Cl2 (g)
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Q3. A. Chloroform (CHCl3) reacts with chlorine (Cl2) to form carbon tetrachloride (CCl4) and hydrogen
chloride (HCl). In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. What is
the maximum yield of CCl4 in moles and in grams? 02
Solution:
Start with the equation:
CHCl3 + Cl2 -------> CCl4 + HCl
Given that there are 25g of chloroform & 25g of Cl 2 ,
from the chemical eq, it tells us that 1 mole of CHCl3 reacts with 1 mole of Cl2 gas
25/(12+1+35.3x3) = 0.2092 mole CHCl3 reacting with 25/35.5x2 = 0.3521 mole Cl2 , which means CHCl3
is a limiting reactant. Maximum yield of CCl4 in this case will be affected by the amount of chloroform
used, so ( 0.2092 mole x 12+35.5x4 = 32g)
Molar yield of CCl4= 0.2092
Yield in gm= 32g
Q3. B. For the given equation: 4FeS2 + 11O2 = 2Fe2O3 + 8SO2, if 300 g of FeS2 is burned in 200
g of O2, 143 g Fe2O3 results. Calculate the % yield for Fe2O3?
03
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Q4. A. Draw the Galvanic cell and name its specific parts.
02
(Correct way to draw Galvanic cell)
Q4. B. The e.m.f. of a cell measured by means of a hydrogen electrode against a standard calomel
electrode at 298K is 0.4188V. If the pressure of hydrogen gas was measured at 1 atm, calculate the pH of
the unknown solution, given the potential of the reference calomel electrode is 0.2415V.
03
-
Pt|H2 (1atm)|H(aH+)||Cl (aCl-)|Hg2Cl2 Hg
E=RT/F ln aH+
Eref= 0.2415V
E=(-2.303RT/F) (-log aH+)=-0.0591pH=EL
Emf of the cell is
E=ER-EL= Eref-Eleft= Eref-(-0.0591pH)
pH= E-Eleft/ 0.0591= 3.0
Q5. A. For the 1st order of the reaction, the rate constant is found to be 7.0x10-7 and 9x10-4 at 57oC.
Calculate the energy of activation of the reaction.
03
T1=273+7=280 & T2=273+57=330
Log k2/k1=Ea/2.303 (T2-T1/T1T2)
3.11=Eax2.83x 10-5
Ea=3.11/2.83 x 10-5
= 1.098x105 Jmol-1
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Q5. B. Write down the chemical reaction as an example for each:
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Heterogeneous catalysis,
Heterogeneous catalysis: A catalytic process in which the catalyst and the reactants are in different
phases is called heterogeneous catalysis. This process is also called contact or surface catalysis.
Homogeneous catalysis,
A catalytic process in which the catalyst is in the same phase as the reactant is called homogenous
catalysis.
Postive catalysis,
The catalyst increases the rate of a reaction.
Negative catalysis,
Alcohol, Acetanilide: Prevents oxidation of Na2SO3 by air
H3PO4: Prevents decomposition of H2O2
Autocatalysis,
In this type of catalysis, one of the products of the reaction catalyses the reaction. In the oxidation of
oxalic acid by KMnO4, Mn2+ ion formed is known to accelerate the reaction. So, when KMnO4 solution is
run into warm solution of oxalic acid (+ dil. H2SO4), initially there is a time lag before decolourisation
occurs; as more KMnO4 is added, the decolourisation becomes almost instantaneous.
Enzyme catalysis, http://en.wikipedia.org/wiki/Enzyme_catalysis
Induced catalysis,
When one reaction influences the rate of other reaction, which does not occur under ordinary conditions,
the phenomenon is known as induced catalysis.
Examples of induced catalysis:
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(1)Sodium arsenite solution is not oxidised by air. If, however, air is passed through a mixture of the
solution of sodium arsenite and sodium sulphite, both of them undergo simultaneous oxidation. The
oxidation
of
sodium
sulphite,
thus,
induces
the
oxidation
of
sodium
arsenite.
(2)The reduction of mercuric chloride (HgCI2) with oxalic acid is very slow, but potassium permanganate
is reduced readily with oxalic acid. If, however, oxalic acid is added to a mixture of KMnO4 and HgCI2,
both are reduced simultaneously. The reduction of potassium permanganate, thus, induces the reduction
of HgCI2.
Acid-base catalysis.
Q6.A. Write down the any six differences (on the basis of properties) between TRUE Solution and
COLLOIDAL solution.
03
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Q6.B. Classify the following into SUSPENTION, TRUE Solution and COLLOIDAL solution. (Milk, Sugar in
water, Clay in water, Blood, Boot polish, Face cream, Foam, Sand in water.)
02
Q7.A. Predict the aromatic character of the following organic compounds on the basis of Moleculare
Orbital Theory.
03
Ar=Aromatic, A.Ar=Anti aromatic, N.Ar= Non Arometic
Q7.B. Calculate the hybridization of PF5, SF6, AlCl3, CO2, and explain.
02
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Q8. A. Determine the heat of reaction for the reaction:
02
4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) = 2NO(g) ∆H = 180.6 kJ
N2(g) + 3H2(g)=2NH3(g) ∆H = -91.8 kJ
2H2(g) + O2(g) = 2H2O(g) ∆H = -483.7 kJ
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Q8. B. Find the pH of a solution formed by dissolving 0.100 mol of HC2H3O2 with a Ka of 1.8x10-8 and
0.200 mol of NaC2H3O2 in a total volume of 1 L.
02
Q8.C. Calculate the [H+] of a 0.300 M acetic acid solution.
01
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