Question Set C

cbse 12th Standard Chemistry Electrochemistry Model Questions Papers Study Material - QB365

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Questions:

Generic 3

6 x 3 = 18

The decomposition of phosphine,
4PH₃(g) ⟶ P₄(g)+6H₂(g)
has the rate law, rate = k [PH₃]
The rate constant is 6.0 x 10^-4 s^-1 at 300 K and activation energy is 3.05 x10⁵ J mol^-1 Calculate the value of rate constant at 310 K
(Given, R = 8.314 JK^-1mol^-1).

The boiling point elevation of 0.30 g acetic acid in 100 g benzene is 0.0633 K. Calculate the molar mass of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?
(Given, K_b for benzene = 2.53 K kg mol^-1).

Using valence bond theory explain the geometry and magnetic behaviour by [Cr(NH₃)₆]³⁺ (At.no.Cr = 24)
Write the IUPAC name of ionization isomer of [Ni(NH₃)₆]⁶⁺

Determine the values of equilibrium constant (K) and \(\triangle G ^{o}\) for the following reaction :
Ni (s) + 2 Ag+ (aq) \(\longrightarrow\) Ni²⁺(aq) + 2 Ag (s), E⁰= 1.05V (1F = 96500 C mol^-1)

Why blue colour of copper sulphate solution gets discharged when zine rod is dipped in it? Given
\({ E }^{ 0 }_{ { Cu }^{ 2+ }/Cu }=-0.34V,{ E }^{ 0 }_{ { Zn }^{ 2+ }/Zn }=-0.76V\)

Give reasons for each of the following:
(i) Transition metal fluorides are ionic in nature, whereas bromides and chlorides are usually covalent in nature.
(ii) Size of trivalent lanthanoid cations decreases with increase in the atomic number.
(iii) Chemistry of all the lanthanoids is quite similar.

Generic 2

6 x 2 = 12

The reaction 2NO₂ + F₂ \(\rightarrow\) 2NO₂ F involves the following steps:
\(\mathrm{NO}_{2}+\mathrm{NO}_{2} \rightleftharpoons{ }^{K}{ } \mathrm{~N}_{2} \mathrm{O}_{4}(\text { Fast })\)
N₂O₄ + F₂ \(\rightarrow\) 2 NO₂F (Slow)
Write the rate law. Calculate the overall order of the reaction and what is the rate determining step?

Write the formula of compound in which transition metal is in +6 oxidation state.

The conductivity of a 0.20 M solution of KCl at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.

10.

Account fpr different magnetic behaviour of hexacyanoferrate (III) and hexafluoroferrate (III).

11.

Though copper, silver and gold have completely filled sets of d-orbitals yet they are considerde as transition metals. Why ?

12.

Which solution has higher concentration, 1 molar or 1 molal solution of the same solute ? Give reason.

Multiple Choice Question

15 x 1 = 15

13.

For a binary ideal liquid solution, the variation in total vapour pressure versus composition of solution is given by which of the curves?

(a)

(b)

(c)

(d)

14.

Consider the following plot between In k and 1/T,

In this plot, the intercept and slope respectively are

(a)

\(-\frac{E_{a}}{R} ; \ln A\)

(b)

\( \ln A ;-\frac{E_{\alpha}}{R}\)

(c)

\(\frac{E_{a}}{R} ;-\ln A\)

(d)

\( \frac{E_{a}}{R} ; A\)

15.

Which is/are rechargable batteries?

(a)

Nickel-cadmium

(b)

Lead-storage

(c)

Both (a) and (b)

(d)

None of these

16.

Which of the following statements is/are true for the diagram?

(a)

The escaping tendency of molecule decreases for each component

(b)

Vapour pressure of the solution decreases

(c)

Solution shows negative deviation from Raoult's law

(d)

All of the above

17.

The coloured spot of KMnO₄ on any article can be bleached by

(a)

SO₂ + H⁺

(b)

C₂\({ O }_{ 4 }^{ 2- }\) + H⁺

(c)

H₂O₂ + H⁺

(d)

all of these

18.

Which of the following reactions are disproportionation reactions?
\((i)\ { Cu }^{ + }\longrightarrow { Cu }^{ 2+ }+{ Cu }\)
\((ii)\ { 3MnO }_{ 4 }^{ - }+{ 4H }^{ + }\longrightarrow { 2MnO }_{ 4 }^{ - }+{ MnO }_{ 2 }+{ 2H }_{ 2 }O\)
\((iii)\ { 2KMnO }_{ 4 }\longrightarrow { K }_{ 2 }{ MnO }_{ 4 }+{ MnO }_{ 2 }+{ O }_{ 2 }\)
\((iv)\ { 2MnO }_{ 4 }^{ - }+{ 3Mn }^{ 2+ }+{ 2H }_{ 2 }O\longrightarrow { 5MnO }_{ 2 }+{ 4H }^{ + }\)

(a)

(i),(ii)

(b)

(i),(ii),(iii)

(c)

(ii),(iii),(iv)

(d)

(i),(iv)

19.

How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca²⁺ ion?

(a)

One

(b)

Two

(c)

Six

(d)

Three

20.

Which of the following shall from an octahedral complex?

(a)

d⁴(low spin)

(b)

d⁸(high spin)

(c)

d⁶(low spin)

(d)

all of these

21.

Which of the following statements are correct concerning redox properties ?
(i) A metal M for which E⁰ for the half life reaction \({ M }^{ n+ }+{ ne }^{ - }\rightleftharpoons M\) is very negative will be a good reducing agent.
(ii) The oxidizing power of the halogen decreases from chlorine to iodine
(iii) The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide

(a)

(i), (ii) and (iii)

(b)

(i) and (ii)

(c)

(i) only

(d)

(ii) and (iii) only

(e)

(iii) only

22.

Two faradays of electricity are passed through a solution of CuSO₄. The mass of copper deposited at the cathode (at mass of Cu = 63.5 amu)

(a)

2 g

(b)

127 g

(c)

0 g

(d)

63.5 g

23.

The coordination compound of which one of the following compositions will produce two equivalents of AgCl on reaction with aqueous silver nitrate solution?

(a)

CoCl₃.3NH₃

(b)

CoCl₃.6NH₃

(c)

CoCl₃.4NH₃

(d)

CoCl₃.5NH₃

24.

For the reaction A + B \(\longrightarrow\) C + D, doubling the concentration of both the reactants increases the reaction rate by 8 times and doubling the concentration of only B simply doubles the reaction rate. The rate law is given as

(a)

r = k[A]^1/2 [B]^1/2

(b)

r = k[A] [B]²

(c)

r = k[A]²[B]

(d)

r = k[A] [B}

25.

The molar mass of the solute sodium hydroxide obtained from the measurement of osmotic pressure of its aqueous solution at 27^oC is 25 g mol^-1. Therefore, its ionization percentage in the solution is

(a)

(b)

(c)

(d)

26.

If two substances A and B have \({ p }_{ A }^{ o }:{ p }_{ B }^{ o }=1:2\) and have mole fraction in solution 1 : 2, then mole fraction of A in vapours is

(a)

0.33

(b)

0.25

(c)

0.52

(d)

0.2

27.

At a given temperature, osmotic pressure of a concentrated solution of a substance .............. .

(a)

is higher than that at a dilute solution

(b)

is lower than that of a dilute solution

(c)

is same as that of a dilute solution

(d)

can not be compared with osmotic pressure of dilute solution.

Generic 5

4 x 5 = 20

28.

(a) Define:
(i) Interstitial compounds
(ii) Misch-metal
(b) Write one difference between transition elements and p-block elements with reference to variability of oxidation states.
(c) What are-inner transition elements? Give some applications of inner transition elements.

29.

(a) For the reaction 2N₂O₅(g) ⟶4NO₂(g) +O₂(g), the rate of formation of NO₂(g) is 2.8 x 10^{- 3} Ms^-1. Calculate the rate of disappearance of N₂O₅(g).
(b) The rate of a reaction increases four times when the temperature changes from 300 k to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.

30.

(i) Name the element of 3d transition series which shows maximum number of oxidation states. Why does it show so?
(ii) Which transition metal of 3d series has positive \(E°(M^{ 2+ }/M)\) value and why?
(iii) Out of \(Cr^{ 3+ }\) and \(Mn^{ 3+ }\), which is a stronger oxidizing agent and why?
(iv) Name a member of the lanthanoid series which is well known to exhibit a +2 oxidation state.
(v) Complete the following equation:
\(MnO_{ 4 }^{ - }8H^{ + }5e^{ - }\longrightarrow \)

31.

(a) What is the rate of reaction? Write two factors that affect the rate of reaction
(b) The rate constant of a first-order reaction increase from \(4\times { 10 }^{ -2 }\ to\ 8\times { 10 }^{ -2 }\)when the temperature changes from 27°C to 37°C. Calculate the energy of activation \(\left( { E }_{ \alpha } \right) \) .
\(log\ 2=0.301,\ log\ 3=0.477,\ log\ 4=0.6021\)
or
(a) For a reaction A+B\(\longrightarrow \) , the rate is given by
\(Rate\ =\ k\left[ A \right] \left[ { B } \right] _{ 2 }\)
(i) How is the rate of reaction affected if the concentrated of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(c) A first order reaction takes 23.1 minutes for 50% completion. calculate the time required for 75% completion of this reaction.
\(log\ 2=0.301,\ log\ 3=0.477,\ log\ 4=0.6021\)

Case Study Questions

2 x 4 = 8

32.

The concentration of a solute is very important in studying chemical reactions because it determines how often molecules collide in solution and thus indirectly determines the rate of reactions and the conditions at equilibrium. There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. Concentration can be expressed in terms of molarity, molality, parts per million, mass percentage, volume percentage, etc.
(i) The molarity (in mol L^-1) of the given solution will be
(a) 1.56
(b) 1.89
(c) 0.263
(d) 1.44
(ii) Which of the following is temperature dependent?
(a) Molarity
(b) Molality
(c) Mole fraction
(d) Mass percentage
(iii) Which of the following is true for an aqueous solution of the solute in terms of concentration?
(a) 1 M = 1 m
(b) 1M > 1m
(c) 1M < 1 m
(d) Cannot be predicted

33.

Read the passage given below and answer the following questions:
Standard electrode potentials are used for various processes:
(i) It is used to measure relative strengths of various oxidants and reductants.
(ii) It is used to calculate standard cell potential.
(iii) It is used to predict possible reactions.
A set of half-reactions (in acidic medium) along with their standard reduction potential, E^o (in volt) values are given below
\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-} ; \quad E^{\circ}=0.54 \mathrm{~V}\)
\(\mathrm{Cl}_{2}+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} ; \quad E^{\circ}=1.36 \mathrm{~V}\)
\(\mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} ; \quad E^{\circ}=1.50 \mathrm{~V}\)
\(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; \quad E^{\circ}=0.77 \mathrm{~V}\)
\(\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} ; E^{\circ}=1.23 \mathrm{~V}\)
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Which of the following statements is correct?

(a) CI^- is oxidised by O₂	(b) Fe²⁺is oxidised by iodine
(c) I^- is oxidised by chlorine.	(d) Mn²⁺ is oxidised by chlorine

(ii) Mn³⁺ is not stable in acidic medium, while Fe³⁺is stable because

(a) O₂ oxidises Mn²⁺ to Mn³⁺

(b) O₂ oxidises both Mn²⁺ to Mn³⁺ and Fe²⁺to Fe³⁺

(c) Fe^3-oxidises H₂O to O₂

(d) Mn³⁺ oxidises H₂O to O₂

(iii) The strongest reducing agent in the aqueous solution is

(a) I^-

(b) Cl^-

(c) Mn2⁺

(d) Fe²⁺

(iv) The emf for the following reaction is
\(\mathrm{I}_{2}+\mathrm{KCl} \rightleftharpoons 2 \mathrm{KI}+\mathrm{Cl}_{2}\)

(a) -0.82 V

(b) +0.82 V

(c) -0.73 V

(d) +0.73 V

Assertion and reason

2 x 1 = 2

34.

Assertion: Ifblood cells are placed in pure water, they swell and burst.
Reason: Due to osmosis, the movement of water molecules into the cell, dilutes the salt content.
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion .
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

35.

Assertion: [Al(NH₃)₆]³⁺ does npt exist in aqueous solution.
Reason: NH₃is a neutral ligand.
Codes:
(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.
(c) Assertion is correct statement but reason is wrong statement.
(d) Assertion is wrong statement but reason is correct statement.

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Answers Key:

Generic 3

6 x 3 = 18

Given, T₁= 300K, T₂= 310K,
E_a= 3.05 x 10⁵ Jmol^-1
k₁= 6.0 x 10^-4 s^-1, k₂= ?
\(log\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) =\frac { { E }_{ a } }{ 2.303R } \left( \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }.{ T }_{ 2 } } \right) \)
\(log\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) =\frac { 3.05\times { 10 }^{ 5 } }{ 2.303\times 8.314 } \left( \frac { 310-300 }{ 300\times 310 } \right) \)
\(log\left( \frac { { k }_{ 2 } }{ { k }_{ 1 } } \right) =\frac { 3.05\times 10^{ 5 }\times 10 }{ 2.303\times 8.314\times 300\times 310 } \)
k₂/k₁= antilog (1.7128)
k₂/k₁= 5.162 x 101 = 51.62
\(\frac { { k }_{ 2 } }{ 6.0\times 10^{ -4 } } \) = 51.62 or k₂= 3.0972 x 10-2s^-1

Use the formula
\(M_2=\frac{K_b\times W_2 \times 1000}{\Delta T_b \times W_1}\)
i = 0.5. Here i < 1. therefore the solute, acetic acid is associated in benzene.

The comp41ex [Cr(NH₃)₆ ]³⁺is formed by d²sp³ hybridization therefore, it has octahedral geometry. Since it has three unpaired electrons, therefore, it is paramagnetic in nature.
[Cr(NH₃)₆]³⁺
Cr = 24 = 4s¹3d⁵
Cr³⁺ = 3d³
(ii) The IUPAC name of ionization isomer of is triamminechloridenickel (II) nitrate.

3.41 \(\times\) 10³⁵, -2 .02 \(\times\) 10⁵ J

E⁰ cell =1.10 V,i.e.,+ve. Reaction takes place changing blue CuSO₄ to colourless ZnSO₄

(i) F is more electronegative than CI and Br, therefore, fluorides are ionic; whereas chlorides and bromides are covalent.
(ii) It is due to poor shielding effect off-electrons, effective nuclear charge increases, so, ionic size decreases.
(iii) It is due to similar ionic size which is due to lanthanoid contraction, they resemble in their properties.

Generic 2

6 x 2 = 12

N₂O₄ + F₂ \(\rightarrow\) 2 NO₂F is the rate determining step.
\(\frac{d x}{d t}=\) k [N₂O₄] [F₂ ], from slow step ...(i)
NO₂ + NO₂ \(\stackrel{\mathrm{K}}{\rightleftharpoons}\) N₂O₄
\(K=\frac{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}{\left[\mathrm{NO}_{2}\right]^{2}}\) from fast step
[N₂O₄] = K[NO₂]
Substituting value of [N₂O₄] in eqⁿ (i) we get
\(\frac{d x}{d t}=k \times K\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{~F}_{2}\right]\)
\(\frac{d x}{d t}=k^{\prime}\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{~F}_{2}\right]\)
The overall order is 2 + 1 = 3, i.e. it is a third order reaction.

K₂MnO₄

Molar conductivity is calculated as,
\(\Lambda ^{ C }_{ m }=\frac { k\times 1000 }{ Molarity } \)
Molar conductivity
\(\Lambda ^{ C }_{ m }=\frac { k\times 1000 }{ Molarity } =\frac { 0.0248\times 1000 }{ 0.20 } \)
=124 S cm^-2 mol^-1

10.

In [Fe(CN )₆]^3- , CN^- is a strong ligand. Hence, electrons in the d-subshell pair up and there is only one unpaired electron left.In [FeF₆]^3- ,is a weak ligand. Hence, electrons in the 3 d subshell do not pair up. There are five unpaired electrons in the 3d-subshell. (The hybridization involved is sp³d² _ outer orbital complex is formed). Hence, [FeF₆ ]^3- has greater magnetic moment than [Fe(CN ]^3- .

11.

These metals in their common oxidation states have incompletely filled d- orbitals, e.g., Cu²⁺ has 3d⁹ and Au³⁺ has 5d⁸ configuration.

12.

In aqueous solution, 1 M has higher concentration than 1 m. This is because 1 m sol = 1 mol in 1000g of solvent or 1000 cc of solvent. (\(\because \) d = 1g/cc). But 1 M sol = 1 mol in 1000cc of solution which contains less than 1000 cc of the solvent because solute is also present in the solution. Hence , 1000 cc of the solvent will contain more than 1 mole of the solute.

Alternatively, 1 M solution means 1 mole of the solute in 1000 cc of the solution whereas 1 m solution means 1 mole of the solute in 1000 g of water ( = 1000 cc of water). Total volume of 1 m solution > 1000 cc due to presence of extra 1 mole of the solute. Hence, number of moles/cc in 1 m solution will be less than that in 1 M solution. In other words, 1 M is more concentrated than 1 m.

In non - aqueous solution, 1 M can be greater than, less than or equal to 1 m, depending upon the density of the solution.

Multiple Choice Question

15 x 1 = 15

13.

14.

(b)
\( \ln A ;-\frac{E_{\alpha}}{R}\)

15.

16.

(d)
All of the above

17.

(d)
all of these

18.

(a)
(i),(ii)

19.

(a)
One

20.

21.

(a)
(i), (ii) and (iii)

22.

23.

(d)
CoCl₃.5NH₃

24.

25.

(b)
60

26.

(d)
0.2

27.

\(\pi =CRT,i.e.,\pi \alpha C\)

Generic 5

4 x 5 = 20

28.

(a) (i) Interstitial compounds: Transition metals have voids or interstitials in which C, H, N, B, etc. can fit into, resulting in formation of interstitial compounds. They are nonstoichiometric, i.e. their composition is not fixed, e.g. steel. They are harder and less malleable and ductile.
(ii) Misch-metal: It is the alloy of cerium (about 25%) and various other lanthanoid elements. It contains iron upto 5% and traces of sulphur, carbon, silicon, calcium and aluminium. It is a pyrophoric material, hence it is used in lighter flints.
(b) Transition metals show variable oxidation state differ by 1 whereas in p-block, these differ by 2. In heavier elemens of p-block, lower oxidation state is more stable due to inert pair effect. In d-block heavier elements show higher oxidation states more easily.
(c) Thef-block elements (lanthanoids and actinoids) are known as inner-transition elements.
Applications:
(i) Lanthanoids are used in forming alloys.
(ii) Actinoids are used to produce electricity in nuclear reactors.
(iii) Lanthanoids are used as catalyst in the process of petroleum cracking.
(iv) Actinoids are used for the synthesis of transurenic elements.

29.

For the reaction \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \longrightarrow \mathrm{4NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)
overall rate of reaction is
\( -\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=+\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \)
\(=+\frac{d\left[\mathrm{O}_{2}\right]}{d t} \)
Given,
\(\frac{d\left[\mathrm{NO}_{2}\right]}{d t}=2.8 \times 10^{-3} \mathrm{Ms}^{-1} \)
\(-\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=? \)
\(-\frac{1}{2} \frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_{2}\right]}{d t} \)
Putting the given values in the above equation. we get
\(\frac{-d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=\frac{1}{4} \times 2 \times 2.8 \times 10^{-3} \mathrm{Ms}^{-1}\)
\(= 1.4 \times 10^{-3} \mathrm{Ms}^{-1} \)
(b)
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right)\)
\(\log { 4 } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left( \frac { 1 }{ 300 } -\frac { 1 }{ 320 } \right) \)
\({ E }_{ a }=\frac { 19.147\times 0.6021\times 300\times 320 }{ 20 }\)
\(\)= 55.336 KJ mol^-1

30.

(i) Electronic configuration of Mn²⁺is [Ar]3^d5 which are half filled and hence is stable. Therefore, third ionization enthalpy is very high i.e. third electron cannot be easily removed. In the case of Fe²⁺, the electronic configuration is 3d6. Therefore, Fe²⁺can easily lose one electron to acquire 3^d5 stable configuration. Thus, Mn²⁺is more stable than Fe²⁺ towards oxidation to + 3 states.
(ii) The high enthalpies of atomization of transition elements are due to the participation of electrons (n - 1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons' from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are always involved in the formation of metallic bonds.
(iii) Lanthanoids show limited the number of oxidation states, such as +2, +3 and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show principal oxidation state of +3 but a number of other oxidation states also. For example uranium (Z= 92) exhibits oxidation states of+3, +4, +5, +6 and +7 and neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f and 6d orbitals.

31.

(a) Rate of reaction is defined as the change in concentration of reactants or products per unit time. Factors that affect rate of a reaction
(i) Concentration of reactant
(ii) Temperature
\(log{k_2\over k_1}={E_a\over 2.303R}\left[T_2-T_1\over T_1T_2\right]\)
k₁ = 4 x 10^-2 k²
T₁= 8 x 10^-2,
T₁ = 273 + 27 = 300K
T₂ = 273 + 37 = 310K,
R = 8.314 JK^-1mol^-1
\(log{2\times10^{-2}\over 4\times10^{-2}}={E_a\over 2.303\times8.314}\left[310-300\over 300\times310\right]\)
\(log2={E_a\over 2.303\times8.314}\times{10\over 300\times310}\)
\(0.301={E_a\over 2.303\times8.314}\times{10\over 300\times310}\)
\(E_a={0.301\times2.3038.314\times300\times 310\over 10}\)
= 53598 J mol^-1
= 53.598 kJ mol^-1
Or
(a) Rate = k[A][B]²
(i) becomes four times
(ii) second order
(b) For a first order reaction
\(k={2.303\over t}log{[A]_0\over [A]}\)
\([A]+0=a,[A]=a-{a\times50\over 100}=0.50a, t=23.1min\)
\(k={2.303\over 23.1}log{a\over 0.5a}\)
\(={2.303\over 23.1}log2\)
\(={2.303\over 23.1}\times0.301\)
= 0.0300 min^-1
\([A]=a-{a\times75\over 100}=0.25a\)
t = ?, k = 0.0300 min^-1
\(t={2.303\over k}log{[A]_0\over [A]}\)
\(={2.303\over 0.0300}log{a\over 0.25a}\)
\(={2.303\over 0.0300}log4\)
\(={2.303\over 0.0300}\times0.6021\)
= 46.2min

Case Study Questions

2 x 4 = 8

32.

1.(d) 1.44
2.(a) Molarity
3.(b) 1M > 1m

33.

(i) (c) : The half cell having the higher reduction potential will undergo reduction process.
(ii) (d) : Electrode potential of Mn³⁺ is higher than O₂.
(iii) (a) : Due to least electrode potential value.
(iv) (a) : Half reactions :

\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-}\)	Reduction E^o = 0.54 V
\(2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-}\)	Oxidation E^o = -1.36 V
	-------------------------------
	e.m.f = -0.82 V

Assertion and reason

2 x 1 = 2

34.

(b): Osmosis is a phenomenon in which the solvent flows from a solution of lower concentration to the solution of higher concentration. Here, movement of water molecules takes place in highly concentrated salt content of blood cell to such an extent that the cell membrane ruptures.

35.

(b): The complex ion[Al(NH₃)₆]³⁺ undergoes the change into new complex ion [Al(H₂O)₆]³⁺ in aqueous medium due to higher heat of hydration of aluminium ion on account of its small size.
\(\left[\mathrm{Al}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+}+6 \mathrm{H}_{2} \mathrm{O} \rightarrow\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}+6 \mathrm{NH}_{3}\)