DELHI PUBLIC SCHOOL, KALINGA
CALCULATION BASED QUESTIONS FOR PHYSICAL CHEMISTRY
Q1. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in
oxygen gives 3.38g carbon dioxide, 0.690 g of water and no other products. A volume of
10.0L (measured at STP) of this welding gas is found to weigh 11.6g. Calculate.
1. Empirical formula.
2. Molar mass of the gas.
3. Molecular formula.
Ans:
Amount of carbon in 3.38g CO =
× 3.38g = 0.9218g
Amount of hydrogen in 0.690g H O =
× 0.690g = 0.0767g
As compound only C and H,
Therefore, total mass of the compound,
= 0.9218 +0.0767g = 0.9985g
% of C in the compound =
× 100 = 92.32
% of H in the compound =
× 100 = 7.68
Calculation of Empirical Formula.
12
2
44
2
2
18
0.9218
0.9985
0.0767
0.9985
Elemen % by mas Atomic mas Moles of the elemen Simplest molar rat Simplest whole no. molar ra
t
s
s
t
io
tio
= 7.69
C
92.32
12
1
1
= 7.68
H
7.68
1
1
1
92.32
12
7.68
1
Empirical formula = CH
10.0L of the gas at STP weight = 11.6g
∴ 22.4L of the gas at S.T.P. will weight =
× 22.4 = 25.984g ≈ 26g
-1
∴ Moral mass = 26g mol
Emprirical formula mass of CH = 12 + 1 = 13
∴
11.6
10.0
∴ n =
Molecular mass
E.F. mass
∴ Molecular formula
= 2 × CH = C2 H2
Q2. Determine the molecular formula of an oxide of iron, in which the mass per cent of iron
and oxygen are 69.9 and 30.1, respectively.
Note: Given that the molar mass of the oxide is 159.69g mol-1
Ans:
% of iron by mass = 69.9% [Given]
% of oxygen by mass = 30.1% [Given]
Atomic mass of iron = 55.85amu.
Atomic mass of oxygen = 16.00amu.
Relative moles of iron in iron oxide = %mass of iron by mass/Atomic mass of iron = 69.9/55.85 = 1.25
Relative moles of oxygen in iron oxide = %mass of oxygen by mass/Atomic mass of oxygen = 30.01/16=1.88
Simplest molar ratio = 1.25/1.25 : 1.88/1.25
⇒ 1 : 1.5 = 2 : 3
∴ The empirical formula of the iron oxide is Fe2O3.
Mass of Fe2O3 = (2 × 55.85) + (3 × 16.00) = 159.7g mol-1 n = Molar mass/Empirical formula mass = 159.7/159.6 =
1(approx)
Thus, Molecular formula is same as Empirical Formula i.e. Fe2O3.
Q3. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20g are dissolved in
enough water to make a final volume up to 2L?
Ans:
Molarity (M) of a solution is given by,
1/15
Number of moles of solute
=
Volume of solution in Litres
Mass of suger/molar mass of sugar
=
2L
20g/[(12×12)+(1×22)+(11×16)]g
=
2L
20g/342g
=
2L
0.0585 mol
=
2L
= 0.02925 mol L
∴
–1
Molar concentration of sugar = 0.02925 mol L–1
Q4. Calculate the concentration of nitric acid in moles per litre in a sample which has a density,
1.41g mL–1 and the mass per cent of nitric acid in it being 69%.
Ans:
Mass percent of nitric acid in the sample = 69% [Given]
Thus, 100g of nitric acid contains 69g of nitric acid by mass.
Molar mass of nitric acid (HNO3).
= {1 + 14 + 3(16)}g mol–1
= 1 + 14 + 48
= 63g mol–1
∴ Number of moles in 69g of HNO3.
69g
=
63g mol
−1
= 1.095 mol
Volume of 100g of nitric acid solution,
Mass of solution
=
density of solution
100g
=
1.41g mL
−1
= 70.92mL = 70.92 × 10
−3
L
Concentration of nitric acid,
1.095mole
=
70.92×10
−3
L
= 15.44mol/L
∴
Concentration of nitric acid = 15.44mol/L.
Q5. Calculate the mass of sodium acetate (CH3COONa) required to make 500mL of 0.375
molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol-1.
Ans:
0.375 M aqueous solution means that 1000ml of the solution contain sodium acetate = 0.375 mole.
mole
∴ 500ml of the solution contain sodium acetate =
Molar mass of sodium acetate = 82.0245g mol-1
∴ Mass of sodium acetate required =
mole, × 82.0245g mol-1 = -15.380g.
0.375
2
0.375
2
Q6. In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
100 atoms of A + 100 molecules of B.
Ans:
A limiting reagent determines the extent of a reaction.
It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting
the amount of products formed.
According to the given reaction, 1 atom of A combines with 1 molecule of B.
Thus, all 100 atoms of A will combine with all 100 molecules of B.
Hence, the mixture is stoichiometric where no limiting reagent is present.
Q7. A sample of drinking water was found to be severely contaminated with
chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was
15ppm (by mass).
Determine the molality of chloroform in the water sample.
Ans:
2/15
100g of the sample contains 1.5 × 10–3g of CHCl3.
⇒ 1000g of the sample contains 1.5 × 10–2g of CHCl3.
∴ Molality of chloroform in water
=
1.5×10
−1
g
Molar mass of CHCl3
= 1.5 × 10
−3
%
Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5)
= 119.5g mol–1
–2
∴ Molality of chloroform in water = 0.0125 × 10 m
= 1.25 × 10–4m
Q8. Use the data given in the following table to calculate the molar mass of naturally occuring
argon isotopes:
Isotope
36
Ar
38
Ar
40
Ar
Isotopic molar mass
35.96755g mol–1
37.96272g mol–1
39.9624g mol–1
Abundance
0.337%
0.063%
99.600%
Ans:
Molar mass of argon,
= [(35.96755 ×
+ (39.9624 ×
0.337
100
90.60
100
) + (37.96272 ×
)]g mol
100
)
−1
= [0.121 + 0.024 + 39.802]g mol
= 39.947g mol
0.063
−1
−1
Q9. In a reaction
A + B2 → AB2
Identify the limiting reagent, if any, in the following reaction mixtures.
5mol A + 2.5mol B.
Ans:
A limiting reagent determines the extent of a reaction.
It is the reactant which is the first to get consumed during a reaction, thereby causing the reaction to stop and limiting
the amount of products formed.
1mol of atom A combines with 1mol of molecule B.
Thus, 2.5mol of B will combine with only 2.5mol of A.
As a result, 2.5mol of A will be left as such.
Hence, B is the limiting reagent.
Q10. Dinitrogen and dihydrogen react with each other to produce ammonia according to the
following chemical equation:
N2(g) + H2(g) → 2NH3(g)
Calculate the mass of ammonia produced if 2.00 × 103g dinitrogen reacts with 1.00 × 103g
of dihydrogen.
Ans:
Balancing the given chemical equation,
N2(g) + H2(g) → 2NH3(g)
From the equation, 1 mole (28g) of dinitrogen reacts with 3 mole (6g) of dihydrogen to give 2 mole (34g) of ammonia.
⇒ 2.00 × 103g of dinitrogen will react with
× 2.00 × 10 g dihydrogen i.e.,
6g
3
28g
2.00 × 103g of dinitrogen will react with 428.6g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 103g
Hence, N2 is the limiting reagent.
∴ 28g of N2 produces 34g of NH3.
Hence, mass of ammonia produced by 2000g of N2 =
× 2000g
34g
28g
= 2428.57g.
3/15
Q11. Calculate the number of atoms in the following:
52g of He.
Ans:
4g of He = 6.022 × 1023 atoms of He
atoms of He
∴ 52g of He =
= 7.8286 × 1024 atoms of He.
6.022×10
23
×52
4
Q12. What will be the mass of one 12C atom in g?
Ans:
1mol of 12C atom = 6.022 × 1023 atom = 12g
Thus, 6.022 × 1023 atoms of 12C have mass = 12g
12C will have mass =
∴ 1 atom of
12
6.022×10
g = 1.9927 × 10
−23
23
g
Q13. The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of
acetic acid in its 0.05M solution. Calculate the concentration of acetate ion in the solution
and its pH.
Ans:
Method 1
1. CH COOH ↔ CH COO + H K = 1.74 × 10
2. H O + H O ↔ H O + OH K = 1.0 × 10
Since Ka >> Kw:
−
3
+
3
+
2
2
−
−14
3
w
CH3 COOH
Ci
=
−5
a
+
H2 O
↔
CH3 COO
−
+
H3 O
+
0.05
0
0
0.05 − .05α
0.05α
0.05α
(.05α)(.05α)
Ka =
=
=
(.05−0.05α)
(.05α)(0.05α)
.05(1−α)
.05α
2
1−α
1.74 × 10
1.74 × 10
0.05α
2
D = b
2
−5
−5
= √
1−α
+ 1.74 × 10
−5
−5
α = 0.05α
α − 1.74 × 10
−5
−25
−5
)
2
− 4(.05)(1.74 × 10
+ .348 × 10
−5
)
−5
Ka
c
1.74×10
−5
.05
34.8×10
−5
×10
10
= √3.48 × 10
−6
= CH3 COOH ↔ CH3 COO
α = 1.86 × 10
[CH3 COO
=
2
− 4ac
= 3.02 × 10
α = √
2
− 1.74 × 10
= (1.74 × 10
α = √
0.05α
=
0.93×10
−
−
+ H
+
−3
] = 0.05 × 1.86 × 10
−3
−3
1000
= .000093
Method 2
Degree of dissociation,
α = √
Ka
c
c = 0.05M
Ka = 1.74 × 10–5
Then, α = √
1.74×10
−5
.05
α = √34.8 × 10
α = √3.48 × 10
−5
−4
4/15
α = 1.8610
−2
CH3 COOH ↔ CH3 COO
−
+ H
+
Thus, concentration of CH3COO– = c.α
= .05 × 1.86 × 10
= .093 × 10
−2
−2
= .00093M
Since[oAc
[H
+
−
] = [H
+
] = .00093 = .093 × 10
pH = − log[H
+
−2
.
]
= − log(.093 × 10
∴
],
−2
)
pH = 3.03
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.
Q14. The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion
in 0.05M solution of phenol? What will be its degree of ionization if the solution is also
0.01M in sodium phenolate?
Ans:
Ionization of phenol:
C6 H5 OH
Initial conc.
At equilibrium
Ka =
Ka =
[C6 H5 O
−
][H3 O
+
+
H2 O
↔
C 6 H5 O
−
+
H3 O
0.05
0
0
0.05 − x
x
x
+
]
[C6 H5 OH]
x×x
0.05−x
As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.
∴
x = √1 × 10
= √5 × 10
= 2.2 × 10
−10
× 0.05
−12
−6
M = [H3 O
+
]
Since [H O ] = [C H O ],
+
−
3
[C6 H5 O
−
6
5
] = 2.2 × 10
−6
M.
Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa.
C6 H5 ONa
Conc.
→
C 6 H5 O
−
+
+
Na
0.01
Also,
C6 H5 OH
Conc.
−
Ka =
H2 O
↔
0.05 − 0.05α
[C6 H5 O
[H3 O
+
−
C 6 H5 O
−
+
0.05α
H3 O
+
0.05α
] = 0.01 + 0.05α; 0.01M
] = 0.05α
[C6 H5 O
−
][H3 O
+
]
[C6 H5 OH]
(0.01)(0.05α)
Ka =
1.0 × 10
0.05
−10
α = 1 × 10
= .01α
−8
Q15. Reaction between N2 and O2– takes place as follows:
2N2 (g) + O (g) ⇌ 2N2 O(g)
2
If a mixture of 0.482mol N2 and 0.933mol of O2 is placed in a 10L reaction vessel and
allowed to form N2O at a temperature for which K = 2.0 × 10 , determine the
composition of equilibrium mixture.
−37
c
Ans:
Let the concentration of N2O at equilibrium be x.
The given reaction is:
2N2(g)
+
O2(g)
↔
2N2 O(g)
Initial Conc.
0.482 mol
0.933 mol
0
At equiluibrium
(0.482 − x)mol
(0.933 − x)mol
x mol
5/15
Therefore, at equilibrium, in the 10L vessel:
0.482−x
[N2 ] =
10
0.933−x/2
, [O2 ] =
, [N2 O] =
10
x
10
The value of equilibrium constant i.e., K = 2.0 × 10
is very small. Therefore, the amount of N2 and O2 reacted is
also very small. Thus, x can be neglected from the expressions of molar concentrations of N2 and O2.
Then,
−37
c
0.482
[N2 ] =
10
Now,
= 0.0482mol L
−1
and[O2 ] =
0.933
10
= 0.0933mol L
−1
2
[N2 O(g) ]
Kc =
2
[N2(g) ] [O2(g) ]
⇒ 2.0 × 10
⇒
x
(
=
2
= 2.0 × 10
100
⇒ x
−37
2
[N2 O] =
x
10
= 6.6 × 10
=
)
2
2
(0.0482) (0.0933)
−37
= 43.35 × 10
⇒ x = 6.6 × 10
x
10
× (0.0482)
2
× (0.0933)
−40
−20
6.6×10
−20
10
−21
Q16. Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
2NO (g) + Br
2
(g) ⇋ 2NOBr (g)
When 0.087mol of NO and 0.0437mol of Br2 are mixed in a closed container at constant
temperature, 0.0518mol of NOBr is obtained at equilibrium. Calculate equilibrium amount
of NO and Br2.
Ans:
The given reaction is:
2NO(g)
+
Br2(g)
2 mol
↔
2NOBr(g)
1 mol
2 mol
Now, 2mol of NOBr are formed from 2mol of NO. Therefore, 0.0518mol of NOBr are formed from 0.0518mol of NO.
Again, 2mol of NOBr are formed from 1mol of Br.
Therefore, 0.0518mol of NOBr are formed from
mol of Br, or 0.0259mol of NO.
The amount of NO and Br present initially is as follows:
[NO] = 0.087mol [Br2] = 0.0437mol
Therefore, the amount of NO present at equilibrium is:
[NO] = 0.087 – 0.0518
= 0.0352mol
And, the amount of Br present at equilibrium is:
[Br2] = 0.0437 – 0.0259
= 0.0178mol
0.0518
2
Q17. The equilibrium constant for the following reaction is 1.6 ×105 at 1024K
H (g) + Br (g) ⇌ 2HBr (g) Find the equilibrium pressure of all gases if 10.0 bar of HBr
is introduced into a
sealed container at 1024K.
2
Ans:
Given,
K for the reaction i.e., H
+ Br
↔ 2HBr
is 1.6 × 10 .
Therefore, for the reaction 2HBr ↔ H
+ Br
, the equilibrium constant will be,
5
p
2 (g)
2(g)
(g)
′
2 (g)
2(g)
1
Kp =
=
(g)
Kp
1
1.6×10
5
= 6.25 × 10
−6
Now, let p be the pressure of both H2 and Br2 at equilibrium.
2HBr(g)
Initial conc.
At equilibrium
↔
H2 (g)
+
Br 2(g)
10
0
0
10 − 2p
p
p
Now, we can write,
6/15
pHBr ×p2
p
′
= Kp
2
HBr
p×p
(10−2p)
p
= 6.25 × 10
2
= 2.5 × 10
10−2p
p = 2.5 × 10
−2
p + (5.0 × 10
(1005 × 10
−3
−6
−3
− (5.0 × 10
−3
)p = 2.5 × 10
)p = 2.5 × 10
p = 2.49 × 10
−3
−2
bar
)p
−2
−2
= 2.5 × 10
−2
(approximately)
Therefore, at equilibrium,
[H2 ] = [Br2 ] = 2.49 × 10
−2
bar
[HBr] = 10 − 2 × (2.49 × 10
−2)
bar
= 9.95 bar = 10 bar(approximately)
Q18. Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium
is represented as:
CH3 COOH (1) + C H5 OH (1) ⇌ CH3 COOC2 H5 (1) + H O (1)
2
2
At 293K, if one starts with 1.00mol of acetic acid and 0.18mol of ethanol, there is 0.171mol
of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
Ans:
Let the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in
excess.
The given reaction is:
CH3 COOH(l)
Initial conc.
At equilibrium
1
V
+ C2 H5 OH(l)
0.18
M
V
M
↔ CH3 COOC2 H5(l)
0
0
1−0.171
0.18−0.171
0.171
V
V
V
=
0.829
V
M
=
0.009
V
+ H2 O
M
0.171
V
M
M
Therefore, equilibrium constant for the given reaction is:
[CH3 COOC2 H5 ][H2 O]
Kc =
[CH3 COOH][C2 H5 OH]
0.171
=
V
0.829
V
×
×
= 3.92
0.171
V
0.009
= 3.919
V
(approximately)
Q19. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per
following endothermic reaction:
CH4 (g) + H O ⇌ CO (g) + 3H
2
2
(g)
1. Write as expression for Kp for the above reaction.
2. How will the values of Kp and composition of equilibrium mixture be affected by.
1. increasing the pressure
2. increasing the temperature
3. using a catalyst?
Ans:
1. K
pCO ×p
p
=
3
H
pCH ×pH
4
2
2
O
2.
1. By Le Chatelier’s principle, on increasing pressure, equilibrium will shift in the backward direction where decreases
number of moles.
2. As the given reaction is endothermic, by Le Chatelier ‘s principle, equilibrium will shift in the forward direction
with increasing temperature.
3. Equilibrium composition will not be disturbed by the presence of catalyst but equilibrium will be attained quickly.
Q20. At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium.
2SO2 (g) + O (g) ⇌ 2SO3 (g)
2
What is Kc at this temperature?
Ans:
For the given reaction,
7/15
Δn = 2 − 3 = −1
T = 450K
R = 0.0831 bar L bar K
Kp = 2.0 × 10
10
bar
−1
mol
−1
−1
We know that,
Kp = Kc (RT)Δn
⇒ 2.0 × 10
10
bar
−1
= Kc (0.0831 L bar K
2.0×10
⇒ Kc =
(0.0831 L bar
= (2.0 × 10
10
= 74.79 × 10
= 7.48 × 10
= 7.48 × 10
bar
10
11
11
−1
−1
K
M
−1
bar
mol
mol
−1
× 450K)
−1
−1
−1
×450K)
−1
)(0.0831 L bar K
L mol
L mol
−10
−1
−1
mol
−1
× 450K)
−1
−1
−1
Q21. The value of Kc for the reaction 3O (g) ⇌ 2O (g) is 2.0 ×10–50 at 25°C. If the
equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3?
2
Ans:
The given reaction is:3O
Then, K =
c
[O3(g) ]
[O2(g) ]
It is given that K
Then, we have,
2.0 × 10
−50
⇒ [O3(g) ]
⇒ [O3(g) ]
2
2
↔ 2O3(g)
2
3
C
= 2.0 × 10
[O3(g) ]
=
2(g)
3
[1.6×10
−50
and [O2(g) ] = 1.6 × 10
−2
.
2
−2 3
]
= 2.0 × 10
−50
= 8.192 × 10
⇒ [O3(g) ] = 2.86 × 10
× (1.6 × 10
−2
)
3
−56
−28
M
Hence, the concentration of O is 2.86 × 10
3
−28
M.
Q22. Calculate the pH of the resultant mixtures:
10mL of 0.2M Ca(OH)2 + 25mL of 0.1M HCl
Ans:
Moles of H3 O
Moles of OH
+
−
25×0.1
=
=
1000
Thus, excess of OH
[OH
−
] =
.0015
35×10
−3
= .0025mol
10×0.2×2
1000
−
= .0040mol
= .0015mol
mol/L = .0428
pOH = − log[OH]
= 1.36
pH = 14 − 1.36
= 12.63
(not matched)
Q23. Calculate the pH of the following solutions:
1mL of 13.6M HCl is diluted with water to give 1litre of solution.
Ans:
For 1mL of 13.6M HCl diluted with water to give 1L of solution:
13.6 × 1mL = M2 × 1000mL
(Before dilution) (After dilution)
13.6 × 10–3 = M2 × 1L
M2 = 1.36 × 10–2
[H+] = 1.36 × 10–2
pH = – log (1.36 × 10–2)
= (– 0.1335 + 2)
= 1.866 ∼ 1.87
Q24. Calculate the pH of the following solutions:
0.3g of NaOH dissolved in water to give 200mL of solution.
8/15
Ans:
For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
NaOH → Na
+
+ OH
(aq)
1000
[NaOH] = 0.3 ×
200
−
(aq)
= 0.5M
−
[OHaq ] = 1.5M
Then, [H
+
10
] =
= 6.66 × 10
−14
1.5
−13
pH = − log(6.66 × 10
−13
)
= 12.18
Q25. Neon gas is generally used in the sign boards. If it emits strongly at 616nm, calculate
1. The frequency of emission.
2. Distance traveled by this radiation in 30s.
3. Energy of quantum.
4. Number of quanta present if it produces 2J of energy.
Ans:
Wavelength of radiation emitted = 616 nm = 616 × 10–9m (Given)
1. Frequency of emission (v)
v =
c
λ
Where,
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of (v) :
8
v =
3.0×10 m/s
616×10
−9
8
m
= 4.87 × 10 × 109 × 10–3s–1
ν = 4.87 × 1014s–1
Frequency of emission (ν) = 4.87 × 1014s–1
2. Velocity of radiation, (c) = 3.0 × 108ms–1
Distance travelled by this radiation in 30s
= (3.0 × 108ms–1) (30s)
= 9.0 × 109m
3. Energy of quantum (E) = hν
(6.626 × 10–34Js) (4.87 × 1014s–1)
Energy of quantum (E) = 32.27 × 10–20J
4. Energy of one photon (quantum) = 32.27 × 10–20J
Therefore, 32.27 × 10–20J of energy is present in 1 quantum.
Number of quanta in 2J of energy
Q26. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can
be represeted as v = 3.29 × 10
15
(Hz)[
1
3
2
−
1
n
2
]
Calculate the value of n if the transition is observed at 1285nm. Find the region of the
spectrum.
Ans:
Wavelength of transition = 1285nm
= 1285 × 10–9m (Given)
v = 3.29 × 10
Since v =
8
=
(
1
3
2
−
1
n
2
)
(Given)
c
λ
3.0×10 ms
1285×10
15
−1
−9
m
v = 2.33 × 10
14
s
−1
Substituting the value of ν in the given expression,
3.29 × 10
1
9
1
9
−
1
n
2
15
=
(
1
9
−
2.33×10
3.29×10
− 0.7082 × 10
1
n
2
) = 2.33 × 10
14
14
15
−1
=
1
n
2
9/15
⇒
1
n
2
1
n
= 1.1 × 10
2
= 4.029 × 10
n = √
−1
− 0.7082 × 10
−1
−2
1
4.029×10
−2
n = 4.98
n ≈ 5
Hence, for the transition to be observed at 1285nm, n = 5.
The spectrum lies in the infra-red region.
Q27. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes
transition from an energy level with n = 4 to an energy level with n = 2?
Ans:
The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition
is given by the relation,
E = 2.18 × 10
−18
1
[
n
2
1
−
n
i
2
]
f
Substituting the values in the given expression of E:
E = 2.18 × 10
−18
1
[
4
= 2.18 × 10
= 2.18 × 10
−18
1−4
[
−18
16
2
1
−
2
2
]
]
× ( −
3
16
)
E = – (4.0875 × 10–19J)
The negative sign indicates the energy of emission.
Wavelength of light emitted (λ) =
hc
E
(since E =
hc
λ
)
Substituting the values in the given expression of λ :
λ =
(6.626×10
−34
8
)(3×10 )
4.0875×10
−19
λ = 4.8631 × 10
= 486.3 × 10
−9
−7
m
m
= 486nm
Q28. What is the maximum number of emission lines when the excited electron of a H atom in n
= 6 drops to the ground state?
Ans:
When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:
Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.
The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by
n(n−1)
2
Given,
n=2
Number of spectral lines =
6(6−1)
2
= 15
Q29. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106ms–1, calculate the de
Broglie wavelength associated with it.
Ans:
According to de Broglie’s equation,
λ =
h
mv
Where,
λ = wavelength associated with the electron
h = Planck’s constant
m = mass of electron
v = velocity of electron
Substituting the values in the expression of λ :
λ =
6.626×10
(9.10939×10
−31
−34
Js
6
kg)(2.19×10 ms
−1
)
10/15
= 3.32 × 10
= 332 × 10
−10
−12
m = 3.32 × 10
−10
100
m ×
100
m
λ = 332pm
∴
Wavelength associated with the electron = 332pm
Q30. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit?
Compare your answer with the ionization enthalpy of H atom (energy required to remove
the electron from n = 1 orbit).
Ans:
−21.8×10
En =
2
n Jatom
−19
−1
For ionization from 5th orbit, n = 5, n = ∞
1
∴ △E = E2 − E1 = −21.8 × 10
= 21.8 × 10
−19
1
× (
n1
= 21.8 × 10
−19
× (
1
5
= 8.72 × 10
−20
2
n2
−
2
1
−
1
∞
2
2
−19
× (
1
n2
2
−
1
n1
2
)
)
)
J
For ionization from 1st orbit, n = 1, n = ∞
1
∴ △E
△E
′
= 21.8 × 10
′
△E
=
21.8×10
8.72×10
−19
× (
1
1
2
−
2
1
∞
) = 21.8 × 10
−19
J
−19
−20
= 25
Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in
the ground state.
Q31. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple
of the de Broglie wavelength associated with the electron revolving around the orbit.
Ans:
Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is
given by:
mvr = n
h
2π
. . . (1)
Where,
n = 1, 2, 3, …
According to de Broglie’s equation:
λ =
h
mv
or mv = . . . (2)
Substituting the value of ‘mv’ from expression (2) in expression (1):
h
λ
hr
λ
= n
h
2π
or 2πr = nλ. . . (3)
Since 2πr represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the
Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron
revolving around the orbit.
′
′
Q32. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of
energy?
Ans:
Energy (E) of a photon = hν
Energy (En) of ‘n’ photons = nhν
⇒ n =
En λ
hc
Where,
–12m
λ = wavelength of light = 4000 pm = 4000 × 10
c = velocity of light in vacuum = 3 × 108m/s
h = Planck’s constant = 6.626 × 10–34Js
Substituting the values in the given expression of n:
n =
(1)×(4000×10
(6.626×10
−34
−12
)
8
= 2.012 × 10
16
)(3×10 )
Hence, the number of photons with a wavelength of 4000 pm and energy of 1J are 2.012 × 1016.
11/15
Q33. The mass of an electron is 9.1 × 10–31kg. If its K.E. is 3.0 × 10–25J, calculate its
wavelength.
Ans:
From de Broglie’s equation,
h
λ =
mv
Given,
Kinetic energy (K.E) of the electron = 3.0 × 10–25J
Since K.E = mv
1
2
2
∴ Velocity (v) = √
= √
2(3.0×10
−25
9.10939×10
2K.E
m
J)
−31
kg
= √6.5866 × 10
v = 811.579 ms
4
−1
Substituting the value in the expression of λ :
Q34. Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Ans:
Radius of Bohr’s nth orbit for hydrogen atom is given by,
rn = (0.0529nm) n2
For,
n=5
r5 = (0.0529nm) (5)2
r5 = 1.3225nm
Q35. Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 ×
10–10s.
Ans:
Frequency (v) =
1
Period
Wavelength (λ) =
c
1
=
9
2.0×10
−10
= 3.0 × 10
v
8
s = 5 × 10 s
ms
−1
5×10
Wave number (v) =
∼
1
λ
1
=
2
9
s
−1
−1
2
= 6.0 × 10 m
= 16.66 m
−1
6.0×10 m
Q36. The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18J atom–1.
What is the energy associated with the fifth orbit?
Ans:
Energy associated with the fifth orbit of hydrogen atom is calculated as:
E5 =
−(2.18×10
(5)
−18
)
2
=
−2.18×10
−18
25
E5 = –8.72 × 10–20J
Q37. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3 OH(I) +
3
2
O2 (g) −
−
→ CO2 (g) + 2H2 O(I); Δr H
C(graphite) + O2(g) −
−
→ CO2(g) ; Δc H
H2 (g) +
1
2
O2(g) −
−
→ H2 O(I); Δf H
⊖
⊖
⊖
= −726kJ mol
= −393kJ mol
= −286kJ mol
−1
−1
−1
.
Ans:
The reaction that takes place during the formation of CH3OH(l) can be written as:
1
Cs + 2H2 O + 2 O2(g) −
−
→ CH3 OH(I) . . . (1)
The reaction (1) can be obtained from the given reactions by following the algebraic calculations as:
Equation (ii) + 2 × equation (iii) – equation (i)
= (–393kJ mol–1) + 2(–286kJ mol–1) – (–726kJ mol–1)
= (–393 – 572 + 726)kJ mol–1
∴ Δf H
⊖
[CH3 OH(I) ] = −239kJ mol
−1
Q38. Calculate the enthalpy change for the process
CCI4 (g) −
→ C(g) + 4CI(g)
12/15
and calculate bond enthalpy of C – Cl in CCl4(g).
Δvap H
Δf H
⊖
Δa H
Δa H
⊖
(CCI4 ) = 30.5kJ mol
−1
(CCI4 ) = −135.5kJ mol
⊖
⊖
(C) = −715.0kJ mol
(CI2 ) = 242kJ mol
−1
.
−1
,
.
where Δ H is enthalpy of atomisation
⊖
a
−1
Ans:
The chemical equations implying to the given values of enthalpies are:
1. CCl −→ CCl Δ H = 30.5kJ mol
⊖
4(l)
2. C
3. Cl
4. C
−
−
→ C(g) Δa H
(s)
2(g)
(g)
−1
vap
4(g)
⊖
= 715.0kJ mol
−
−
→ 2Cl(g) Δa H
⊖
−1
= 242kJ mol
−1
+ 4Cl(g) −
−
→ CCl4(g) Δf H = −135.5kJ mol
−1
−
→ C
+ 4Cl
Enthalpy change for the given process CCI
, can be calculated
using the following algebraic calculations as:
Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)
4(g)
ΔH = Δa H
⊖
(C) + 2Δa H
= (715.0kJ mol
–1
⊖
(Cl2 ) − Δvap H
) + 2(242kJ mol
∴ ΔH = 1304kJ mol
–1
⊖
(g)
(g)
− Δf H
)– (30.5kJ mol
–1
)– (– 135.5kJ mol
–1
)
−1
Bond enthalpy of C–Cl bond in CCl4(g)
=
1304
4
kJ mol
= 326kJ mol
−1
−1
Q39. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at –10.0°C.
–1 at 0°C.
Δ
H = 6.03kJ mol
Cp[H2O(l)] = 75.3J mol–1 K–1
Cp[H2O(s)] = 36.8J mol–1 K–1
fus
Ans:
Total enthalpy change involved in the transformation is the sum of the following changes:
1. Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
2. Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
3. Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.
Total ΔH = Cp [H2 OCI]ΔT + ΔHfreezing + Cp [H2 O(s) ]
Q40. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and
9.7kJ mol–1 respectively. Find the value of ΔrH for the reaction:
N2 O4 (g) + 3CO(g) −
−
→ N2 O(g) + 3CO2 (g)
Ans:
Δr H
for a reaction is defined as the difference between Δ H value of products and Δ H value of reactants.
f
f
Δr H = ∑ Δf H(products)– ∑ Δf H(reactants)
For the given reaction,
N2 O(g) + 3CO(g) −
−
→ N2 O(g) + 3CO2(g)
Δr H = [{Δf H(N2 O) + 3Δf H(CO2 )}
− {Δf H(N2 O4 ) + 3Δf H(CO)}]
Substituting the values of Δ H for N2O, CO2, N2O4, and CO from the question, we get:
f
Δr H = [{81kJ mol
− {9.7kJ mol
−1
−1
+ 3(−393)kJ mol
+ 3(−110)kJ mol
Δr H = −777.7kJ mol
−1
−1
}
}]
−1
Hence, the value of ΔrH for the reaction is −777.7kJ mol
−1
.
Q41. For the reaction,
2A(g) + B(g) → 2D(g)
ΔU
⊖
= −10.5kJ and ΔS
⊖
= −44.1kJ
−1
.
Calculate ΔG for the reaction, and predict whether the reaction may occur spontaneously.
⊖
Ans:
13/15
For the given reaction,
2A(g) + B(g) → 2D(g)
Δng = 2 − (3)
= –1 mole
Substituting the value of ΔU in the expression of ΔH :
⊖
ΔH
⊖
= ΔU
⊖
+ Δng RT
= (–10.5kJ) – (–1) (8.314 × 10–3kJ K–1 mol–1) (298K)
= –10.5kJ – 2.48kJ
ΔG
⊖
= +0.16kJ
Since ΔG for the reaction is positive, the reaction will not occur spontaneously.
⊖
Q42. Calculate the number of kJ of heat necessary to raise the temperature of 60.0g of
aluminium from 35°C to 55°C. Molar heat capacity of Al is 24J mol–1 K–1 .
Ans:
Mass of Al = 60g
Rise in temperature,ΔT = 55– 35 = 20°C
Molar heat capacity of Al = 24J mol K
Specific heat capacity of Al = Jg K
∴ Energy required m × c × ΔT
= 60 ×
× 20 =
= 1066.67J
−1
24
−1
−1
−1
27
24
28800
27
27
= 1.068kJ or 1.07kJ
Q43. In a process, 701J of heat is absorbed by a system and 394J of work is done by the system.
What is the change in internal energy for the process?
Ans:
According to the first law of thermodynamics,
ΔU = q + W(i)
Where,
change in internal energy for a process
q = heat
W = work
Given,
q = +701J (Since heat is absorbed)
W = –394J (Since work is done by the system)
Substituting the values in expression (i), we get
ΔU =
ΔU = 701J + (– 394J)
ΔU = 307J
Hence, the change in internal energy for the given process is 307J.
Q44. Enthalpy of combustion of carbon to CO2 is –393.5kJ mol–1. Calculate the heat released
upon formation of 35.2g of CO2 from carbon and dioxygen gas.
Ans:
Formation of CO2 from carbon and dioxygen gas can be represented as:
C(s) + O2(g) −
−
→ CO2(g) Δf H = −393.5kJ mol
−1
Heat released on formation of 44g CO2 = –393.5kJ mol–1
Heat released on formation of 35.2g CO2
=
−393.5kJ mol
44g
−1
× 35.2g
=– 314.8kJ mol
–1
Q45. For an isolated system, ΔU = 0, what will be ΔS?
Ans:
Change in internal energy (ΔU) for an isolated system is zero for it does not exchange any energy with the
surroundings. But entropy tends to increase in case of spontaneous reaction. Therefore, ΔS > 0 or positive.
Q46. The equilibrium constant for a reaction is 10. What will be the value of ΔG ? R =
8.314JK–1 mol–1, T = 300K.
⊖
14/15
Ans:
ΔG
⊖
= −RT ln k = −2.303RT log K.
R = 8.314Jk
ΔG
⊖
−1
mol
−1
; T = 300K; K = 10
= −2.303 × 8.314Jk
= −5527J mol
−1
−1
mol
−1
= −5.527kJ mol
× (300k) × log 10
−1
.
Q47. Given
N2 (g) + 3H2 (g) −
−
→ 2NH3 (g); Δr H
⊖
=– 92.4kJ mol
–1
What is the standard enthalpy of formation of NH3 gas?
Ans:
Standard enthalpy of formation of a compound is the change in enthalpy that takes place during the formation of 1
mole of a substance in its standard form from its constituent elements in their standard state.
Re-writing the given equation for 1 mole of NH3(g),
1
2
N2(g) +
∴
=
=
3
2
H2(g) −
−
→ NH3(g)
Standard enthalpy of formation of NH3(g)
1
2
1
2
Δr H
⊖
(−92.4kJ) mol
= −46.2kJ mol
−1
−1
Q48. The reaction of cyanamide, NH2CN(s), with dioxygen was carried out in a bomb
calorimeter, and ΔU was found to be –742.7kJ mol–1 at 298K. Calculate enthalpy change
for the reaction at 298K.
NH2 CN(g) +
3
2
O2 (g) −
−
→ N2 (g) + CO2 (g) + H2 O(l)
Ans:
Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + Δng RT
Where,
change in internal energy
Δn = change in number of moles
For the given reaction,
ΔU =
g
Δng = ∑ ng (products)– ∑ ng (reactants)
= (2– 1.5) moles
Δng = 0.5 moles
And,
ΔU =
– 742.7kJ mol
–1
T = 298K
R = 8.314 × 10
–3
kJ mol
–1
K
–1
Substituting the values in the expression of ΔH :
ΔH = (– 742.7kJ mol
(8.314 × 10
–3
kJ mol
–1
–1
) + (0.5 mol)(298K)
K
–1
)
=– 742.7 + 1.2
ΔH =– 741.5kJ mol
–1
15/15