Model Test Paper

Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

Since the reaction shifts to backward direction on cooling, this means that the backward reaction is an exothermic reaction. Therefore, the forward reaction is an endothermic reaction and ∆H > 0. On the other hand, enthalpy cannot be zero(C option) because this above-given process is not isothermal(constant temperature).

In chemistry, a hydride is the anion of hydrogen, H⁻, or more commonly it is a compound in which one or more hydrogen centres have nucleophilic, reducing, or basic properties. In compounds that are regarded as hydrides, the hydrogen atom is bonded to a more electropositive element or groups.

All the elements except beryllium combine with hydrogen upon heating to form their hydrides, MH₂.

2M + H₂→ 2M⁺H^- (M= Li, Na, K, Rb, Cs)

M’ + H₂→ M’H₂ (M’ = Mg, Ca, Sr, Ba, Ra)

Be cannot form hydride by this reaction because of the highest amount of ionization energy among alkali and alkali earth metals.

BeH₂, however, can be prepared by the reaction of BeCl₂ with LiAlH₄

BeCl₂ + LiAlH₄→ 2BeH₂ + LiCl + AlCl₃

In 1931, German chemist and physicist Erich Hückel proposed a rule to determine if a planar ring molecule would have aromatic properties. This rule states that if a cyclic, planar molecule has 4n+2π electrons, it is aromatic. This rule would come to be known as Hückel's Rule.

From the above options A, B, C cannot be correct because of not being satisfied with huckle's rule.

Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

Here, A is a mass number and Z is an atomic number of X(any element).

From the above notation, to be the pair of isotope Z should be similar. Only A and B have pairs of isotopes and one the other hand, C and D have different atomic number.

In the set of A - NH₃ and H₂O are not the electrophiles as they have lone pair of electrons.

In the set of B – AlCl₃, SO₃, NO₂⁺ are the electrophiles as AlCl₃ is electron-deficient which tends to seek electron and later, SO₃, NO₂⁺ both are electrophiles.

In the set of C - NO₂⁺, CH₃⁺, CH₃—C⁺=O are all electrophiles as because of positive charges.

In the set of D – all are not electrophiles as one of them have a negating charge which is a neutrophile indeed.

Rule- For quantities created from measured quantities by multiplication and division, the calculated result should have as many significant figures as the measured number with the least number of significant figures.

Here, the least significant figure is of two. Therefore, there would be two significant figures present in the answer to the following calculations.

O2⁻²(aq) + 2H₂O(l)→ H₂O₂(aq) + 2OH⁻(aq)

This is a Bronsted-Lowry acid-base reaction where peroxide ion acts as a base, accepting two H⁺ from water molecules. Alkaline and alkaline earth peroxides contain this ion, so they react with water to produce hydrogen peroxide.

2O₂^- (aq) + 2H₂O(l) → H₂O₂(aq) + 2OH^-(aq)

IUPAC name of the compound is according to the IUPAC rules. Here, the rules which will be applied are as follows-

Find and name the longest possible chain.

The numbering of the identified longest possible chain is based on the priorities.

The suffix in the name of a compound is based on the most priority group in that compound.

Based on these rules, IUPAC name of the compound given is 4-Methylhept-5-ene-2-one

The greenhouse effect is the trapping of the sun's warmth in a planet's lower atmosphere, due to the greater transparency of the atmosphere to visible radiation from the sun than to infrared radiation emitted from the planet's surface. Greenhouse gases are the substances responsible for the greenhouse effect.

The greenhouse gases in Earth's atmosphere are water vapour, carbon dioxide, methane, nitrous oxide and ozone.

According to molecular orbital theory electronic configurations of O₂⁺ and O₂^– species are as follows

The higher bond order of O₂⁺ shows that it is more stable than O₂^–. Both the species have unpaired electrons. So both are paramagnetic in nature.

The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.

To understand the type of reaction we need to calculate the enthalpy of the forward reaction.

[Standard enthalpy of formation of CaO + Standard enthalpy of formation CO₂] – [Standard enthalpy of formation of CaCO₃]

= [-635.1kj mol^-1– 393.5kj mole^-1]– [-1206.9kj mole^-1]

= -1028.6kj mole^-1 + 1206.9kj mole^-1

= 178.3kj mole^-1

This above reaction is an endothermic reaction.

If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat.

Hence, this above reaction would be in the forward direction while increasing the temperature.

Given, pH =2.85 and C =0.08 mol dm^-3

Now since HOCl is a weak acid its dissociation will be given as-

We know that;

pH = -log [H⁺]

-2.85 log [H⁺]

[H⁺] = antilog (-2.85)

[H⁺] = 1.41 × 10^-3

We also know that, for a weak monobasic acid-

H⁺ =

[H⁺]² =K_aC (On squaring both sides)

K_a = 2.5 × 10^-5

Hence the ionization constant of HOCl will be 2.5 × 10^-5

Pb valency in PbO is +2 and in PbO₂ it is +4.

PbO is a basic oxide and simple acid-base reaction takes place between PbO and HNO₃.

On the other hand in PbO₂ lead is in + 4 oxidation state and cannot be oxidised further. Because if see lead’s electronic configuration [Xe] 4f¹⁴5d¹⁰6s²6p², maximum electron lead can give will be four and not more than that because d-orbital is full filled one and thus not letting it donate anymore. Therefore no reaction takes place. Thus, PbO₂ is passive, only PbO reacts with HNO₃.

2PbO + 4HNO₃→ 2Pb (NO₃)₂ + 2H₂O (Acid base reaction)

5 volume solution of H₂O₂ means that 1L of this H₂O₂ will give 10L of oxygen at STP

2H₂O₂(l) → O₂ (g) + H₂O(l)

2*34 g 22.4L at STP

68g

22.4L of O₂ at STP is produced from H₂O₂ = 68g

5L of O₂ at STP is produced from H₂O₂ = = 15.18g

Therefore, the strength of H₂O₂ in 5 volume H₂O₂ = 15.18g/L

Given, mass (m) = 100g = 0.1 kg

Wavelength (λ) = 100 km/h = m/s = 27.78 m/s

According to the de Broglie equation, λ =

Where λ is wavelength,

h is Planck's constant, h = 6.626 10^-34 Js

= 6.626 10^-34 Kgm²s^-1

m is the mass of a particle, moving at velocity v.

Substituting the values, λ = m

λ = 2.385 10^-34 m

Since, the wavelength is very small due to large mass of ball therefore, wave nature was not observed and ball of 100g does not move like a wave when it is thrown by a bowler.

The outermost electronic configuration of nitrogen is 1s²2s²2p³.

Nitrogen has stable half filled configuration and therefore it has no tendency to accept electron therefore, energy has to be supplied in order to add an electron to it. Thus, nitrogen has positive electron gain enthalpy.

On the other hand, the outermost electronic configuration of oxygen is 1s²2s²2p⁴. It does not have half filled or fully filled stable configuration so it shows tendency to accept electron and thus, it has negative electron gain enthalpy.

Although first ionisation enthalpy of oxygen is lower than that of nitrogen because O has 4 electrons in 2p subshell and it can lose one electron to acquire half filled stable electronic configuration and therefore, it has low ionization enthalpy. On the other hand N already have stable half filled electronic configuration so it would not readily lose its electron and thus, N has higher ionisation enthalpy than O.

Formal charge = valence electrons- (lone pair electrons + 1/2 bounded electrons)

(i) Formal charge calculations:

For N: 5 – 1/2 (8) = +1

For O: 6 – 6 – 1/2 (2) = -1

(ii) Formal charge calculations:

For N: 5 – 1/2 (8) = +1

For O without H: 6 - 6 - (1/2)2 = -1

For O without H on left side of N:6-4-1/2 (4)= 0

For O with H: 6 - 4 - (1/2)4 = 0

(iii) Formal charge calculations:

For S: 6 - (1/2)8 = +2,

For O without H: 6 - 6 - (1/2)2 = -1,

For O with H: 6 - 4 - (1/2)4 = 0.

Those conditions are constant volume and constant pressure. This can be explained as given below-

By first law of thermodynamics:

q_v = ∆U + (–w)

q_v = ∆U + p∆V (since, –w = p∆V)

∆V = 0 because volume is constant.

q_v = ∆U + 0

Therefore, q_v = ∆U = change in internal energy

Now, At constant pressure

q_p = ∆U + p∆V

Also, ∆U + p∆V = ∆H

Therefore, q_p = ∆H = change in enthalpy.

Thus, at constant volume and constant pressure heat change is state function because it is equal to change in internal energy and change in enthalpy respectively which are state functions and they are independent of path.

Let S be the solubility.

Initially, assume 1 mole of Al(OH)₃ was present. From which S moles of it was dissolved to give S and 3S moles of Al³⁺ and 3OH^- respectively. This can be understood from the following equation.

Given, = 2.7 × 10^-11

We know,

= [Al³⁺][OH^-]³

= (S)(3S)³ = 27S⁴

S⁴ = = 1 ×

S = 1 × 10^-3 mol L^-1

Molar mass of Al(OH)₃ = 27 + 163 + 13 = 78 g/mol

Solubility of Al(OH)₃ in g L^-1 = 1 × × 78 g/L

= 78 × = 7.8 × 10^-2 g L^-1

Hence, solubility of Al(OH)₃ is 7.8 × 10^-2 g L^-1.

Now to find pH of the solution:

From above equation,

[OH^-] = 3S = 3 × 1 × = 3 ×

We know,

We also know, pH + pOH = 14

Therefore, pH of the solution is 11.47

(Oxidation no. of every atom is underlined by red.)

(a) Na₂S₂O₃

(b) Na₂S₄O₆

(Oxidation no. of every atom is underlined by red.)

(a) Na₂S₂O₃

(b) Na₂S₄O₆

(ii) The reactivity of isotopes of hydrogen is not same towards oxygen because of their different bond dissociation enthalpy. Deuterium and tritium have higher bond dissociation enthalpy than hydrogen thus their reactivity is also low.

(i) Beryllium sulphate and magnesium sulphate are readily soluble in water whereas the sulphates of barium, calcium and strontium are only sparingly soluble because as we move down the group the solubility of sulphates decreases.

On moving down the group the lattice dissociation enthalpy and hydration enthalpy both decreases but the decrease in hydration enthalpy is more than that of lattice dissociation enthalpy. So water molecules are not able to break the metal sulphate bond and thus on moving down the group the sulphates become less soluble.

(ii) The temperature is maintained around 393K during the preparation of plaster of paris because at a temperature higher than 393K the whole water of crystallisation is lost and the result anhydrous CaSO₄ formed is called dead burnt plaster as it loses the properties of setting of water.

2CaSO₄. H₂O → 2CaSO₄ + H2O

(Plaster of pars) (Dead burnt plaster)

(i) Propene can be converted to propane by hydrogenation that is addition of H₂ in the presence of Ni/Pd catalyst.

CH₃—CH = CH₂→ CH₃CH₂CH₃

(propene) (propane)

(ii) Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes.

CH₃ CH₂ CH₂ Cl → CH₃CH₂CH₃ + HCl

(propane)

(iii) Sodium salts of carboxylic acids on heating with soda lime (mixture of sodium hydroxide and calcium oxide) give alkanes containing one carbon atom less than the carboxylic acid. This process of elimination of carbon dioxide from a carboxylic acid is known as decarboxylation.

CH₃ CH₂ CH₂ COO^– Na⁺ + NaOH → CH₃CH₂CH₃

(propane)

C. A and R both are correct and R is the correct explanation of A.

The first ionization enthalpies of the alkali metals are considerably low and further decrease down the group.

This is because the effect of increasing size due to increase in no. of orbitals outweighs the increasing nuclear charge, and the outermost electron is very well screened from the nuclear charge. Thus, it becomes easy to remove an electron from outermost orbital while moving down the group and this is the reason of decreasing ionisation enthalpy.

C. A and R both are correct and R is the correct explanation of A.

A nitro group can be introduced into the benzene ring by heating benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid which is called nitrating mixture.

This nitrating mixture forms a very good electrophile called nitronium ion and then it attacks on benzene ring to form nitrobenzene.

B. A is false but R is correct.

The upper stratosphere consists of considerable amount of ozone (O₃), which protects us from the harmful ultraviolet (UV) radiations coming from the sun. Ozone is not destroyed by solar radiations rather it is destroyed because of release of chlorofluorocarbon compounds (CFCs), also known as freons.

Due to CFCs ozone layer is depleting and allows excessive UV radiations to reach the surface of earth. CFCs mix with the normal atmospheric gases and reach the stratosphere. In stratosphere, they get broken down by powerful UV radiations, releasing chlorine free radical which then reacts with ozone and results in depletion of ozone.

CF₂Cl₂^UV→ Cl^. + F₂Cl

Cl^. + O₃ → ClO^. + O₂

(a) Suppose gas is at point ‘A’ and temperature there is T₁. Now, on increasing the temperature of the gas above critical temperature (T_c) keeping the volume constant, the gas will reach at point ‘F’ and consider there is temperature T₂ and volume V₁ with pressure p1.

Now on compressing the gas up to Volume V₂. During this compression the pressure and volume of the gas will move along the curve FG and will reach at point G. Now, on

cooling the gas as soon as gas will reach the point ‘H’ located on isotherm of T_c, it will liquify without passing through equilibrium state. The gas will not pass through two phases because V₂ of the gas is less than V_c i.e. molecules are closer to each other. Gas is at a higher pressure than critical pressure. Cooling slows down the molecular motion and intermolecular forces can hold the molecules together.

(b) The given liquids in increasing order of their viscosities are as follows-

Benzene < water < ethane-1,2 diol

Ethane-1, 2-diol shows more extent of hydrogen bonding than water while in benzene hydrogen bonding is absent. This factor of hydrogen bonding makes the liquids viscous.

(a)

(i) As BCl₃ is an electron deficient molecule with B having only 6 electrons and therefore it acts as Lewis acid and can accept a pair of electron to complete its octet. This is done on the basis of Lewis concept which states that any electron deficient or positively charged species acts as Lewis acid.

E.g. : NH₃ + BCl₃→ H₃N : BCl₃

(ii) Monobasic acids are those which donate one proton to a base in an acid base reaction. Boric acid in water, accepts OH^- ion acting as lewis acid and water releases H⁺ ions. So instead of donating one proton, it accepts one pair of electron and thus it is a monobasic acid and the reaction is as follows:

B(OH)₃ + H₂O → H⁺ + [ B(OH)₄]^-

(b)

(i) A = B₂H₆ B = B₂H6.2NH₃ C=B₃N₃H₆

B₂H₆ reacts with excess NH₃ to give B₂H6.2NH₃ which is formulated as [BH₂(NH₃)₂]⁺[BH₄]^-. This on heating gives B₃N₃H₆ (borazine) which is called inorganic benzene.

(ii) Reactions involved in these processes are-

3B ₂H₆ +6NH₃ → 3[BH₂ (NH₃ )₂ ]⁺ [BH₄ ]^- → 2B₃N₃H₆ +12H

(a)

(b)

(i) Carbon number ‘2’ in CH₃CH₂Cl has more positive charge than that in CH₃CH₂Br because Cl is more electronegative than Br and that’s why C-Cl bond is more polar than C-Br bond. Cl keeps the shared electron pair towards itself as it is more electronegative. So the carbon adjacent to Cl acquires more positive charge s compared to carbon adjacent to Br.

(ii) CH₃–CH = CH–CH = CH₂ (I) is more stable than

CH₃–CH = CH–CH₂–CH = CH₂ (II) because I shows resonance effect whereas there is no conjugation in II. We already know that more the resonating structure more is the stability. I have 2 resonating structures and thus, it is more stable.

CH₃–CH = CH–CH = CH₂ ↔ CH₃=CH - CH=CH - CH₂