Illustration 1: A 8.0 g sample contained Fe3O4, Fe2O3 and inert materials. It was treated with an excess of aqueous KI solution in acidic medium, which reduced all the ion to Fe+2 ions. The resulting solution was diluted to 50.0 cm3 and a 10.0 cm3 of it was taken. The liberated iodine in this solution required 7.2 cm3 of 1.0 M Na2S2O3 for reduction to iodide. The iodine from another 25.0 cm3 sample was extracted, after which the Fe+2 ions was titrated against 1.0 M KMnO4 in acidic medium. The volume of KMnO4 solution used was found to be 4.2 cm3. Calculate the mass percentages of Fe3O4 and of Fe2O3 in the original mixture.
Solution: This problem can be done by two methods.
In the first method, we break up Fe3O4 as an equimolar mixture of FeO and Fe2O3.
Fe3O4 is FeO. Fe2O3
Equivalents of Na2S2O3 = = 7.2 ´ 10-3
Equivalents of I2 in 10 cc = 7.2 ´ 10-3
Equivalents of I2 in 50 cc = 7.2 ´ 10-3 ´ 5 = 3.6 ´ 10-2
Since equivalents of I2 is equal to that of KI which in turn is equal to the total equivalents of Fe2O3 (Fe2O3 in the free state and Fe2O3 combined with FeO).
\ equivalents of total Fe2O3 = 3.6 ´ 10-2
Equivalents of KMnO4 solution = = 2.1 ´ 10-2
Since KMnO4 reacts with the total Fe2+ (Fe2+ in FeO and Fe2+ that was produced by the action of KI on Fe2O3)
\ equivalents of total Fe+2 in 50 mL = 2.1 ´ 10-2 ´ 2 = 4.2 ´ 10-2
Since equivalents of Fe2+ produced from Fe2O3 is equal to the equivalents of Fe2O3
\ Equivalents of FeO = 4.2 ´ 10-2 — 3.6 ´ 10-2
= 6 ´ 10-3
\ moles of FeO = 6 ´ 10-3
moles of Fe2O3 combined with FeO = 6 ´ 10-3
total moles of Fe2O3 =
(because when Fe2O3 ® Fe2+ ‘n’ factor is 2).
= 1.8 ´ 10-2
moles of Fe2O3 in the free state = 1.8 ´ 10–2 — 6 ´ 10–3
= 1.2 ´ 10-2
mass of Fe3O4 = 6 ´ 10-3 ´ 232 = 1.392 g
mass of Fe2O3 = 1.2 ´ 10-2 ´ 160 = 1.92 g
percentage Fe3O4 = 17.4%
percentage of Fe2O3 = 24%
Here we take Fe3O4 as a single entity.
Equivalents of Na2S2O3 = = 7.2 ´ 10-3
Equivalents of I2 in 50 cc = 7.2 ´ 10-3 ´ 5 = 3.6 ´ 10-2
\ Equivalents of Fe3O4 + Fe2O3 = 3.6 ´ 10-2
Let us assume that the moles of Fe3O4 is x and that of Fe2O3 is y. Since on reacting with KI both Fe3O4 and Fe2O3 give Fe2+ ‘n’ factor for both is two.
\ 2x + 2y = 3.6 ´ 10-2 ——– (1)
Equivalents of KMnO4 = = 2.1´10-2
Equivalents of Fe2+ in 50 mL = 2.1 ´ 10-2 ´ 2 = 4.2 ´ 10-2
Moles of Fe2+ in 50 mL = 4.2 ´ 10-2
Since the moles of Fe3O4 are x, moles of Fe2+ produced from Fe3O4 will be 3x and that produced from Fe2O3 will be 2y.
\ 3x + 2y = 4.2 ´ 10-2 ——— (2)
(2) – (1) gives x = 6 ´ 10-3 \ y = 1.2 ´ 10-2
Solving this, percentage of Fe3O4 is 17.4% and Fe2O3 is 23.75%.
Illustration 2: A sample of hard water contains 96 ppm of and 183 ppm of , with Ca+2 as the only cation. How many moles of CaO will be required to remove HC from 1000 kg of this water? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca+2 ions? (Assume CaCO3 to be completely insoluble in water). If the Ca+2 ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH? (one ppm means one part of the substance in one million part of water, mass/mass.) [IIT–JEE ‘97]
Solution: In 1000 kg of water the mass of HC = 183 g
moles of HC = = 3
moles of = = 1
\ total moles of Ca2+ in the solution = 1 + 1.5 = 2.5
Ca(HCO3)2 + CaO ® 2CaCO3 + H2O
moles of CaO required to be added to remove all HC = 1.5.
Now the Ca2+ in the solution will be only associated with . Therefore moles of Ca+2 left in the solution = 1.
ppm of Ca2+ = 1 ´ 40 = 40 ppm
moles of Ca2+ ions in 1 L of H2O = 10-3
moles of H+ ions that Ca+2 will exchange with = 2 ´ 10-3
\ pH = -log(2´10-3) = 2.7
Illustration 3: Only gases remain after 18g carbon is treated with 25 lit air at 17°C and 5 atm pressure. Determine the concentrations of NaHCO3 and Na2CO3 produced by the reaction of the mixture of gases with 0.75 lit of 1 M NaOH. (Assume that air contains, by volume, 80%, N2, 19%O2 and 1% CO2)
Solution: 25 L of air contains = 0.998 mol O2
and = 0.0525 moles of CO2
18gC = = 1.5 mol carbon
Let x = no. of moles of CO produced, and y = no. of moles of CO2 produced
x + y = 1.5 and 0.5x + y = 0.998
\ x = 1.004 and y = 0.496
The 0.496 mol CO2 produced from C and O2 plus the 0.0525 mol originally present make 0.548 mol CO2 total.
NaOH + CO2 ¾® NaHCO3
2NaOH + CO2 ¾® Na2CO3 + H2O
with -0.548 mol CO2 and 0.75 mol NaOH, let a = no. of moles of NaHCO3 and b = no. of moles of Na2CO3.
\ a + b = 0.548 and 2b + a = 0.75
\ a = 0.346 mol NaHCO3
or =0.46 M NaHCO3
b = 0.202 mol Na2CO3
or = 0.269 M Na2CO3
Illustration 4: Hydrogen peroxide solution (20 ml) reacts quantitatively with a solution of KMnO4 (20 ml) acidified with dilute H2SO4. The same volume of the KMnO4 solution is just decolourised by 10 ml of MnSO4 in neutral medium simultaneously forming a dark brown precipitate of hydrated MnO2. The brown precipitate is dissolved in 10 ml of 0.2M sodium oxalate under boiling condition in the presence of dilute H2SO4. Write the balanced equations involved in the reactions and calculate the molarity of H2O2.
Solution: Meq. of sodium oxalate = 10 ´ 0.2 ´ 2 = 4 meq. = meq. of MnO2 = Meq.
\Meq. of H2O2 = 4 meq.
(n – factor = 2)
Let the molarity of H2O2 = x (M)
then x ´ 20 = 2,
Balanced chemical equations are:
5H2O2+2MnO4–+ 6H+ ® 2Mn2++8H2O + 5O2
2MnO4–+3Mn2++2H2O ® 5MnO2+4H+
MnO2+C2O42– + 4H+® Mn2+ + 2H2O + 2CO2
Illustration 5: A container holds 3L of N2(g) and H2O (l) at 29°C. The pressure is found to be
1 atm. The water in the container is instantaneously split into hydrogen and oxygen by electrolysis, according to the reaction, H2O(l) ¾®H2(g) + . After the reaction is complete, the pressure is 1.86 atm. What mass of water was present in the container. The aqueous tension of water at 29°C is 0.04 atm.
Solution: This problem can be done in two ways. One is the right way and the other is wrong way. Let’s see the first method and guess whether it is right or wrong.
Total pressure in the beginning = = 1 atm
\ = (1–0.04) = 0.96 atm.
When water is split into H2 + O2, the total pressure is 1.86 atm.
\ = 1.86
0.96 + = 1.86
\ = (1.86 – 0.96) ´ = 0.6 atm.
= 0.0726 moles
\ Mass of water = 0.0726 ´18 = 1.306 g
This method looks perfectly OK. But there is one small error made. The error is so small that it might be difficult to point out. Find the error on your own and only after several trials you should look at correct solution that follows now.
The error that has happened is that the pressure of 1 atm is no doubt the pressure of N2 and H2O vapours but the electrolysis is done only to the H2O (liquid) and that too instantaneously. This implies that the H2O (vapour) that was in equilibrium with the liquid initially would be still left after the complete electrolysis of liquid water. \ 1.86 atm is the pressure of H2,O2,N2 and H2O (vapour).
= = 0.0694 moles
\= 0.0694 moles
\ mass of H2O (l) = 0.0694 ´ 18 = 1.25 g
Illustration 6: Assume that air contains 79% N2 and 21% O2 by volume. Calculate the density of moist air at 25°C and 1 atm pressure when relative humidity is 60%. The vapour pressure of water at 25°C is 23.78 mm of Hg. Relative humidity is given by percentage relative humidity
Solution: First of all it is important to understand the distinction between vapour pressure and partial pressure of vapour.
When a liquid is in equilibrium with its vapours, the vapours exert a pressure on the surface of the liquid called vapour pressure. This is the maximum possible pressure exerted by the vapours of the liquid at that tempreature. If we assume that vapours of a liquid exist in a container without the liquid, then the vapours behave just like any other gas, obeying all the gas laws. The pressure exerted by the vapours now can be anything which is less than or equal to that of vapour pressure. This is called partial pressure of vapour. If the partial pressure somehow becomes more than vapour pressure the vapours start liquefying to form liquid till vapours would be in equilibrium with liquid, exerting the vapour pressure.
To this problem let us look at what we are supposed to calculate. We need to find the density of moist air. From the ideal gas equation, PV =
Þ r =
Therefore, we need to find the molecular weight of moist air. To find the molecular weight of a mixture of gas, we need the molar composition or mole fraction of the gas mixture.
Partial pressure of water (vapour) = = 0.0187 atm.
\ mole fraction of water vapour = = 0.0187
Pressue of (N2+O2) = (1–0.0187) atm = 0.9813 atm
Let the pressure of N2 be 79 x, then pressure of O2 is 21 x ( they are in that molar ratio)
\ 79 x + 21 x = 0.9813 atm
x = 0.009813 atm
\ = 79 x = 0.7752 atm
\ = 0.2061 atm.
Mole fraction of N2 = = 0.7752
Mole fraction of O2 = = 0.2061
\Effective molecular weight of moist air
= (0.0187 ´18) + (0.7752 ´28) + (0.2061 ´32) = 28.63 g/mol
\ r = = 1.1716 g / L
Illustration 7: One way of writing the equation of state for a real gas is
Where B is a constant. Derive an approximate expression for B in terms of vander waal’s constant ‘a’ and ‘b’
or P =
Multiply by [V] then, PV =
or PV = RT
or PV = RT
\ PV =RT
or PV = RT
B = b –
Problem 8: Two containers A and B have the same volume. Container A contains 5 moles of O2 gas. Container B contains 3 moles of He and 2 moles of N2. Both the containers are separately kept in vacuum at the same temperature. Both the containers have very small orifices of the same area through which the gases leak out . Compare the rate of effusion of O2 with that of He gas mixture.
Solution: Since both the containers are in the same conditions of P,V and T,
As the mixture contains three moles of He and 2 moles of N2, the effective molecular weight of the mixture would be.
\= = 0.652
Though this solution looks OK, there is one big flaw in it. The error is that we have assumed that He and N2 from vessel B would effuse out with the same rate. This assumption was made in because we have taken the composition of the gas mixture coming out of the vessel to be same as that of the mixture that was inside the vessel . It should be duly noted that the two mixtures (inside and the are that effused out) have different compositions. Therefore first we must find the composition of the gas mixture coming out of the vessel B.
This means that initially the ratio of moles of N2 to the moles of He coming out of the vessel are in the molar ratio of 0.252 and not .
Let moles of the He coming out to be x
\ moles of N2 coming out is 0.252 x
\ = 0.2
= 0.8 Þ Mmix = 0.2 ´28 + 0.8´4 = 8.8
\ = 0.52
Illustration 9: 15 ml of an oxide of nitrogen was taken in an eudiometer tube an mixed with hydrogen till the volume was 42 ml. On sparking, the resulting mixture occupied 27 ml. To the mixture, 10 ml of oxygen was added and on explosion, again the volume fell to 19 ml. Find the formula of the oxide of nitrogen that was originally admitted. Both explosions led to the formation of water.
Solution: There are two ways of solving this problem
This is method that most commonly thought of by student
Let the oxide of nitrogen be NxOy. On mixing with H2 the volume was 42 ml.
\ the volume of H2 added was 27 ml
On sparking, the reaction would be
NxOy+yH2 ¾¾® yH2O+
Since on adding 10 ml of O2, combustion still takes place, it implies that hydrogen was in excess (since N2 will not react)
\ NxOy(g) + yH2(g) ¾¾® yH2O(I) +
Initially 15 ml 27 ml 0 0
After the reaction 0 (27-15y)
\ Total volume after the first explosion= (27-15y) +
\ \ 2=2y …(1)
Do not immediately write the formula as N2O, because this is a molecule and therefore N2O is its empirical formula and not molecular formula.
In the next explosion, hydrogen reacts with 10 ml of O2 after the volume fell of 10 ml. Since we do not know whether O2 or H2 is in excess, we consider all the possibilities.
Hydrogen is in excess
H2(g)+ ¾® H2O(I)
Initially (27-15y) 10
Finally (27-15y-20) 0
\ Final volume of gases = Volume of remaining hydrogen + nitrogen
This equation violates the first equation namely, (x=2y)
Therefore H2 cannot be in excess.
(ii) Hydrogen is exactly consumed by O2
\ 27-15y =10
We know that x and y have to be integers. Therefore hydrogen cannot completely consume oxygen.
(iii) oxygen is in excess
H2(g) + ¾® H2O(I)
Initially 27-15y 10
Finally 0 10-
\ Final volume
on solving: x=2 and y=1
\ Formula is N2O
In this method we repeat method 1 till we get x=2y.
Now, we know that H2 was left in excess after reacting with NxOy because we are told in the problem that on adding O2 again explosion occurs.
If H2 were to be in excess after reacting with NxOy, it automatically implies y=1. Because if y=1, than 15 ml of NxOy would have 15 ml O. Now 15 ml of O requires 15 ml of H2 to form water Since H2 is 27 ml, it immediately implies that H2 is excess. If y=2, then 15 ml of NxOy will have 30 ml of O and 30 ml of O would require 30 ml of H2 which is more than what have.
\ y cannot be equal or grater than 2.
\ y =1
\ Formula is N2O.
Illustration 10: The dissociation I2 2I utilizes one photon per iodine molecule dissociated. The maximum l for this is 4995 Å . Calculate number of moles of I2 dissociated per kJ of photon energy.
Solution: Q E = nhn where n = no. of photons
= 2.51`3 ´ 1023 no. of photons or no. of I2 molecules
= = 0.417 moles of I2
Illustration 11: Suppose 10-17 J of light energy is needed by the interior of the human eye to see an object. How many photons of green light (l = 550 nm) are needed to generate this minimum amount of energy?
Solution: Let the no. of photons needed to see the object = n
Q E = nhn =
\ n =
= = 28 (approx)
Illustration 12: A beam of electron accelerated with 4.64V is passed through a tube containing mercury vapours. As a result of absorption, electronic changes occured with mercury atoms and light was emitted. If the full energy of single electron was converted into light, what was the wave number of emitted light?
Solution: Q KE = mv2
From De-broglie’s Principles
l = …(1)
Putting the value of
h = 6.625 ´ 10–34 Js
me = 9.11 ´ 10–31 kg
K.E. = e.v.
Where charge e– = 1.9 ´ 10–19C
Putting all these values in equation (1) and on solving
l = Å
= Å = Å
= 5.69 ´ 10–10m
\ wave number = ´ 1010 meter–1
= 1.76 ´ 109m–1
Illustration 13: Light of wavelength 470 nm falls on the surface of potassium metal, electrons are emitted with a velocity 6.4 ´ 104 m/s.
- a) What is the kinetic energy of emitted electron?
- b) What is the minimum amount of energy required to remove an electron from K atom?
Solution: a) K.E. of the emitted electrons
= mv2, where m = mass of electron
= 9.11 ´ 10–31 kg
v = velocity of emitted electrons
= 6.4 ´ 104 m/s
\ K.E. = ´ 9.11 ´ 10–31kg ´ (6.4 ´ 104 m/s)2
= ´ 9.11 ´ 10–31 ´ 6.4 ´ 6.4 ´ 108J
= 186.57 ´ 10–23J = 1.86 ´ 10–21J
- b) \ hn = w0 + (K.E.)maximum …(1)
Where w0 = work function = minimum amount of energy required to remove electron.
From equation (1)
= w0 + (KE)maximum
Þ = w0 + 1.86 ´ 10–21J
Þ 422.8 ´ 10–21J = w0 + 1.86 ´ 10–21J
Þ w0 = 421.01 Joules
Illustration 14: 1.8g hydrogen atoms are excited to radiations. The study of spectra indicates that 27% of the atoms are in 3rd energy level and 15% of atoms in 2nd energy level and the rest in ground state. IP of H is 21.7 ´ 10-12 erg.
Calculate (a) Number of atoms present in III and II energy level. (b) Total energy evolved when all the atoms return to ground state.
Solution: a) Q 1.8gm hydrogen atoms = 1.8 ´ 6.023 ´ 1023 atoms of H
No. of atoms present in 3rd energy level
= 1.8 ´ 6.023 ´ 1023 ´
= 2.927 ´ 1023 atoms
Similarly no. of atoms present in 2nd energy level
= 1.8 ´ 6.023 ´ 1023 ´ 0.15 atoms of H
= 1.626 ´ 1023 atoms of H
- b) When all H-atom jumps from 3rd energy level to ground state, the amount of energy released,
E1 = 13.6 ´ ´ 2.927 ´ 1023 eV
= 35.384 ´ 1023 eV
Similarly when all H-atom present in orbit number 2nd, jumps to ground state, let the amount of energy released,
E2 = 13.6 ´ 1.626 ´ 1023 eV
= 16.585 ´ 1023 eV
Since energy is additive, so, total energy released
ET = E1 + E2
= 51.969 ´ 1023 eV = 83.15 ´ 104 J
Illustration 15: 5 ml of ethylacetate was added to a flask containing 100 ml of 0.1 N HCl placed in a thermostat maintained at 30°C. 5 ml of the reaction mixture was withdrawn at different intervals of time and after chilling, titrated against a standard alkali. The following data were obtained :
|Volume of alkali used in ml||9.62||12.10||1.10||14.75||21.05|
Show that hydrolysis of ethyl acetate is a first order reaction.
Solution: The hydrolysis of ethyl acetate will be a first order reaction if the above data confirm to the equation.
Where V0, Vt and V¥ represent the volumes of alkali used at the commencement of the reaction, after time t and at the end of the reaction respectively, Hence
V¥ – V0 = 21.05 – 9.62 = 11.43
Time V¥ – Vt k1
75 min 21.05 – 12.10 = 8.95
119 min 21.05 – 13.10 = 7.95
183 min 21.05 – 14.75 = 6.30
A constant value of k shows that hydrolysis of ethyl acetate is a
first order reaction
Illustration 16: An experiment requires minimum b-activity produced at the rate of 346 b-particles per minute. The half-life period of 42Mo99, which is a b-emitter is 66.6 hrs. Find the minimum amount of 42Mo99 required to carry out the experiment in 6.909 hrs.
Solution: The minimum b-activity required is 346 b-particles per minute. Since the experiment has to be completed in 6.909 hours. Moreover, we are asked to determine the minimum amount of required to carryout the experiment
Rate of emission of b-particles = 346 particles per minutes
Total b-particle required in 6.909 hours
= 346 ´ 6.909 ´ 60 = x = 143244 = x
Q lt = 2.303 log
Solving, a = 2.07 ´ 106 b-particles
= 3.43 ´ 10–18 moles = 3.48 ´ 10–18 ´ 99 gm
Illustration 17: The nucleidic ratio, to in a sample of water is 8 ´ 10-18 : 1. Tritium undergoes decay with a half – life period of 12.3 years. How many tritium atoms would 10 g of such a sample contain 40 years after the original sample is collected ?
Solution: a) Since = 2n
Where n = no. of half lives = = 4
\ = 24 = 16
\ Remaining no. of atoms of tritium, after 49 years
N = ´ N0 = ´ 8 ´ 10–18 = 5 ´ 10–19
But atom remains unchanged = 1
Hence ratio of to normal = 5 ´ 10–19 : 1
- b) 10 gm H2O = moles of H2O
= moles of H-atom
= ´ 6.023 ´ 1023 H-atom
= ´ 6.023 ´ 1023 ´ 8 ´ 10–18 Tritium atoms
N0 = 5.35 ´ 106 tritium atoms
N = N0 ´ 2–n = 5.35 ´ 106 ´ tritium atoms
N = 0.3344 ´ 106 tritium atoms
Illustration 18: 84Po218 (t1/2 = 3.05 min) decays to 82Pb214 (t1/2 = 2.68 min) by a-emission, while Pb214 is a b emitter. In an experiment starting with 1g atom of pure Po218 how much time would be required for the change in the no. of nuclei of 82Pb214 with time to reach a maximum.
Solution: From question
84Po218 82Pb214 83X214
Let 84Po218 changes into 82Pb214 with rate constant k1
and similarly 82Pb214 changes into 83X214 with rate constant k2
Therefore k1 = = = 0.227 min—1
k2 = = = 0.2585 min–1
For maximum conc. of 82Pb214,
(k2 – k1)t = ln = ln =0.1299
t = = 4.12 min
Illustration 19: trans-1,2- Dideuterocyclopropane (A) undergoes a first order decomposition. The observed rate constant at a certain temperature, measured in terms of the disappearance of A, was 1.52 ´ 10-4s-1. Analysis of the reaction products showed that the reaction followed two parallel paths, one leading to dideuteropropene (B) and the other to cis-1,2-dideuterocyclopropane (C). B was found to constitute 11.2% of the reaction product, independent of the extent of reaction. What is the order of the reaction for each of the paths, and what is the value of the rate constant for the formation of each of the products?
Solution: Since, from unit of rate constant, i.e. sec–1, it is clear that it is the 1st order reaction
k(overall) = 1.52 ´ 10–4s–1
Let B is forming with rate constant = k1
and C is forming with rate constant = k2
For parallel path, k = k1 + k2
= 11.2% Þ k1 = 11.2% ´ k
= 0.112 ´ 1.52 ´ 10–4s–1
= 0.17024 ´ 10–4s–1 = 1.7024 ´ 10–5s–1
Therefore = 88.8%
\ k2 = 88.8% ´ k = 0.888 ´ 1.52 ´ 10–4 s–1 = 1.349 ´10–4s–1
Illustration 20: A 32P radionuclide with half-life T = 14.3 days is produced in a reactor at a constant rate q = 2.7 ´ 109 nuclei per sec. How soon after the beginning of production of that radionuclide will its activity be equal to A = 1.0 ´ 109 dis/s ?
Solution: Since = q – A, where A = lN
\ = q – lN
Integrating both sides with proper upper limit and lower limit
= = [t – 0]
on solving, = lt
On solving, t = 8.24 ´ 105sec
= 9.54 days
Illustration 21: In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl5 were introduced and equilibriated at a temperature at 250oC. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, a of PCl5 and KP for the reaction
PCl5 PCl3 + Cl2 at this temperature.
Solution: PCl5(g) PCl3(g) + Cl2(g)
Initial moles 5 0 0
At equilibrium 5-x x x
Total moles = 5 + x + 4 (including moles of Argon)
= 9 + x
Since total moles = = 11.99 12
\ x = 3
\ a = = 0.6
KP = = 1.75 atm.
The degree of dissociation of PCl5 would have been 0.6 even in the absence of Argon. As one can see, the total pressure of gases constituting equilibrium is equal to 3.11 atm. The observed equilibrium pressure is 4.678 atm which means by the addition of 4 moles of Argon, the total pressure increases. This implies that addition of Argon has been done at constant volume which does not result in any change in degree of dissociation.
Illustration 22: 256 g of HI was heated in a sealed bulb at 444°C till the equilibrium was attained. The acid was found to be 22% dissociated at equilibrium. Calculate equilibrium constants for synthesis and dissociation of HI?
Solution: 2HI (g) H2 (g) + I2 (g)
Initial moles 0 0
Moles at equilibrium 2–2a a a (given a= 0.22)
Equilibrium constant for dissociation KC== = 0.0199
\ Equilibrium constant for synthesis = = 50.25
(H2 + I2 2HI )
[Note: The equilibrium reaction considered for dissociation is
2HI H2 + I2 and not HI ½ H2 + ½ I2 because for synthesis, the reaction is with 1 mole each of H2 and I2 i.e., H2 + I2 2HI]
Illustration 23: Kp for the reaction N2O4 (g) 2NO2 (g) is 0.66 at 46°C. Calculate the percent dissociation of N2O4 at 46°C and a total pressure of 0.5 atm. Also calculate the partial pressure of N2O4 and NO2 at equilibrium.
Solution: This problem can be solved by two methods.
Method (1): Let the number of moles of N2O4 initially be 1 and a is the degree of dissociation of N2O4.
Initial moles 1 0
Moles at equilibrium 1–a 2a
Total moles at equilibrium = 1–a + 2a = 1+a
\ Kp =
a = 0.5 i.e., 50% dissociation
Hence, partial pressure of N2O4 = 0.167 atm.
and partial pressure of NO2 = 0.333 atm.
Method (2): Let the partial pressure of NO2 at equilibrium be p atm, than the partial pressure of N2O4 at equilibrium will be (0.5–p)atm.
\ Kp =
p2 + 0.66 p–0.33 = 0
On solving p = 0.333 atm.
\ = 0.333 atm and = 0.167 atm.
Illustration 24: Kp for the reaction PCl5 PCl5 + Cl2 at 250°C is 0.82. Calculate the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.
Solution: Let 1 mole of PCl5 be taken initially. If ‘x’ moles of PCl5 dissociate at equilibrium, its degree of dissociation = x
|Total moles||1 – x + x + x = 1 + x|
P = 5 atm Kp = 0.82
x = 0.375 (0r 37.5%)
Now the new pressure P = 10 atm
Let y be the new degree of dissociation. As the temperature is same (250°C), the value of Kp will remain same. i.e. the same manner. Proceeding in the same manner.
Kp = Þ y = or y = 0.275 (or 27.5%)
Note: by increasing pressure, degree of dissociation has decreased, i.e., the system shifts to reverse direction. Compare the result by applying Le Chatelier’s principle.
Illustration 25: The equilibrium constant KP of the reaction
2SO2(g) + O2(g) 2SO3(g)
is 900 atm–1 at 800°C. A mixture containing SO3 and O2 having initial partial pressures of 1 atm and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800°C.
Solution: It can be seen that as SO2 is not present initially, so equilibrium cannot be established in the forward direction. Therefore it is established from reverse direction. Let n be the increase in partial pressure of O2. Then at equilibrium the partial pressures of SO2, O2 at SO3 are (0+2n), (2+n) and (1–2n) in atm respectively.
2SO2(g) + O2(g) 2SO3(g)
(0+2n) (2+n) (1–2n)
As n is small (because equilibrium constant for the reverse reaction is very small i.e., 1/900), it can be neglected in comparison to 2 and also
1–2n can be taken approximately to 1.
\ 900 =
Solving for n, we get n = 0.0118
Hence, = 2+n = 2.0118 atm
= 2n = 0.0236 atm
= 1–2n = 1–0.0236 = 0.9764 atm
Illustration 26: The pH of blood is maintained by proper balance of H2CO3 and NaHCO3 concentrations. What volume of 5 M NaHCO3 solution should be mixed with a 10 mL sample of blood which is 2 M in H2CO3 in order to maintain a pH of 7.4?
Ka(H2CO3) = 7.8 ´ 10–7.
Solution: Let x ml 5 M NaMCO3 be added
\ No. of millimole of NaHCO3 added = 5x
No. of millimole of H2CO3 taken = 10 ´ 2 = 20
Ka = 7.8 ´ 10–7
\ pKa = 6.10
For acidic buffer mixture
pH = pKa +
7.4 = 6.10 +
\ x = 80
So volume of 5 M NaHCO3 solution to be added is 80 mL
Illustration 27: A certain acid base indicator is red in acid solution and blue in basic solution. At pH = 5, 75% of the indicator is present in the solution in its blue form. Calculate Ka for the indicator and pH range over which the indicator changes from 90% red – 10% blue to 90% blue – 10% red.
Solution: HIn In– + H+
Ka = 10–5 ´ = 3 ´ 10–5
The above equation, after taking negative log of both sides may be put as
pH = pKa + log
pH for 90% red – 10% blue is as calculated below
pH = – log (3 ´ 10–5) + log
= 4.52 – 0.95 = 3.47
pH for 10% red – 90% blue is as calculated below
pH = 4.52 +
= 4.52 + 0.95 = 5.47
Thus the required pH range is 3.47 – 5.47
Illustration 28: The solubility products of BaSO4 and BaCrO4 at 250C are 1 ´ 10-10 and 2.4 ´10-10 respectively. Calculate the simultaneous solubility of BaSO4 and BaCrO4.
Solution: Let the solubilities of BaSO4 and BaCrO4 be x and y mol L–1 respectively
BaSO4(s) Ba2+ + SO42–
x + y x
BaCrO4(s) Ba2+ + CrO42–
x + y y
\ y = 2.4x
Putting the value of y
10 ´ 10–10 = (x + 2.4x)x = 3.4x2
x = = 592 ´ 10–6 mol L–1
Putting the value of x
y = 2.4 ´ 5.42 ´ 10–6 = 1.3 ´ 10–5 mol L–1
Illustration 29: 20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. What is the pH of the solution? Calculate the additional volume of 0.2 M NaOH required to make pH of the solution 4.74. Ka of CH3COOH = 1.8 ´ 10–5
Solution: Acetic acid taken = 5.0 ´ 0.2 = 10 millimole
NaOH added = 20 ´ 0.2 = 4 millimole
Salt formed = 4 millimole
Acid remaining free = 10 –4 = 6 millimole
Total volume of solution = 20 + 50 = 70 mL
pKa= –log(1.8 ´ 10–5) = 4.74
Resulting mixture is an acid buffer mixture for which
pH = pKa +
= 4.74 +
= 4.74 + = 4.57
To make the pH of solution equal to 4.74, we have to make [salt] = [acid]. To do so 5 mL 0.2 M NaOH will have to be added further.
Illustration 30: If at 25°C, the equivalence point of a titration of 50 mL of a solution of a weak monobasic acid occurs when 30 mL of 0.1 M NaOH has been added and the pH of solution is 6.0 after the addition of 18 mL of NaOH solution. Calculate Ka for the acid.
Solution: Since weak acid is monobasic and NaOH is also monoacidic so they must react in 1:1 molar proportion.
Amount of NaOH added for the end point = 30 ´ 0.1 = 3 millimole. So, amount of acid present in 50 mL solution = 3 millimole. During titration when 18 mL 0.1 i.e. 1.8 millimole NaOH is added, the salt formed will be 1.8 millimole and acid remaining free will be 3 – 1.8 i.e. 1.2 millimole. The mixture will be an acid buffer mixture for which
pH = pKa +
6.0 = pKa +
\ pKa = 5.94
- 10 gm sample of iron (II) oxalate containing inert impurity required 200 ml of 0.1 M K2Cr2O7 solution in presence of dil. HCl for complete oxidation. The resulting solution and the gas (if any evolved) is then treated with excess volume of 2.8 N NaOH and the Fe(OH)3 so ppted is removed by filtration. The filtrate after the removal of all Fe3+ion required 20 ml of 0.6 N HCl in presence of phenolphthalein indicator to reach at end point. What volume of NaOH was used and what is percentage purity of sample.
- 1.5 gm of an impure sample of (NH4)2SO4 was boiled with excess of caustic soda solution in a flask and the ammonia evolved was passed into 200 ml of semi-normal H2SO4 solution. The partially neutralized acid was made to 500 ml with distilled water 25 ml of this diluted acid required 40.8 ml of deci-normal caustic soda for complete neutralization. Calculate the percentage purity of (NH4)2 SO4.
- 4.2g of pure sodium bicarbonate was heated in air. The residue obtained was dissolved in water and the volume made upto 500 ml. 20 ml. of this solution required for complete neutralization 16 ml. of a solution of sulphuric acid, prepared by dissolving 4 ml of a sample of conc. acid of sp. gr 1.76 to one liter. Calculate the % strength of the acid in the sample.
- A 10 gm solid sample of 60% purity (NH4)2SO4 and NH4Cl along with some inert impurities was dissolved in water to prepare one litre solution. 25 ml of this solution was boiled with 25 ml of Ca(OH)2 until all the NH3 was evolved. The excess of Ca(OH)2 was neutralised by 24.3 ml of H2SO4. Calculate the percentage of each component in the sample
- 19.6 g of ferrous ammonium sulphate were dissolved and made upto 500 ml with acidified water. 25 ml of this solution required 20 ml and 27.5 ml of the solutions A and B of KMnO4 respectively. How many ml of A must be added to 1 litre of B to make 0.14N KMnO4 solution?
- Calculate the percentage of available Chlorine in the given bleaching powder, 5g of which were suspended in water equal to 500 ml. 20 ml of this suspension when treated with KI liberated Iodine which required 20 ml of decinormal Na2S2O3 solution.
- A polyvalent metal weighing 0.100 g and having atomic weight 51.0, reacted with dil. H2SO4 to give 43.9 ml of hydrogen at S.T.P. This solution containing the metal in the lower oxidation state, was found to required 58.8 ml of 0.1N permanganate for complete oxidation. What are the valencies of the metal ?
- A chemist is preparing to analyse sample that will contain not more than 0.5 g of uranium. His procedure calls for preparing uranium as U4+ ion and oxidising it by MnO4– in acid solution.
5U+4 + 2MnO4–‑ + 6H2O ¾® 5UO22+ + 2Mn2+ + 4H3O+
It he wants to react total U4+ sample with a maximum of 50 ml of KMnO4 solution what concentration should he choose.
- Hydroxyl amine reduces iron (III) according to the equation:
2NH2OH + 4Fe3+ ® N2O (g) + H2O + 4Fe2+ + 4H+.
Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is :
nO4– + 5Fe2+ + 8H+ ® Mn2+ + 5Fe3+ + 4H2O.
A 10 ml. sample of hydroxylammine solution was diluted to 1 litre. 50 ml. of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 ml. of 0.02 M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylammine in one litre of the original solution.
- 1.000 g of a moist sample of a mixture of potassium chloride and potassium chlorate was dissolved in water and made upto 250 ml. 25 ml of this solution was treated with SO2 to reduce the chlorate to chloride and excess SO2 was removed by boiling. The total chloride was precipitated as silver chloride. The weight of the ppt was 0.1435 g. In another experiment, 25 ml of the original solution was heated with 30 ml of 0.2N solution of ferrous sulphate and unreacted ferrous sulphate required 37.5 ml of 0.08N solution of an oxidizing agent for complete oxidation. Calculate the molar ratio of the chlorate to the chloride in the given mixture. Fe+2 reacts with ClO3– according to the equation :
ClO3– + 6Fe+2 + 6H+ ® Cl– + 6Fe+3 + 3H2O
- Iodic acid (HIO3) reacts with an aqueous solution of sulphur-dioxide according to the reaction.
2HIO3 + 5SO2 + 4H2O ® 5H2SO4 + I2
In an experiment 20 cc of a solution of iodic acid was allowed to react with an excess of an aqueous solution of sulphur-dioxide. The excess sulphur-dioxide and the iodine formed were removed by heating the solution. The H2SO4 present in the solution required 35.5 cc of 0.16N sodium hydroxide for neutralization. Calculate the amount of HIO3 present per litre of the solution.
- A 50 ml 1.92% (w/v) solution of a metal ion Mn+ (atomic weight = 60) was treated with 5.332g hydrazine hydrate (90% pure) and the mixture was saturated with CO2 gas when entire metal gets precipitated as a complex [M(NH2 – NHCOO)n]. The complex was filtered off and the filterate was titrated with KIO3 in the presence of conc. HCl according to following equation.
N2H4 + + 2H+ + Cl– ¾® ICl + 3H2O + N2
The volume of KIO3 solution needed for the end point to arrive at, was 480 ml. Find the value of n.
- Chile salt peter a source of NaNO3 also contains NaIO3. The NaIO3 can be used as a source of Iodine produced in the following reactions:
IO3– + 3HSO3– ¾® I– + 3H+ + 3SO42–
5I– + IO3– + 6H+ ¾® 3I2 + 3H2O
One litre of chile salt peter solution containing 5.80 gm NaIO3, is treated with stoichiometric quantity of NaHSO3. Now additional amount of same solution is added to the reaction mixture to bring about the second reaction. How many grams of NaHSO3 are required in step 1 and what additional volume of chile salt peter must be added in step II to bring in complete conversion of I– to I2?
- 1.249 g of a sample of pure BaCO3 and impure CaCO3 containing some CaO was treated with dil.HCl and it evolved 168 ml of CO2 at NTP. From this solution, BaCrO4 was precipitated, filtered and washed. The precipitate was dissolved in dilute sulphuric acid and diluted to 100 ml. 10 ml of this solution, when treated with KI solution, liberated iodine which required exactly 20 ml of 0.05N Na2S2O3. Calculate the percentage of CaO in the sample.
- CuSO4 reacts with KI in acidic medium to liberate I2
2CuSO4 + 4KI ¾® Cu2I2 + 2K2SO4 + I2
Mercuric periodate Hg5(IO6)2 reacts with a mixture of KI and HCl according to the following equation.
Hg5(IO6)2 + 4KI + 24HCl ¾® 5K2HgI4 + 8I2 + 24KCl + 12H2O
The liberated iodine is titrated against Na2S2O3 solution 1 ml of which is equivalent to 0.0499 gm of CuSO4.5H2O. What volume in mL of Na2S2O3 solution will be required to react with I2 liberated from 0.7245 gm of Hg5(IO6)2?
Molecular weight of Hg5(IO6)2 = 1448.5
Molecular weight of CuSO4.5H2O = 299.5
- A car tyre has a volume of 10 litres when inflated. The tyre is inflated to a pressure of 3 atm. at 17° C with air. Due to driving the temperature of the tyre increases to 47°C (a) What would be the pressure at this temperature ? (b) How many litres of air measured at 47°C and a pressure of 1 atm should be let out to restore the tyres 3 atm at 47° C.
- The total pressure of a mixture of H2 and O2 is 1.00 bar. The mixture is allowed to react to form water, which is completely removed to leave only pure H2 at a pressure of 0.35 bar. Assuming ideal gas behaviour and that all pressure measurements were made under the same temperature and volume conditions, calculate the composition of the original mixture.
- Two gases in adjoining vessels are brought into contact by opening a stop cock between them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other measured 0.1 lit and contained O2 at 600 torr and 220 K. The reaction to form N2O4 (solid) exhausts the limiting reactant completely.
- a) Neglecting the vapour pressure of N2O4 what is the pressure of the gas remaining at 220 K after completion of the reaction?
- b) What weight of N2O4 is formed?
- a) The compressibility factor for CO2 at 273K and 100atm pressure is 0.2005. Calculate the volume occupied by 0.2 mole of CO2 gas at 100 atm using (i) ideal gas (ii) real gas nature
- b) Calculate the pressure exerted by 1 mole of methane (CH4) in a 250 ml container at 300 K using van der Waal’s equation. What pressure will be predicted by ideal gas equation ?
a = 2.253L2 atm. mol-2,
b = 0.04281 lit mol-1
R = 0.0821 L atm mol-1K.
- A vessel contains 10 g of I2(s) and N2 at a pressure of 10 atm at 25°C. The volume of the vessel is one litre. If this vessel is connected to a 40 litre empty vessel and the temperature of the first vessel is increased to 250°C and of the second to 200°C. Calculate the final pressure in vessels.
- In a basal metabolism experiment timed at 6 minutes, a patient exhaled 52.5L of air measured over water at 20° C. Vapor pressure of water at 20° C is 17.5 torr. Barometric pressure was 760 torr. The inhaled and exhaled air contained on a dry basis 20.3% and 16.8% by volume of oxygen. Neglecting other factors, such as, gas solubility in water, volume difference of inhaled and exhaled air, calculate the rate of oxygen consumption in ml. (STP) per minute by the patient.
- 10 ml of ammonia were enclosed in an eudiometer and subjected to electric sparks. The sparks were continued till there was no further increase in volume. The volume after sparking measured 20 ml. Now 30 ml. of O2 were added and sparking was continued again. The new volume then measured 27.5 ml. All volumes were measured under identical conditions of temperature and pressure. V.D. of ammonia is 8.5. Calculate the molecular formula of ammonia. Nitrogen and Hydrogen are diatomic.
- An open flask contains air at 27°C Calculate the temperature to which the air should be heated so that
- a) One third of air measured at 27° C escapes.
- b) One third of air measured at the final temperature escapes.
- 15 ml of an oxide of nitrogen was taken in an eudiometer tube and mixed with hydrogen till the volume was 42 ml. On sparking, the resulting mixture occupied 27 ml. To the mixture, 10 ml. of oxygen was added and on explosion again the volume fell to 19 ml. Find the formula of the oxide of nitrogen that was originally admitted. Both explosions led to the formation of water only.
- At 250 ml O2 over water at 30°C and a total pressure of 740 torr is mixed with 300 ML of N2 over water at 250°C and a total pressure of 790. What will be the total pressure it the mixture is in 500 ML vessel over water at 350°C? given Aqueous ten sins of water at 250°C, 30°C and 350°C are 23.8, 31.8 and 42.2 toor respectively
- Two flasks of equal volume connected by a narrow tube (of negligible volume) are at 300 K and contain 0.7 mol of H2 at 0.5 atm. One flask is immersed into a bath kept at 400 K while the other remains at 300 K. Calculate the final pressure and the amount of H2 in each flask.
- If 250 mL of O2 over water at 30oC and a total pressure of 740 torr is mixed with 300 mL of N2 over water at 25oC and a total pressure of 780 torr, what will be the total pressure if the mixture is in a 500 mL vessel over water at 35oC? Given: Aqueous tensions of water at 25oC, 30oC and 35oC are 23.8, 31.8 and 42.2 torr, respectively.
- The vapour pressure of a substance of low volatility can be measured by passing an inert gas over a sample of the substance and analyzing the composition of the resulting gas mixture. When Nitrogen was passed over mercury at 23° C the mixture analyzed to 0.7 mg. of Hg., 50.25 mg. of N2, at a total pressure of 745 torr. Calculate the vapour pressure of mercury at this temperature.
- A mixture of NH3(g) and N2H4(g) is placed is sealed container at 300. The total pressure is 0.5 atm. The container heated t0 1200 K, at which time both substance decompose completely according to the equation
2NH3(g) ¾¾® N2(g) + 3H2(g)
N2H4(g) ¾¾® N2(g) + 2H2(g)
After decomposition is complete, the total pressure at 1200 K is found to be 4.5 atom. Find the amount percent of N2H4(g) is the original mixture
- The pressure in bulb dropped from 2000 to 1500 mm of Hg in 47 minutes, when the contained O2 leaked through a small hole, the bulb was then completely evacuated. A mixture of O2 and another gas of mol. wt 79 in the molar ratio 1:1 at a total pressure of 4000 mm of Hg was introduced. Find the mole ratio of the two gases remaining in the bulb after a period of 74 minutes.
- If the radius of the first Bohr orbit in a hydrogen atom can be determined with an uncertainty of 1% of its actual value, what will be the uncertainty in the velocity of electron? Compare the value with the velocity of electron in the first orbit. Given that
me = 9.1 ´ 10–31 kg and e0 = 8.854 ´ 10–12 C2N–1 m–2.
- The wavelength of Ka characteristic X-ray of iron and potassium are 1.931 ´ 10–8 and 3.737 ´ 10–8 cm respectively. What is the atomic number and name of the element of which characteristic Ka wavelength is 2.289 ´ 10–8 cm.
- An hydrogen like atom (atomic number = Z) is in a higher excited state of quantum number n. This excited atom make a transition to the first excited state by successively emitting by two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of n and Z.
- Light of wavelength 2000Å falls on an aluminium surface (work function of Al is 4.2eV). Calculate (a) the kinetic energy of the slowest and fastest emitted photo electrons. (b) stopping potential. (c) Cutt-off wavelength for Al.
- O2 undergoes photochemical dissociation into one normal oxygen atom and one oxygen atom, 1.967 eV more energetic than normal. The dissociation of O2 into two normal atoms of oxygen requires 498 kJ mole-1. What is the maximum wavelength effective for photochemical dissociation of O2?
- The dye acriflavine, when dissolved in water, has its maximum light absorption at 4530 Å and its maximum fluorescence emission at 5080°. The number of fluorescence quanta is, on the average, 53% of the number of quanta absorbed. Using the wavelengths of maximum absorption and emission, what percentage of absorbed energy is emitted as fluorescence?
- The ionisation energy of a H like Bohr atom is 4 Rydberg. (a) calculate the wavelength radiated when electron energy jumps from the first excited state to ground state. (b) What is the radius of I orbit of this atom? Given 1 Rh = 2.18 ´ 10-18 J.
- The minimum energy required to overcome the attractive forces between electron and the surface of Ag metal is 7.52 ´ 10-19 J. What will be the maximum kinetic energy of electron ejected out from Ag which is being exposed to UV light of l= 360 A°?
- Estimate the difference in the energy between 1st and 2nd Bohr orbit for a hydrogen atom. At what minimum atomic number a transition from n = 2 to n = 1 energy level would result in the emission of x-rays with l = 3.0 ´ 10-8 m? Which hydrogen like species does this atomic number correspond to?
- Find the quantum number ‘n’ corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths 108.5 and 30.4 nm.
- According to Mosely, frequency of X-ray and the charge present on the nucleus of atom of the element has the following relation
= a( Z – b)
Where n = frequency
Z = atomic number
a, b are constant
- a) If the straight line is at an angle of 45° with intercept 1, on axis, calculate the frequency n when atomic number Z = 50.
- b) What is the atomic number when n = 2500 sec–1.
- An electron in a H-atom in its ground state absorbs 1.50 times as much as energy as the minimum required for its escape (13.6eV) from the atom. Calculate the wavelength of the emitted electrons.
- To what series does the spectral lines of atomic hydrogen belong if its wave member is equal to the difference between the wave numbers of the following two lines of the Balmer series: 486.1 and 410.2 nm. What is the wavelength of that line?
- A stationary He+ ion emitted a photon corresponding to the first line (Ha) of the Lyman series. That photon liberated a photoelectron from a stationary H atom in ground state. What is the velocity of photo electron?
- The IP1 of H is 13.6 eV. It is exposed to electromagnetic waves of 1028 Å and gives out induced radiations. Find the wavelength of these induced radiations.
- The nuclei of two isotopes of same substance A-236 & A-234 are present in 4:1 ratio in an ore obtained from some other planet. The half-lives are 30 minute and 60 minute respectively. Both isotopes are a-emitter. Calculate after how much time their activity will be identical and also calculate the time required to bring the ratio of their atoms 1:1.
- Two reactions (i) A —® products and (ii) B —® products, follow first order kinetics. The rate of the reaction (i) is doubled when the temperature is raised from 300 K to 310 K. The half life for this reaction at 310 K is 30 min. At the same temperature B decomposes twice as fast as A. If the energy of activation for the reaction (ii) is half that of reaction (i), calculate the rate constant of the reaction (ii) at 300 K.
- At 25°C inversion of sucrose proceeds with constant half-life of 500 min at pH = 5 and with half life 50 minutes at pH = 4 for any concentration of sugar. Assuming the rate of inversion is given by
What are values of a and b ?
- The half – life for a first – order decomposition reaction at 650 K is 363 min and its energy of activation is 217.58 kJ mol-1.
- a) What fraction of the molecules at 650 K have sufficient energy to react ?
- b) Calculate the time required for the compound to be 75 % decomposed at 723 K.
- A mixture of 239Pu and 240Pu has a specific activity of 6 ´ 109 disintegration per sec (dps). The half-lives of the isotopes are 2.44 ´ 104 and 6.58 ´ 103y, respectively. Calculate the isotopic composition of this sample.
- To investigate the beta-decay of Mg-23 radionuclide, a counter was activated at the moment t = 0, it registers N1 beta particles by a moment t1= 2 sec and by a moment
t2 =3t1, the number of registered particles was 2.44 times greater. Find the mean time of the given nuclei.
- A sample of , as iodide ion was administered to a patient in a carrier consisting of 0.10 mg of stable iodide ion. After 4 days, 67.7% of the initial radioactivity was detected in the thyroid gland of the patient. What mass of the stable iodide ion had migrated to the thyroid gland ? [t1/2 of 131I = 8.0 days]
- A certain organic compound A decomposes by two parallel first order mechanisms
|If K1 : K2 = 1 : 9 and K1 = 1.3 ´ 10-5 s-1|
Calculate the concentration ratio of C to A if an experiment is started with only A and allowed to run for one hour.
- A small amount of solution containing Na24 radionuclide with activity A = 2 ´ 103 disintegrations per second was injected in the bloodstream of a man. The activity of 1 cm3 of blood sample taken t = 5 hour later turned out to be A¢ = 16 disintegrations per minute per cm3. The half-life of the radionuclide is T = 15 hours. Find the volume of the man’s blood.
- For the following sequential reaction
- a) Find out the time for maximum rate of formation of B. Given that K1 = 1.5 ´ 10–4sec–1 and K2 = 1.5 ´ 10–7 sec–1.
- b) Find out the concentration of C, when the initial concentration of A is 1.5 M and given that K1 = 1.2 ´ 10–5 sec–1 and K2 = 1.1 ´ 10–2 sec–1, t = 1 day
- A first order reaction, A ® B, requires activation energy of 70 kJ mol-1. When a 20% solution of A was kept at 25oC for 20 minutes, 25% decomposition took place. What will be the percent decomposition in the same time in a 30% solution maintained at 40oC? Assume that activation energy remains constant in this range of temperature.
- A solution contains a mixture of isotopes I1 (t1/2 = 14 days) and I2 (t1/2= 25 days). Total activity is 1 curie at t = 0. The activity reduces by 50% in 20 days, then find
- a) the initial activity I1 and I2
- b) The ratio of their initial no. of nuclei.
- From the following data for the reaction between A and B,
|[A], mol lit-1||[B], mol lit-1||Initial rate mol lit-1 s-1, at|
|300 K||320 K|
|2.5 ´ 10-4
5.0 ´ 10-4
1.0 ´ 10-3
|3.0 ´ 10-5
6.0 ´ 10-5
6.0 ´ 10-5
|5.0 ´ 10-4
4.0 ´ 10-3
1.6 ´ 10-2
|2.0 ´ 10-3
- i) the order of the reaction with respect to A and with respect to B,
- ii) the rate constant at 300 K,
iii) the energy of activation, and
- iv) the pre-exponential factor.
- In presence of an acid, N-chloroacetamide slowly changes into p-chloroacetamide. The progress of the reaction can be measured by titrating iodine liberated with Na2S2O3 solution. The results were obtained as follows:
Time (in hours) 1 2 6 ¥
Amount in p-chloroacetamide (in ml) 13 22.5 39.3 45
Find the fraction of N-chloroacetamide decomposed after 3 hours.
- What amount of heat is evolved by a curie of Rn (an a-emitter) in
- a) one hour
- b) its mean-life?
Given that kinetic energy of an a-particle is 5.5 MeV and l – 2 ´ 10–6 sec–1 for reaction.
- H2 and I2 are mixed at 400°C in a 1.0 L container and when equilibrium established, the following concentrations are present: [HI] = 0.49 M, [H2] = 0.08 M and [I2] = 0.06 M. If now an additional 0.3 mol of HI are added, what are the new equilibrium concentrations, when the new equilibrium H2 + I2 2HI is re-established?
- A mixture of SO3, SO2 and O2 gases is maintained in a 10 litre flask at a temperature at which the equilibrium constant (KC) for the reaction: 2SO2(g) + O2(g) 2SO3(g)
- i) If the number of moles of SO2 and SO3 in the flask are equal, how many moles of O2 are present?
- ii) If the number of moles of SO3 in the flask is twice the number of moles of SO2, how many moles of O2 are present?
- A saturated solution of iodine in water contains 0.33g of I2 in one litre. More than this can dissolve in a KI solution because of the following equilibrium. I2 + I– I–3. A 0.1 M KI solution actually dissolves 12.5g of iodine per litre, most of which is converted to I3–. Assuming that the concentration of I2 in all saturated solutions is the same, calculate the equilibrium constant for the above reaction. What is the effect of adding water to a clear saturated solution of I2 in the KI solution?
- When PCl5 is introduced into an evacuated vessel, at equilibrium at 250°C and at 2 atm pressure, the equilibrium mixture contains 40.7% by volume of Cl2. If the gas mixture is allowed to expand such that the equilibrium pressure is 0.2 atm, calculate the partial pressure of Cl2 and the % by volume of Cl2 in the mixture.
- At 1000 K, the pressure of CO2 in equilibrium with CaCO3 and CaO is equal t0 3.9´10-2 atm. The equilibrium constant for the reaction
C(s) + CO2 2CO(g)
is 1.9 at the same temperature when pressures are in atmospheres. Solid carbon CaO and CaCO3 are mixed and allowed to come to equilibrium at 1000 K in closed vessel. What is the pressure of Co at equilibrium?
- Solid NH4F on rapid heating in a closed vessel at 357°C develops a constant pressure of 275 MM of Hg owing to the partial decomposition of NHgI into NH3 and It I– but the pressure gradually increases further (When excess solid residually remains in the vessel) owing to the dissociation of HI. Calculate the find pressure developed under equilibrium.
NH4I(s) NH3(g) + HI(g)
2HI(g) H(g) + I2(g) Kc= 0.15 at 257°C
- 500 ml of 0.15 M AgNO3 are added to 500 ml of 1.09 M ferrous sulphate solution and the reaction is allowed to reach equilibrium at 25°C. 25 ml of the equilibrium mixture requires 30 ml of 0.0832 M KmnO4 for oxidation. Calculate K for the reaction :
Ag+ + Fe2+ Ag(s) + Fe3+
- N2O4 dissociates as N2O4 2NO2. At 550°C and one atmosphere, % decomposition of N2O4 is 50.3%. At what pressure and same temperature, the equilibrium mixture will have the ratio of N2O4 : NO2 as 1 : 8
- For the equilibrium : LiCl.3NH3(s) LiCl.NH3(s) + 2NH3(g) ; Kp = 9 atm2 at 40°C. A 5L vessel contains 0.1 mole of LiCl.NH3. How many moles of NH3 (minimum) should be added to the flask at this temperature to drive the backward reaction to completion ?
- Ammonium hydrogen dissociates as follows
NH4HS(s) H2S(g) + NH3(g)
Let solid NH4HS is placed in evacuated flask at certain temperature, it will dissociate until the total gas pressure is 500 torr. Calculate the following
- a) Equilibrium constant for the dissociation reaction
- b) It the additional NH3 is introduced into the equilibrium mixture without charge in temperature until Partial pressure of H2S under these condition
- i) What is the partial pressure of H2S under these conditions
- ii) What is the total pressure in the flask
- A mixture of N2 and H2 in the molar ratio of 1 : 3 was equilibrated at 400°C and at a pressure of 270 atm. The equilibrium mixture contained 40% of ammonia by volume. Calculate the pressure required at the same temperature to give an equilibrium mixture containing 10% of ammonia by volume, when the initial molar ratio of Nitrogen to Hydrogen was 3 : 1.
- In the preparation of quickline from limestome; CaCO3(s) CaO(s) + CO2(g)
The Kp values fitting an empirical equation valid in between 850 to 950°C.
log Kp = 7.282 –
Where T is absolute temperature. If the reaction is carried out in quiet air, what temperature would carried out in quiet air, what temperature used be predicted for complete decomposition of limestone.
- For the reaction at 298 K
A(g) + B(g) C(g) + D(g)
DH° = – 29.8 Kcal and DS° = – 0.1 Kcal K–1 what is the value of equilibrium constant?
- One mole of N2 and 3 moles of PCl5 are placed in a 100L vessel heated to 227°C. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour,
- a) Calculate the degree of dissociation of PCl5.
- b) Calculate Kp for the reaction : PCl5 (g) PCl3(g) + Cl2(g).
- When gaseous NO and NO2 are mixed, the following equilibria are readily established
- i) 2NO2 N2O4 (g) KP = 6.8 atm
- ii) NO + NO2 N2O3 .
In an experiment, when NO and NO2 were mixed in the ratio of 1 : 1, the final total pressure was 3.9 atm and the partial pressure of NO2 was 0.3 atm. Calculate the equilibrium constant for the second equilibrium.
- Calculate the [NH4+ ] ion concentration needed to prevent Mg (OH)2 from precipitating in a litre of solution which contains 0.01 mole NH3 and 0.001 mole of Mg2+.
Kb(NH3) = 1.8 ´ 10 -5, Ksp (Mg (OH)2) = 1.12 ´ 10-4.
- An aqueous solution of a metal bromide MBr2 (0.05M) is saturated with H2S. What is the minimum pH at which MS will precipitate.
KSP MS = 6´10-21; [H2S] = 0.1M
K1 = 10 -7 and K2 = 1.3 ´ 10-13 for H2S
- 25 mL of a weak base BOH was titrated with 0.5 M HCl. The pH of solution upon addition of 10 mL acid was 8.6 and that upon addition of 25 mL acid was 8. Calculate pH of solution when (i) 0 mL (ii) 20 mL and (iii) 50 mL of acid has been added.
- Calculate the [OH–] and [H3PO4] of a solution prepared by dissolving 0.1 mol of Na3PO4 in sufficient water to make 1L of solution.
K1 = 7.1 ´ 10-3
K2 = 6.3 ´ 10-8
K3 = 4.5 ´ 10 -13
- 12.5 mL solution of 0.5 M H3XO4is titrated with 0.5M NaOH. Calculate pH at the following stages.
- a) When 1st step neutralisation is complete
- b) When 2nd step neutralisation is complete.
- c) When 3rd step neutralisation is complete.
K1, K2 and K3 for H3XO4 are 1.0 ´ 10–4, 2.0 ´ 10–7 and 4.0 ´ 10–11 respectively.
- You are given a solution of 0.5M acetic acid. To what volume should 1L of the solution be diluted in order to (a) double the pH (b) double the OH– concentration ?
Ka (acetic acid ) = 1.8´10-5.
- What is the pH of the solution when 0.2 mole of HCl is added to 1L of a solution containing :
- i) 1M each of acetic acid and acetate ions.
- ii) 1M each of acetic acid and acetate ions.
Ka (acetic acid ) = 1.8 ´ 10-5.
- An acid-base indicator which is actually a weak monobasic acid exists 50% dissociated in an acid solution of pH = 4.60. The acid colour and basic colour of the indicator are quite distinct and say they are colour A and colour B respectively. Colour A predominates over colour B when concentration of acidic form of the indicator is at least 10 times that of the conjugate base of the indicator. However, colour B predominates over colour A only when concentration of basic form is about 20 times be a suitable one for indicating the end point of a titration carried out with a mixture of two weak acids (20 mL 0.1 M HA + 30 mL 1.5 M HB) vs. 0.5 M NaOH. Ka of HA and that of HB are 1 ´ 10–5 and 2 ´ 10–10 respectively. Also find pH range of the indicator and pH at the end point of the titration.
- When 20 mL solution of 4 M NaHCO3 solution is mixed with 100 mL sample of blood containing 12.4 g of H2CO3 the pH of the mixture was found to be 6.4 What volume of 5 M NaHCO3 solution should be mixed with a 10 mL of the same blood sample which is 1.0 M in H2CO3 in order to maintain a pH of 8.0.
- 50 mL of a weak monobasic acid was titrated with 0.2M NaOH and volume of alkali needed for the end point was 20 mL. The pH of solution upon addition of 12 mL alkali was 5.17. 100 mL solution of this acid at concentration of 0.2 M and also containing M3+ ion at a given concentration of 0.1 M should be mixed with what volume of 0.5 M NaOH so as to produce a buffer solution in which metal hydroxide may begin to precipitate?
- In pure formic acid following auto ionisation equilibrium exists
HCOOH(l) + HCOOH(l) HCOOH2+ + HCOO–
If the formic acid at 27°C is 0.004% auto ionised. Calculate rate constant of the intra molecular proton transfer of formic acid, when the rate constant of proton transfer between conjugate acid of formic acid and its conjugate base is 2 ´ 102 M2. (Density of formic acid = 1.22 g/mL).
- A weak base BOH was titrated against a strong acid. The pH at 1/4th equivalence point was 9.24. Enough strong base (6 m.eq) was now added to completely convert the salt. The total volume was 50 ml. Find the pH at this point.
- The solubility product of AgCl at 250C is 1.0´10-10. A solution of Ag+ ion at a concentration of 4´10-3 M just fails to yield a precipitate of AgCl with a concentration of 1´10-3 M of Cl– ion when the concentration of NH3 in the solution is 2´10-2 M. Calculate the magnitude of the equilibrium constant for the reaction Ag(NH3)2+ Ag+ + 2NH3
- The solubility product of Ag2C2O4 at 250C is 1.29´10-11. A solution of K2C2O4 containing 0.152 moles in 500 ml water is shaken at 250C with excess Ag2CO3 till the following equilibrium is reached :
Ag2CO3 + K2C2O4 Ag2C2O4 + K2CO3
At equilibrium, the solution contains 0.0358 moles of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3.
- The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298K. Given that solubility of SO2 in water at 298K is 1.3653 mol L–1 and the pKa of H2SO3 is 1.92, estimate the pH of rain on that day.
- 114.28ml, 57.6gm
- 80.96 %
- 87 %
- (NH4)2SO4 = 26.5%; NH4Cl = 33.5%
- 461.54 mL
- 35.5 %
- 2 & 5
- 39.6 g/mL
- 10 g/L
- n = 3
- 9.14gm, 20 mL
- 40 mL
- a) 3.31 atm b) 3.098
- x(H2) = 0.78, x(O2) = 0.22
- 0.30 atm, 0.402g
- a) 0.448 litre, 8.987 ´ 10–3 lit
- 0.4254 atm
- 0.279 liter per minute
- a) 177° C, b) 127° C
- 871 torr
- 0.57 atm, 0.4 mol at 300 K and 0.3 mol at 400 K
- 8.71 torr.
- 1.449 ´ 10-3 torr.
- O2 : gas :: 1:1.236
- 2.19 ´ 106ms–1
- Atomic Number = 24, Cr
- n = 6, Z = 3
- a) 2,0 eV; (b) 2V’; (c) 2970Å
- 1740.5 Å
- (a) 303.89 Å (b) 2.645 ´ 10-9 cm.
- 47.68 ´ 10-19 J
- 10.2 eV; Z = 2, He+ ion
- a) n = 2401 sec–1; (b) Z = 51
- Brackett series, l = 2.63 ´ 10-4 cm.
- 3.09 ´ 108 cm sec-1.
- I1 = 1029 Å; I2 = 1216 Å, I3 = 6568 Å.
- 180 minute and 120 minute
- 0.0327 min-1
- a = 1, b = 2
- a) 4 ´ 10-18; b) 12.5 min
- ; Pu240 – 61%
- tav= 15.9 hours
- Concentration ratio = 0.537
- 6 litre
- a) 25.59 days, (b) 1.325 M
- a) = 0.3669, = 0.6331
- b) = 0.3245
- i) 2, 1
- ii) 2.67 ´ 108 M-2 s-1
iii) 5.533 ´ 104 J mol-1
- iv) 1.145 ´ 1018 Ms-1.
- a) 11.45 J, (b) 1.03 ´ 104J
- HI = 0.724 M, H2 = 0.113 M, I2 = 0.093 M
- (i) 0.10 mole (ii) 0.40 mole
- 707, Reverse reaction is favoured by the addition of water.
- PCl3 = 0.097 atm % Cl2 = 48.66% by volume.
- [314.6 mm Hg]
- 0.19 atm
70 (A) Kp= 62500 torr-2 (B) (i) (ii) Ptotal=789 torr
- 203.33 atm
- K = 1
- (A) 0.33 (B) 0.205 atm-1
- 11, 9.18, 4.239
- [OH–] = 3.70 ´ 10-2 M`
[H3PO4] = 6.03 ´ 10–18
- 5.35, 8.55 and 11.5
- (a) 3.7´104L (b) 3.95 L
- (i) 4.568 (ii) 1
- Ph range = 3.6 – 5.9, pH = 11.75
- 20 ml
- 2.24 ´ 10–4