Important Questions
Practice these important questions to strengthen your understanding
Question 1
Draw the hybridised orbitals for the complexes \( [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} \) and \( [\mathrm{NiCl}_4]^{2-} \), indicating the type of hybridisation involved and the electron configuration of the central metal ion.
Answer:
\( [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} \) :
Co\(^{3+}\) electronic configuration: \( 3d^6 \)
Electron pairing occurs due to strong field NH\(_3\) ligands.
Hybridisation: \( d^2sp^3 \) (inner orbital)
Hybrid orbitals formed from two 3d, one 4s, and three 4p orbitals.
All six hybrid orbitals are occupied by ligand electron pairs.
\( [\mathrm{NiCl}_4]^{2-} \) :
Ni\(^{2+}\) electronic configuration: \( 3d^8 \)
No electron pairing due to weak field Cl\(^{-}\) ligands.
Hybridisation: \( sp^3 \) (outer orbital)
Hybrid orbitals formed from one 4s and three 4p orbitals.
Four hybrid orbitals occupied by ligand electron pairs.
Unpaired electrons remain in 3d orbitals.
Co\(^{3+}\) electronic configuration: \( 3d^6 \)
Electron pairing occurs due to strong field NH\(_3\) ligands.
Hybridisation: \( d^2sp^3 \) (inner orbital)
Hybrid orbitals formed from two 3d, one 4s, and three 4p orbitals.
All six hybrid orbitals are occupied by ligand electron pairs.
\( [\mathrm{NiCl}_4]^{2-} \) :
Ni\(^{2+}\) electronic configuration: \( 3d^8 \)
No electron pairing due to weak field Cl\(^{-}\) ligands.
Hybridisation: \( sp^3 \) (outer orbital)
Hybrid orbitals formed from one 4s and three 4p orbitals.
Four hybrid orbitals occupied by ligand electron pairs.
Unpaired electrons remain in 3d orbitals.
Explanation:
Drawing hybrid orbitals involves understanding the electronic configuration and ligand field strength.
For \( [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} \), strong field ligands cause pairing of electrons, allowing the use of inner d orbitals in hybridisation, resulting in \( d^2sp^3 \) hybrid orbitals forming octahedral geometry.
For \( [\mathrm{NiCl}_4]^{2-} \), weak field ligands do not cause pairing, so only s and p orbitals hybridise to form \( sp^3 \) orbitals, leading to tetrahedral geometry.
Electron configurations show paired or unpaired electrons, explaining magnetic properties.
Drawing hybrid orbitals involves understanding the electronic configuration and ligand field strength.
For \( [\mathrm{Co}(\mathrm{NH}_3)_6]^{3+} \), strong field ligands cause pairing of electrons, allowing the use of inner d orbitals in hybridisation, resulting in \( d^2sp^3 \) hybrid orbitals forming octahedral geometry.
For \( [\mathrm{NiCl}_4]^{2-} \), weak field ligands do not cause pairing, so only s and p orbitals hybridise to form \( sp^3 \) orbitals, leading to tetrahedral geometry.
Electron configurations show paired or unpaired electrons, explaining magnetic properties.
Question 2
Describe the effect of a catalyst on the energy profile of a reaction and how it influences the reaction rate.
Answer:
A catalyst provides an alternate reaction pathway with a lower activation energy (E_a).
On the energy profile diagram, this is shown as a lower peak compared to the uncatalyzed reaction.
By lowering the activation energy, the catalyst increases the fraction of molecules that can overcome this energy barrier.
This results in an increased rate constant and hence a faster reaction rate.
The catalyst itself is not consumed and does not alter the overall energy change (ΔG) of the reaction.
On the energy profile diagram, this is shown as a lower peak compared to the uncatalyzed reaction.
By lowering the activation energy, the catalyst increases the fraction of molecules that can overcome this energy barrier.
This results in an increased rate constant and hence a faster reaction rate.
The catalyst itself is not consumed and does not alter the overall energy change (ΔG) of the reaction.
Explanation:
Catalysts work by providing a different mechanism that requires less energy to proceed.
The energy profile shows this as a reduced activation energy peak.
Lower activation energy means more molecules have enough energy to react.
This increases the rate constant according to the Arrhenius equation.
Students should visualize how catalysts speed up reactions without changing reactants or products energy levels.
Catalysts work by providing a different mechanism that requires less energy to proceed.
The energy profile shows this as a reduced activation energy peak.
Lower activation energy means more molecules have enough energy to react.
This increases the rate constant according to the Arrhenius equation.
Students should visualize how catalysts speed up reactions without changing reactants or products energy levels.
Question 3
Which of the following correctly describes the phase of catalyst in the contact process for sulphuric acid manufacture?
Answer:
Explanation:
In the contact process, the catalyst used is either platinised asbestos or vanadium pentoxide, both of which are solids. The reactants, sulphur dioxide and oxygen, are gases. This makes the catalyst phase solid, distinct from the gaseous reactants, indicating heterogeneous catalysis. This solid catalyst provides a surface for the reaction to occur efficiently.
In the contact process, the catalyst used is either platinised asbestos or vanadium pentoxide, both of which are solids. The reactants, sulphur dioxide and oxygen, are gases. This makes the catalyst phase solid, distinct from the gaseous reactants, indicating heterogeneous catalysis. This solid catalyst provides a surface for the reaction to occur efficiently.
Question 4
Write the balanced equation for the side chain chlorination of toluene to benzal chloride and its hydrolysis to benzaldehyde.
Answer:
Step 1 Side chain chlorination:
C6H5CH3 + 3 Cl2 → C6H5CCl3 + 3 HCl (in presence of light hv)
Step 2 Hydrolysis:
C6H5CCl3 + 3 H2O → C6H5CHO + 3 HCl
Overall:
Toluene reacts with chlorine under UV light to form benzal chloride, which on hydrolysis with water at elevated temperature yields benzaldehyde.
C6H5CH3 + 3 Cl2 → C6H5CCl3 + 3 HCl (in presence of light hv)
Step 2 Hydrolysis:
C6H5CCl3 + 3 H2O → C6H5CHO + 3 HCl
Overall:
Toluene reacts with chlorine under UV light to form benzal chloride, which on hydrolysis with water at elevated temperature yields benzaldehyde.
Explanation:
The side chain chlorination of toluene involves substitution of the benzylic hydrogens by chlorine atoms under UV light.
Three chlorine atoms replace the three benzylic hydrogens to form benzal chloride.
Hydrolysis of benzal chloride involves replacement of the chlorine atoms by hydroxyl groups, which rearrange to form the aldehyde group.
This two-step process is a common industrial method for preparing benzaldehyde.
Balancing the equations ensures conservation of atoms and helps students understand stoichiometry in organic reactions.
The side chain chlorination of toluene involves substitution of the benzylic hydrogens by chlorine atoms under UV light.
Three chlorine atoms replace the three benzylic hydrogens to form benzal chloride.
Hydrolysis of benzal chloride involves replacement of the chlorine atoms by hydroxyl groups, which rearrange to form the aldehyde group.
This two-step process is a common industrial method for preparing benzaldehyde.
Balancing the equations ensures conservation of atoms and helps students understand stoichiometry in organic reactions.
Question 5
Explain why phenol is more acidic than ethanol. Support your answer with resonance structures of phenol and phenoxide ion.
Answer:
Phenol is more acidic than ethanol because the phenoxide ion formed after deprotonation is resonance stabilized.
Step 1: In phenol, the negative charge on oxygen after losing H+ is delocalized over the aromatic ring through resonance.
Step 2: Draw resonance structures showing negative charge moving to ortho and para positions on the ring.
Step 3: This delocalization stabilizes the phenoxide ion, making phenol more acidic.
Step 4: In ethanol, the negative charge on oxygen is localized and not resonance stabilized.
Therefore, phenol is more acidic than ethanol due to resonance stabilization of its conjugate base.
Step 1: In phenol, the negative charge on oxygen after losing H+ is delocalized over the aromatic ring through resonance.
Step 2: Draw resonance structures showing negative charge moving to ortho and para positions on the ring.
Step 3: This delocalization stabilizes the phenoxide ion, making phenol more acidic.
Step 4: In ethanol, the negative charge on oxygen is localized and not resonance stabilized.
Therefore, phenol is more acidic than ethanol due to resonance stabilization of its conjugate base.
Explanation:
Acidity depends on the stability of the conjugate base.
Phenol's conjugate base (phenoxide ion) is resonance stabilized by delocalization of negative charge into the aromatic ring.
This resonance stabilization lowers the energy of phenoxide ion, increasing acidity.
Ethanol's conjugate base lacks such resonance and is less stable.
Drawing resonance structures helps visualize charge delocalization and understand acidity differences.
This concept is fundamental in organic acid-base chemistry.
Acidity depends on the stability of the conjugate base.
Phenol's conjugate base (phenoxide ion) is resonance stabilized by delocalization of negative charge into the aromatic ring.
This resonance stabilization lowers the energy of phenoxide ion, increasing acidity.
Ethanol's conjugate base lacks such resonance and is less stable.
Drawing resonance structures helps visualize charge delocalization and understand acidity differences.
This concept is fundamental in organic acid-base chemistry.
Question 6
What is the temperature range for the diazotisation reaction and why is it critical for the formation of diazonium salts?
Answer:
The temperature range for the diazotisation reaction is 273-278 K. This low temperature is critical because diazonium salts are unstable at higher temperatures and tend to decompose. Maintaining this temperature range ensures the formation and stability of the diazonium salt in solution, allowing it to be used immediately for further reactions.
Explanation:
Diazonium salts are thermally unstable compounds that decompose easily when heated. Performing the diazotisation reaction at 273-278 K (0-5 °C) slows down the decomposition and stabilizes the diazonium ion in solution. This controlled temperature is essential to successfully prepare diazonium salts and use them in subsequent synthetic steps without premature decomposition.
Diazonium salts are thermally unstable compounds that decompose easily when heated. Performing the diazotisation reaction at 273-278 K (0-5 °C) slows down the decomposition and stabilizes the diazonium ion in solution. This controlled temperature is essential to successfully prepare diazonium salts and use them in subsequent synthetic steps without premature decomposition.
Question 7
Which of the following conductivity ranges correctly classifies a solid as a semiconductor?
Answer:
Explanation:
Solids are classified based on their electrical conductivity.
Insulators have very low conductivity (\(10^{-20}\) to \(10^{-10}\) \(\text{ohm}^{-1}\text{m}^{-1}\)).
Semiconductors have intermediate conductivity ranging from \(10^{-6}\) to \(10^{4}\) \(\text{ohm}^{-1}\text{m}^{-1}\).
Conductors have high conductivity above this range.
Thus, the given range correctly identifies semiconductors.
Solids are classified based on their electrical conductivity.
Insulators have very low conductivity (\(10^{-20}\) to \(10^{-10}\) \(\text{ohm}^{-1}\text{m}^{-1}\)).
Semiconductors have intermediate conductivity ranging from \(10^{-6}\) to \(10^{4}\) \(\text{ohm}^{-1}\text{m}^{-1}\).
Conductors have high conductivity above this range.
Thus, the given range correctly identifies semiconductors.
Question 8
Write the balanced chemical equation for the reaction of phosphine with hydriodic acid and explain the nature of the product formed.
Answer:
The balanced chemical equation is:
\( \mathrm{PH_3} + \mathrm{HI} \rightarrow \mathrm{PH_4I} \)
The product \( \mathrm{PH_4I} \) is phosphonium iodide, formed by the protonation of phosphine.
Phosphine acts as a Lewis base donating its lone pair to \( \mathrm{H^+} \) from HI, forming the phosphonium ion \( \mathrm{PH_4^+} \) and iodide ion \( \mathrm{I^-} \).
\( \mathrm{PH_3} + \mathrm{HI} \rightarrow \mathrm{PH_4I} \)
The product \( \mathrm{PH_4I} \) is phosphonium iodide, formed by the protonation of phosphine.
Phosphine acts as a Lewis base donating its lone pair to \( \mathrm{H^+} \) from HI, forming the phosphonium ion \( \mathrm{PH_4^+} \) and iodide ion \( \mathrm{I^-} \).
Explanation:
Phosphine reacts with acids like hydriodic acid by accepting a proton due to the lone pair on phosphorus.
This forms a phosphonium salt, \( \mathrm{PH_4I} \), which contains the phosphonium cation \( \mathrm{PH_4^+} \).
The reaction demonstrates the basic nature of phosphine as a Lewis base.
The product is ionic and stable, showing the acid-base interaction clearly.
Balancing the equation confirms the stoichiometry of the reaction.
Phosphine reacts with acids like hydriodic acid by accepting a proton due to the lone pair on phosphorus.
This forms a phosphonium salt, \( \mathrm{PH_4I} \), which contains the phosphonium cation \( \mathrm{PH_4^+} \).
The reaction demonstrates the basic nature of phosphine as a Lewis base.
The product is ionic and stable, showing the acid-base interaction clearly.
Balancing the equation confirms the stoichiometry of the reaction.
Question 9
For a binary solution of liquids A and B the vapour pressure of pure A is 450 mm Hg and that of pure B is 700 mm Hg at 350 K If the total vapour pressure of the solution is 600 mm Hg calculate the mole fractions of A and B in the liquid phase and the mole fraction of A in the vapour phase
Answer:
Let \(x_A\) and \(x_B\) be the mole fractions of A and B in the liquid phase respectively
Given \(p_A^0 = 450\) mm Hg, \(p_B^0 = 700\) mm Hg, total vapour pressure \(P = 600\) mm Hg
According to Raoult's law:
\[ P = x_A p_A^0 + x_B p_B^0 \]
Since \(x_B = 1 - x_A\), substitute:
\[ 600 = x_A \times 450 + (1 - x_A) \times 700 \]
\[ 600 = 450 x_A + 700 - 700 x_A \]
\[ 600 - 700 = (450 - 700) x_A \]
\[ -100 = -250 x_A \]
\[ x_A = \frac{100}{250} = 0.4 \]
Thus, \(x_B = 1 - 0.4 = 0.6\)
To find mole fraction of A in vapour phase \(y_A\):
\[ y_A = \frac{x_A p_A^0}{P} = \frac{0.4 \times 450}{600} = \frac{180}{600} = 0.3 \]
Therefore, the mole fractions are:
Liquid phase: \(x_A = 0.4\), \(x_B = 0.6\)
Vapour phase: \(y_A = 0.3\)
Given \(p_A^0 = 450\) mm Hg, \(p_B^0 = 700\) mm Hg, total vapour pressure \(P = 600\) mm Hg
According to Raoult's law:
\[ P = x_A p_A^0 + x_B p_B^0 \]
Since \(x_B = 1 - x_A\), substitute:
\[ 600 = x_A \times 450 + (1 - x_A) \times 700 \]
\[ 600 = 450 x_A + 700 - 700 x_A \]
\[ 600 - 700 = (450 - 700) x_A \]
\[ -100 = -250 x_A \]
\[ x_A = \frac{100}{250} = 0.4 \]
Thus, \(x_B = 1 - 0.4 = 0.6\)
To find mole fraction of A in vapour phase \(y_A\):
\[ y_A = \frac{x_A p_A^0}{P} = \frac{0.4 \times 450}{600} = \frac{180}{600} = 0.3 \]
Therefore, the mole fractions are:
Liquid phase: \(x_A = 0.4\), \(x_B = 0.6\)
Vapour phase: \(y_A = 0.3\)
Explanation:
This problem involves applying Raoult's law to find the composition of a binary liquid solution and its vapour phase
First we use the total vapour pressure formula which is the sum of the partial pressures of components weighted by their mole fractions in liquid phase
By substituting the known values and solving for mole fraction of A we find the liquid composition
Then using the mole fraction in liquid and vapour pressures we calculate the mole fraction of A in vapour phase
This stepwise approach helps understand how vapour-liquid equilibrium compositions relate to each other
This problem involves applying Raoult's law to find the composition of a binary liquid solution and its vapour phase
First we use the total vapour pressure formula which is the sum of the partial pressures of components weighted by their mole fractions in liquid phase
By substituting the known values and solving for mole fraction of A we find the liquid composition
Then using the mole fraction in liquid and vapour pressures we calculate the mole fraction of A in vapour phase
This stepwise approach helps understand how vapour-liquid equilibrium compositions relate to each other
Question 10
Explain how the rate of a reaction can be expressed in terms of rate of change of partial pressure for gaseous reactants.
Answer:
For gaseous reactions at constant temperature, concentration is directly proportional to partial pressure.
Therefore, the rate of reaction can be expressed as the rate of change of partial pressure of reactants or products.
Mathematically, rate = - \frac{\Delta P_{reactant}}{\Delta t} or rate = \frac{\Delta P_{product}}{\Delta t}, where P is partial pressure.
This allows expressing reaction rates in units of pressure per unit time, such as atm s⁻¹.
Therefore, the rate of reaction can be expressed as the rate of change of partial pressure of reactants or products.
Mathematically, rate = - \frac{\Delta P_{reactant}}{\Delta t} or rate = \frac{\Delta P_{product}}{\Delta t}, where P is partial pressure.
This allows expressing reaction rates in units of pressure per unit time, such as atm s⁻¹.
Explanation:
In gaseous reactions, concentration and partial pressure are related by the ideal gas law.
Since concentration is proportional to partial pressure at constant temperature, changes in concentration correspond to changes in partial pressure.
Expressing rate in terms of partial pressure is useful especially when measuring gases.
This concept helps connect kinetic data with measurable gas properties.
In gaseous reactions, concentration and partial pressure are related by the ideal gas law.
Since concentration is proportional to partial pressure at constant temperature, changes in concentration correspond to changes in partial pressure.
Expressing rate in terms of partial pressure is useful especially when measuring gases.
This concept helps connect kinetic data with measurable gas properties.
Question 11
For the reaction Hg(l) + Cl₂(g) → HgCl₂(s), if the rate of disappearance of Cl₂ is 2.0 × 10⁻⁵ mol L⁻¹ s⁻¹, what is the rate of formation of HgCl₂?
Answer:
Since the stoichiometric coefficients of Cl₂ and HgCl₂ are both 1, the rate of formation of HgCl₂ equals the rate of disappearance of Cl₂.
Therefore, rate of formation of HgCl₂ = 2.0 × 10⁻⁵ mol L⁻¹ s⁻¹
Therefore, rate of formation of HgCl₂ = 2.0 × 10⁻⁵ mol L⁻¹ s⁻¹
Explanation:
For reactions where stoichiometric coefficients of reactants and products are equal, the rate of disappearance of reactants equals the rate of formation of products.
Here, both Cl₂ and HgCl₂ have coefficient 1.
Therefore, the rate of formation of HgCl₂ is the same as the rate of disappearance of Cl₂.
This is a direct application of the definition of reaction rate based on stoichiometry.
For reactions where stoichiometric coefficients of reactants and products are equal, the rate of disappearance of reactants equals the rate of formation of products.
Here, both Cl₂ and HgCl₂ have coefficient 1.
Therefore, the rate of formation of HgCl₂ is the same as the rate of disappearance of Cl₂.
This is a direct application of the definition of reaction rate based on stoichiometry.
Question 12
Which of the following ethers is an example of a symmetrical ether?
Answer:
Symmetrical ethers have identical alkyl or aryl groups attached to the oxygen atom.
A) CH3–O–CH3 has two methyl groups, so it is symmetrical.
B) C2H5–O–CH3 has ethyl and methyl groups, unsymmetrical.
C) C6H5–O–CH3 has phenyl and methyl groups, unsymmetrical.
D) C2H5–O–C6H5 has ethyl and phenyl groups, unsymmetrical.
Correct answer is A.
A) CH3–O–CH3 has two methyl groups, so it is symmetrical.
B) C2H5–O–CH3 has ethyl and methyl groups, unsymmetrical.
C) C6H5–O–CH3 has phenyl and methyl groups, unsymmetrical.
D) C2H5–O–C6H5 has ethyl and phenyl groups, unsymmetrical.
Correct answer is A.
Explanation:
Ethers are classified based on the groups attached to oxygen.
If both groups are the same, the ether is symmetrical.
If different, it is unsymmetrical.
This question tests recognition of symmetrical vs unsymmetrical ethers by comparing substituents.
Ethers are classified based on the groups attached to oxygen.
If both groups are the same, the ether is symmetrical.
If different, it is unsymmetrical.
This question tests recognition of symmetrical vs unsymmetrical ethers by comparing substituents.
Question 13
Explain the effect of temperature on the Maxwell-Boltzmann distribution curve and its impact on the fraction of molecules able to react.
Answer:
As temperature increases, the Maxwell-Boltzmann distribution curve shifts to the right and flattens.
This means the most probable kinetic energy of molecules increases.
The curve broadens, indicating a wider range of molecular energies.
Consequently, a greater fraction of molecules have kinetic energy equal to or greater than the activation energy.
This increase in energetic molecules leads to more effective collisions and faster reaction rates.
Thus, temperature rise increases the fraction of molecules able to react by providing more energy to overcome the activation barrier.
This means the most probable kinetic energy of molecules increases.
The curve broadens, indicating a wider range of molecular energies.
Consequently, a greater fraction of molecules have kinetic energy equal to or greater than the activation energy.
This increase in energetic molecules leads to more effective collisions and faster reaction rates.
Thus, temperature rise increases the fraction of molecules able to react by providing more energy to overcome the activation barrier.
Explanation:
The Maxwell-Boltzmann distribution shows how molecular energies are spread at a given temperature.
When temperature rises, molecules move faster and have higher energies.
The curve shifts right, meaning more molecules have enough energy to react.
This explains why reaction rates increase with temperature.
It helps students visualize the microscopic basis of temperature effects on kinetics.
The Maxwell-Boltzmann distribution shows how molecular energies are spread at a given temperature.
When temperature rises, molecules move faster and have higher energies.
The curve shifts right, meaning more molecules have enough energy to react.
This explains why reaction rates increase with temperature.
It helps students visualize the microscopic basis of temperature effects on kinetics.
Question 14
Explain the role of catalyst in the decomposition of potassium chlorate.
Answer:
Manganese dioxide (MnO₂) acts as a catalyst in the decomposition of potassium chlorate.
It lowers the activation energy of the reaction.
This increases the rate of decomposition of KClO₃ into KCl and O₂.
The catalyst itself remains unchanged after the reaction.
It lowers the activation energy of the reaction.
This increases the rate of decomposition of KClO₃ into KCl and O₂.
The catalyst itself remains unchanged after the reaction.
Explanation:
In the decomposition of potassium chlorate, MnO₂ provides an alternate pathway with lower activation energy.
This allows the reaction to proceed faster at a given temperature.
The catalyst is not consumed and can be recovered after the reaction.
This example illustrates practical use of catalysts in industrial and laboratory processes.
It helps students connect theory with real-world chemical reactions.
In the decomposition of potassium chlorate, MnO₂ provides an alternate pathway with lower activation energy.
This allows the reaction to proceed faster at a given temperature.
The catalyst is not consumed and can be recovered after the reaction.
This example illustrates practical use of catalysts in industrial and laboratory processes.
It helps students connect theory with real-world chemical reactions.
Question 15
Describe the toxicity of methanol and its biological oxidation pathway in the human body.
Answer:
Methanol is highly toxic to humans. When ingested, it is metabolized in the liver by the enzyme alcohol dehydrogenase to formaldehyde, which is further oxidized by aldehyde dehydrogenase to formic acid. Formic acid accumulation leads to metabolic acidosis and can cause damage to the optic nerve resulting in blindness. The overall pathway is: Methanol \(\xrightarrow{alcohol\ dehydrogenase}\) Formaldehyde \(\xrightarrow{aldehyde\ dehydrogenase}\) Formic acid. Toxic effects include headache, dizziness, nausea, vomiting, visual disturbances, and in severe cases, coma and death.
Explanation:
Methanol toxicity arises because its metabolites, especially formic acid, are highly toxic. The enzymes alcohol dehydrogenase and aldehyde dehydrogenase convert methanol stepwise into these harmful compounds. Formic acid causes metabolic acidosis and optic nerve damage, leading to blindness. Understanding this pathway helps in diagnosing and treating methanol poisoning effectively.
Methanol toxicity arises because its metabolites, especially formic acid, are highly toxic. The enzymes alcohol dehydrogenase and aldehyde dehydrogenase convert methanol stepwise into these harmful compounds. Formic acid causes metabolic acidosis and optic nerve damage, leading to blindness. Understanding this pathway helps in diagnosing and treating methanol poisoning effectively.
Question 16
How is the rate constant related to the activation energy and temperature?
Answer:
The rate constant k is related to activation energy Ea and temperature T by the Arrhenius equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where A is the Arrhenius factor (frequency factor), R is the gas constant, and T is the absolute temperature.
This equation shows that as temperature increases, the exponential term increases, resulting in a larger rate constant.
Similarly, a higher activation energy decreases the rate constant because fewer molecules have sufficient energy to overcome the energy barrier.
\[ k = A e^{-\frac{E_a}{RT}} \]
where A is the Arrhenius factor (frequency factor), R is the gas constant, and T is the absolute temperature.
This equation shows that as temperature increases, the exponential term increases, resulting in a larger rate constant.
Similarly, a higher activation energy decreases the rate constant because fewer molecules have sufficient energy to overcome the energy barrier.
Explanation:
The Arrhenius equation quantitatively relates the rate constant to temperature and activation energy.
It shows that the rate constant increases exponentially with temperature.
This is because higher temperature means more molecules have enough energy to overcome the activation barrier.
Conversely, a larger activation energy reduces the rate constant, slowing the reaction.
This relationship helps predict how reaction rates change with temperature and catalyst presence.
The Arrhenius equation quantitatively relates the rate constant to temperature and activation energy.
It shows that the rate constant increases exponentially with temperature.
This is because higher temperature means more molecules have enough energy to overcome the activation barrier.
Conversely, a larger activation energy reduces the rate constant, slowing the reaction.
This relationship helps predict how reaction rates change with temperature and catalyst presence.
Question 17
Arrange the following compounds in increasing order of their acidity: CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH.
Answer:
Step 1 Identify the compounds and their structures: 1. CH3CH2CH(Br)COOH (3-bromobutyric acid) 2. CH3CH(Br)CH2COOH (2-bromobutyric acid) 3. (CH3)2CHCOOH (isobutyric acid) 4. CH3CH2CH2COOH (butyric acid)
Step 2 Understand acidity factors: The acidity of carboxylic acids is influenced by electron withdrawing groups near the carboxyl group. Bromine is an electron withdrawing group and increases acidity by stabilizing the conjugate base. The closer the bromine to the carboxyl group, the stronger the acid. Alkyl groups like methyl are electron donating and decrease acidity.
Step 3 Analyze each compound: - 2-bromobutyric acid has bromine at the alpha position (adjacent to COOH), so highest acidity. - 3-bromobutyric acid has bromine at beta position, less effect than alpha. - Butyric acid has no substituents, baseline acidity. - Isobutyric acid has methyl groups at alpha carbon, electron donating, so least acidic.
Step 4 Arrange in increasing order of acidity: (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH2CH(Br)COOH < CH3CH(Br)CH2COOH
Step 2 Understand acidity factors: The acidity of carboxylic acids is influenced by electron withdrawing groups near the carboxyl group. Bromine is an electron withdrawing group and increases acidity by stabilizing the conjugate base. The closer the bromine to the carboxyl group, the stronger the acid. Alkyl groups like methyl are electron donating and decrease acidity.
Step 3 Analyze each compound: - 2-bromobutyric acid has bromine at the alpha position (adjacent to COOH), so highest acidity. - 3-bromobutyric acid has bromine at beta position, less effect than alpha. - Butyric acid has no substituents, baseline acidity. - Isobutyric acid has methyl groups at alpha carbon, electron donating, so least acidic.
Step 4 Arrange in increasing order of acidity: (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH2CH(Br)COOH < CH3CH(Br)CH2COOH
Explanation:
Acidity in carboxylic acids depends on the stability of the conjugate base formed after losing a proton. Electron withdrawing groups like bromine increase acidity by stabilizing the negative charge on the conjugate base. The closer the substituent to the carboxyl group, the stronger the effect. Alkyl groups donate electrons and reduce acidity. Thus, 2-bromobutyric acid is the most acidic because bromine is closest to COOH. 3-bromobutyric acid is next, then normal butyric acid, and isobutyric acid is least acidic due to electron donating methyl groups. This reasoning helps students understand how substituents influence acidity through inductive effects.
Acidity in carboxylic acids depends on the stability of the conjugate base formed after losing a proton. Electron withdrawing groups like bromine increase acidity by stabilizing the negative charge on the conjugate base. The closer the substituent to the carboxyl group, the stronger the effect. Alkyl groups donate electrons and reduce acidity. Thus, 2-bromobutyric acid is the most acidic because bromine is closest to COOH. 3-bromobutyric acid is next, then normal butyric acid, and isobutyric acid is least acidic due to electron donating methyl groups. This reasoning helps students understand how substituents influence acidity through inductive effects.
Question 18
132. Explain why molecular solids are generally electrical insulators.
Answer:
Molecular solids are made up of molecules held together by weak intermolecular forces such as dispersion forces, dipole-dipole interactions, or hydrogen bonds.
These molecules do not have free electrons or ions to carry charge.
Because the constituent particles are neutral molecules and there are no mobile charged particles, molecular solids cannot conduct electricity.
Hence, molecular solids are electrical insulators.
These molecules do not have free electrons or ions to carry charge.
Because the constituent particles are neutral molecules and there are no mobile charged particles, molecular solids cannot conduct electricity.
Hence, molecular solids are electrical insulators.
Explanation:
Molecular solids consist of molecules held by weak forces, not ionic or metallic bonds.
Since there are no free ions or electrons, electrical conduction is not possible.
The absence of mobile charged particles means they cannot carry electric current.
This explains why molecular solids are insulators.
Molecular solids consist of molecules held by weak forces, not ionic or metallic bonds.
Since there are no free ions or electrons, electrical conduction is not possible.
The absence of mobile charged particles means they cannot carry electric current.
This explains why molecular solids are insulators.
Question 19
Explain the difference in reactivity of alkyl halides towards nucleophilic substitution reactions based on the nature of alkyl group (primary, secondary, tertiary).
Answer:
The reactivity of alkyl halides towards nucleophilic substitution depends on whether the alkyl group is primary, secondary, or tertiary.
Primary alkyl halides favor S_N2 reactions because the carbon attached to the halogen is less sterically hindered, allowing the nucleophile to attack directly.
Secondary alkyl halides can undergo both S_N1 and S_N2 mechanisms depending on the reaction conditions and nucleophile strength.
Tertiary alkyl halides favor S_N1 reactions because the carbocation intermediate formed is stabilized by alkyl groups through inductive and hyperconjugation effects, but steric hindrance prevents S_N2 attack.
Thus, primary halides react mainly by S_N2, tertiary by S_N1, and secondary can react by either mechanism.
Primary alkyl halides favor S_N2 reactions because the carbon attached to the halogen is less sterically hindered, allowing the nucleophile to attack directly.
Secondary alkyl halides can undergo both S_N1 and S_N2 mechanisms depending on the reaction conditions and nucleophile strength.
Tertiary alkyl halides favor S_N1 reactions because the carbocation intermediate formed is stabilized by alkyl groups through inductive and hyperconjugation effects, but steric hindrance prevents S_N2 attack.
Thus, primary halides react mainly by S_N2, tertiary by S_N1, and secondary can react by either mechanism.
Explanation:
The nature of the alkyl group affects the mechanism and rate of nucleophilic substitution.
In primary alkyl halides, the nucleophile can easily approach the electrophilic carbon due to less steric hindrance, favoring a direct backside attack characteristic of S_N2.
In tertiary alkyl halides, the bulky alkyl groups hinder the nucleophile's approach, making S_N2 difficult; instead, the reaction proceeds via S_N1 involving carbocation formation.
Secondary alkyl halides are intermediate and can follow either pathway depending on conditions.
This explanation helps students understand how structure influences reaction pathways and rates.
The nature of the alkyl group affects the mechanism and rate of nucleophilic substitution.
In primary alkyl halides, the nucleophile can easily approach the electrophilic carbon due to less steric hindrance, favoring a direct backside attack characteristic of S_N2.
In tertiary alkyl halides, the bulky alkyl groups hinder the nucleophile's approach, making S_N2 difficult; instead, the reaction proceeds via S_N1 involving carbocation formation.
Secondary alkyl halides are intermediate and can follow either pathway depending on conditions.
This explanation helps students understand how structure influences reaction pathways and rates.
Question 20
The catalytic activity of a solid catalyst depends largely on
Answer:
B The surface area and availability of free valencies
Explanation:
The catalytic activity of a solid catalyst is primarily influenced by its surface properties rather than bulk properties like volume or color. The surface area determines how many reactant molecules can be adsorbed, and free valencies on the surface provide active sites for the reaction to occur. Solubility is generally not relevant for solid catalysts in heterogeneous catalysis. Therefore, the correct answer is option B.
The catalytic activity of a solid catalyst is primarily influenced by its surface properties rather than bulk properties like volume or color. The surface area determines how many reactant molecules can be adsorbed, and free valencies on the surface provide active sites for the reaction to occur. Solubility is generally not relevant for solid catalysts in heterogeneous catalysis. Therefore, the correct answer is option B.
