IIT JAM Chemistry - MCQ Test 5 - Chemistry MCQ

The reaction rate of substance A, at an initial concentration of 3.24 × 10^–2 M, is nine times greater than its rate at a lower concentration of 1.2 × 10^–3 M. To determine the order of the reaction, we can use the relationship between concentration and rate.

The general rate equation for a reaction can be expressed as:

• Rate = k [A]ⁿ

Here, k is the rate constant, [A] is the concentration, and n is the order of the reaction.

Given the information:

• At concentration [A]₁ = 3.24 × 10^–2 M, Rate₁ = k (3.24 × 10^–2)ⁿ
• At concentration [A]₂ = 1.2 × 10^–3 M, Rate₂ = k (1.2 × 10^–3)ⁿ

From the problem, we know:

• Rate₁ = 9 × Rate₂

Substituting the rate expressions:

• k (3.24 × 10^–2)ⁿ = 9 × k (1.2 × 10^–3)ⁿ

We can simplify by dividing both sides by k:

• (3.24 × 10^–2)ⁿ = 9 × (1.2 × 10^–3)ⁿ

Next, we can express the ratio:

• 3.24 × 10^–2 = 9 × 1.2 × 10^–3

Taking logarithms helps to solve for n:

• n log(3.24 × 10^–2) = log(9) + n log(1.2 × 10^–3)

Rearranging gives:

• n (log(3.24 × 10^–2) - log(1.2 × 10^–3)) = log(9)

In the synthesis of ammonia by the Haber process, the disappearance of nitrogen is linked to the production of ammonia.

When 60 moles of ammonia are produced in one hour, we can calculate the rate of nitrogen's disappearance. The balanced equation for the Haber process is as follows:

• N₂ + 3H₂ ⇌ 2NH₃

This equation shows that 1 mole of nitrogen produces 2 moles of ammonia. Therefore, the relationship between ammonia produced and nitrogen consumed can be expressed as:

• For every 2 moles of NH₃, 1 mole of N₂ is used.

From the production of 60 moles of ammonia, we can determine how much nitrogen was consumed:

• 60 moles of NH₃ corresponds to:
• 30 moles of N₂ (since 60 ÷ 2 = 30).

Next, we calculate the rate of nitrogen's disappearance over one hour:

• 30 moles of nitrogen over 60 minutes gives:
• 30 moles ÷ 60 minutes = 0.5 moles/min.

Thus, the rate of disappearance of nitrogen is 0.5 mol/min.

Covalent Character of Alkali Metal Compounds

The covalent character of alkali metal compounds varies based on the halogen present. Here’s a brief overview of how this character decreases across different compounds:

• MCl has a higher covalent character compared to others.
• This is followed by MI and MBr.
• MF shows the least covalent character.

In summary, the order of decreasing covalent character is:

• MCl > MI > MBr > MF

This trend can be attributed to the size and electronegativity of the halogens:

• Smaller halogens, like F, have a stronger ionic character.
• Larger halogens, like I, allow for more covalent character.

Anomalous properties of lithium

• The melting and boiling points of lithium are relatively high compared to other group I metals.
• Lithium is considerably softer than its counterparts in group I.
• Unlike other group I metals, lithium forms a nitride, specifically Li3N.
• The ion of lithium and its compounds are more heavily hydrated than those of other group I elements.

The option that does not illustrate the anomalous properties of lithium is that it is much softer than the other group I metals.

The chemical formula of sodium carbonate (washing soda) is Na₂​CO₃​. When it is heated it undergoes a thermal decomposition reaction. This takes place above 500 degree C. The reaction of decomposition is Na₂​CO₃​(s)→Na₂​O(s)+CO₂​(g)

 The name of the first reaction product Na₂​O is Sodium Oxide which is solid and used in glass making and the name of the second reaction product is Carbon Dioxide which is a gas.

Solubility of Sulphates in Water

The solubility of sulphates in water varies among different compounds. Here are some key points regarding the solubility of the sulphates mentioned:

• MgSO₄ (Magnesium Sulphate):
  • Highly soluble in water.
  • Commonly used in agriculture and medicine.
• BaSO₄ (Barium Sulphate):
  • Practically insoluble in water.
  • Used in radiology for imaging.
• CaSO₄ (Calcium Sulphate):
  • Slightly soluble in water.
  • Forms gypsum when hydrated.
• BeSO₄ (Beryllium Sulphate):
  • Moderately soluble in water.
  • Less common compared to others.

Among these, MgSO₄ has the highest solubility in water, making it the most suitable choice for applications requiring dissolved sulphates.

Catenation refers to the ability of an element to form long chains or rings of atoms. Among the p-block elements, certain elements exhibit this property more prominently.

Here are the key points regarding the catenation abilities of the mentioned elements:

• Carbon is renowned for its exceptional catenation ability, forming long chains and complex structures.
• Silicon can also exhibit catenation, although not as extensively as carbon.
• Germanium shows some degree of catenation, but it is less stable compared to carbon and silicon.
• Lead, however, does not effectively demonstrate catenation. This is due to its larger atomic size and the presence of metallic bonds that hinder the formation of long chains.

Therefore, among the elements listed, lead is the one that does not show significant catenation.

To calculate the entropy change (ΔS) of the system for the isothermal expansion of an ideal gas, we use the formula:

ΔS=nRln⁡(V_f/V_i)

Where:

• n = 0.50 mole (number of moles)
• R=8.314 J K^-1mol⁻¹ (universal gas constant)
• V_f=5.0 L (final volume)
• V_i=1.0 L (initial volume)

Substituting the values:

ΔS=0.50 x 8.314 x ln⁡(5.0/1.0) 

First, calculate the natural logarithm:

ln⁡(5.0/1.0)=ln⁡(5.0)≈1.609

Now calculate ΔS:

ΔS=0.50 x 8.314 x 1.609≈6.7 J K⁻¹

Final Answer:

a)6.7 J K⁻¹

Ketones react with Mg-Hg in the presence of water to produce a range of organic compounds. The key products of this reaction are:

• Alkene: A hydrocarbon with a carbon-carbon double bond, formed as a result of dehydration.

• Pinacolone: A type of ketone with a specific structure that results from rearrangement during the reaction.

• α-hydroxy ketone: This compound features both a hydroxyl group (-OH) and a ketone group (C=O), indicating that a ketone has been reduced.

• Pinacols: These are diols formed from the reaction, where two hydroxyl groups are present on adjacent carbon atoms.

The formation of pinacols is particularly significant, as it showcases how ketones can undergo reduction and rearrangement to yield more complex structures.

Order of reaction refers to how the rate of a chemical reaction depends on the concentration of reactants. Here are some key points regarding the order of reaction:

• Determination: The order can only be established through experimental methods.
• Rate Law: The order is equal to the sum of the powers of the concentration terms in the differential rate law.
• Stoichiometric Coefficients: The order is not influenced by the stoichiometric coefficients of the reactants.
• Fractional Orders: Contrary to common belief, the order of a reaction can be a fractional value.

Thus, the incorrect statement is that "order cannot be fractional." This misconception arises because some may assume that the order must always be a whole number, but this is not true.

t_1/4 represents the time required for the concentration of a reactant to decrease to three-quarters of its original amount. For a first-order reaction, the relationship between this time and the rate constant K can be expressed mathematically.

The equation for t_1/4 is given by:

• It is inversely proportional to the rate constant K.
• The formula can be simplified to:
• t_1/4 = 0.29 / K.

This means that as the rate constant K increases, the time t_1/4 decreases, indicating that the reaction occurs more quickly.

Decarboxylation is a chemical reaction where a carboxyl group is removed from a compound, often releasing carbon dioxide. This process is significant in organic chemistry, particularly for compounds that can easily lose this group under mild conditions.

Several factors influence the ease of decarboxylation:

• Stability of intermediate: Compounds with stable intermediates typically decarboxylate more readily.
• Presence of electron-withdrawing groups: These groups can increase the acidity of the carboxylic acid, making decarboxylation easier.
• Structure of the compound: Certain structural features can facilitate the reaction.

Among the options provided, the compound that undergoes decarboxylation most readily under mild conditions is option B.

The acidity of carboxylic acids can vary significantly based on their structure and substituents. Here's a breakdown of the acids in question:

• PhCOOH (Benzoic acid)
• o-NO₂C₆H₄COOH (Ortho-nitrobenzoic acid)
• p-NO₂C₆H₄COOH (Para-nitrobenzoic acid)
• m-NO₂C₆H₄COOH (Meta-nitrobenzoic acid)

Factors influencing acidity include:

• Electron-withdrawing groups, like nitro (-NO₂), increase acidity by stabilising the negative charge on the conjugate base.
• The position of these groups is crucial. Ortho and para positions enhance acidity more than the meta position due to resonance effects.

In general, the order of acidity for these acids is:

• Ortho-nitrobenzoic acid (II) has the highest acidity due to strong electron-withdrawing effects from the nitro group.
• Para-nitrobenzoic acid (III) is next, as it benefits from resonance but is slightly less acidic than the ortho isomer.
• Meta-nitrobenzoic acid (IV) is less acidic than both ortho and para due to reduced resonance stabilization.
• Benzoic acid (I) is the least acidic since it lacks any electron-withdrawing substituents.

Thus, the correct order of acidity is:

• II > IV > I > III

To determine the rate constant for a first-order reaction:

• Start with one mole of compound A.
• The reaction is 3/4 complete in one hour.
• This means that 1/4 of the original amount remains after one hour.

For a first-order reaction, the relationship between the concentration and time can be expressed via the formula:

ln([A]₀/[A]) = kt

• [A]₀ = initial concentration (1 mole).
• [A] = concentration remaining after time t (0.25 moles).
• t = time (1 hour = 60 minutes).

Substituting into the equation:

• ln(1/0.25) = k × 60
• ln(4) = k × 60

Calculating ln(4):

• ln(4) ≈ 1.386.
• Thus, 1.386 = k × 60.

Solving for k:

• k = 1.386 / 60.
• k ≈ 0.0231 min^–1.

This value indicates the rate constant for the reaction, which confirms that the correct answer is:

• 0.0231 min^–1.

Nitrogen can form a variety of oxides, which are compounds made of nitrogen and oxygen. The most common nitrogen oxides include:

• Nitric oxide (NO)
• Nitrogen dioxide (NO₂)
• Nitrous oxide (N₂O)
• Dinitrogen tetroxide (N₂O₄)
• Trinitrogen oxide (N₂O₃)

In total, nitrogen can form five significant oxides. These compounds are important in various environmental and industrial processes.

Optical isomers are molecules that are mirror images of each other, often differing in their spatial arrangement. Not all compounds exhibit optical isomerism; some lack the necessary chiral centres. Below are the key points regarding the options provided:

• [Co(NH₃)₃Cl₃]: This complex has a symmetrical structure, which means it does not have optical isomers.
• [Co(en)₃]Cl₃: This compound is chiral due to the presence of the ethylenediamine (en) ligands, resulting in optical isomers.
• [Co(en)₂Cl₂]Cl: This complex also has chirality, leading to the existence of optical isomers.
• [Co(en)(NH₃)₂Cl₂]Cl: The combination of ligands makes this complex chiral, allowing for optical isomers.

In summary, the complex that does not exhibit optical isomerism is [Co(NH₃)₃Cl₃].

In a zero-order reaction:

• The rate is constant and does not depend on the initial concentration of the reactant.
• This means that changing the concentration will not affect the rate of reaction.

Half-life concepts:

• The half-life (_t1/2) of a first-order reaction is independent of the initial concentration.
• In contrast, the half-life of a zero-order reaction is directly related to the initial concentration of the reactant.

Thus, the statement regarding the rate of a zero-order reaction depends on the initial concentration is incorrect.

Among the complex ions listed, [Cr(NH₃)₆]³⁺ is expected to absorb visible light due to its electronic structure and the nature of the ligands involved.

• Chromium ions, especially in the +3 oxidation state, have unpaired electrons.
• The presence of ammonia (NH₃) as a ligand creates a strong field environment, leading to a significant splitting of the d-orbitals.
• This splitting allows for the absorption of visible light, promoting electrons to higher energy levels.

In contrast:

• [Sc(H₂O)₃(NH₃)₃]³⁺ has a filled d-orbital, meaning it does not absorb visible light.
• [Ti(en)₂(NH₃)₂]⁴⁺ and [Zn(NH₃)₆]²⁺ also have filled d-orbitals or low crystal field splitting, making them unlikely to absorb visible light.

Thus, the complex ion that absorbs visible light is [Cr(NH₃)₆]³⁺.

The reaction [Fe(CNS)₆]^3– to [FeF₆]^3– involves the following changes:

• A decrease in magnetic moment.

• An increase in magnetic moment.

• A decrease in co-ordination number.

• An increase in co-ordination number.

The correct answer is: B.

Higher the oxidation state of the metal, greater the crystal field splitting energy . In options (a), (b) and (d), Co is present in +2+2 oxidation state and in (c ) it is present in +3+3 oxidation state and hence has a higher value of CFSE.

An optically active compound with the molecular formula C₈H₁₆ undergoes ozonolysis, resulting in acetone as one of the products. The structure of this compound can be identified by examining the given images.

• Ozonolysis is a reaction where ozone breaks down alkenes and alkynes, producing carbonyl compounds.
• In this case, one of the resulting products is acetone, indicating the presence of specific carbon chain arrangements.
• The correct structure of the optically active compound is shown in option B.