CUET UG Chemistry Booster Test - 3 Factors and Theories
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QUESTION 1 OF 20
Regarding the temperature and rate dynamics:
Statements:
1. The decomposition rate of N₂O₅ is identical at 0°C and 50°C.
2. A 10°C rise generally doubles the rate constant due to doubling the fraction of molecules with energy equal to or greater than Ea.
3. Potassium permanganate is decolourised slower at lower temperatures.
QUESTION 2 OF 20
Analytically, how does an increase in temperature specifically accelerate the reaction rate?
QUESTION 3 OF 20
Arrange the logical steps to derive the integrated Arrhenius equation for comparing two temperatures.
1. Rearrange to form log(k₂/k₁) = (Ea/2.303R) × ((T₂ − T₁)/(T₁T₂))
2. Take the natural logarithm of k = Ae^(-Ea/RT) to get ln k = -Ea/RT + ln A
3. Subtract the equation for T₁ from the equation for T₂
4. Substitute T₁, k₁ and T₂, k₂ into the logarithmic equation
QUESTION 4 OF 20
Match List-I with List-II:
| List I | List II |
|---|---|
| 1. y-axis | a. 1/T |
| 2. x-axis | b. ln A |
| 3. Slope | c. ln k |
| 4. Intercept | d. -Ea/R |
QUESTION 5 OF 20
What is the precise mathematical relationship defining Threshold Energy?
QUESTION 6 OF 20
According to the table of rate constant units, what is the unit of a first order rate constant?
QUESTION 7 OF 20
Identify the thermodynamic reaction condition based on the potential energy diagram if energy is released when the complex decomposes to form products:
QUESTION 8 OF 20
On an energy profile diagram, what dictates the final enthalpy of the reaction?
QUESTION 9 OF 20
The specific statistical framework plotting the fraction of molecules versus kinetic energy was developed by which two scientists?
QUESTION 10 OF 20
Statements on the Maxwell-Boltzmann distribution at changing temperatures:
1. The area under the curve must be constant since total probability must be one.
2. The total area under the entire curve doubles when temperature increases by 10 degrees.
3. Broadening to the right indicates a greater proportion of molecules with much higher energies.
QUESTION 11 OF 20
According to intermediate complex theory, what happens to the catalyst after the intermediate complex decomposes?
QUESTION 12 OF 20
Match Catalyst Properties (List-I) with their effects (List-II).
| List I | List II |
|---|---|
| 1. Gibbs energy | a. Does not change |
| 2. Non-spontaneous reactions | b. Reduces potential energy barrier |
| 3. Equilibrium Constant | c. Does not catalyse |
| 4. Alternate pathway | d. Remains unaltered |
QUESTION 13 OF 20
Based on the passage, the collision theory provides deeper insight specifically into which aspects of a reaction?
QUESTION 14 OF 20
Identify the theory type foundational to Trautz and Lewis's collision theory as mentioned in the passage.
QUESTION 15 OF 20
During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, what is the exact amount of water remaining at completion?
QUESTION 16 OF 20
Arrange the molecular formulas sequentially as they appear from left to right in the chemical equation for the inversion of cane sugar.
1. C₁₂H₂₂O₁₁
2. H₂O
3. C₆H₁₂O₆ (Glucose)
4. C₆H₁₂O₆ (Fructose)
QUESTION 17 OF 20
When a catalyst lowers the activation energy for the forward reaction by providing an alternate pathway, what is its effect on the reverse reaction?
QUESTION 18 OF 20
Identify the type of complex described as having a "transitory existence".
QUESTION 19 OF 20
Identify the correct statements regarding the criteria for effective collisions.
Statements:
1. Proper orientation is required to facilitate breaking of old bonds and formation of new bonds.
2. The steric factor accounts for the fact that molecules must be properly oriented during collision.
3. Molecules must have kinetic energy strictly less than threshold energy.
4. Effective collisions require proper orientation.
QUESTION 20 OF 20
In the formation of methanol from bromoethane, if molecules bounce back without forming products, it is termed:
Test Complete!
Answer Review
1 Regarding the temperature and rate dynamics:
Statements:
1. The decomposition rate of N₂O₅ is identical at 0°C and 50°C.
2. A 10°C rise generally doubles the rate constant due to doubling the fraction of molecules with energy equal to or greater than Ea.
3. Potassium permanganate is decolourised slower at lower temperatures.
�� Higher temperature increases reaction rate. �� More molecules acquire activation energy. �� Reactions proceed slower at lower temperatures.
According to the Arrhenius equation, reaction rate increases with temperature because a greater fraction of molecules possess energy equal to or greater than the activation energy (Ea). As a rule of thumb, a 10°C rise in temperature approximately doubles the rate constant for many reactions. Potassium permanganate decolourisation occurs more slowly at lower temperatures because fewer molecules possess sufficient activation energy for effective collisions. Statement 1 is incorrect because the decomposition rate of N₂O₅ is not identical at 0°C and 50°C. Higher temperature results in a faster decomposition rate. Therefore, Statements 2 and 3 are correct. Hence, the correct answer is A.
- �� Option B → Statement 1 is incorrect.
- �� Option C → Statement 1 is incorrect and Statement 2 is correct.
- �� Option D → Includes incorrect Statement 1.
Used: Elimination
- Application
- Identify incorrect statements using temperature dependence of reaction rates.
- Final Logic
- Higher temperature increases reaction rate; therefore Statement 1 is false.
- Hotter → Faster
2 Analytically, how does an increase in temperature specifically accelerate the reaction rate?
�� Temperature increases molecular kinetic energy. �� More molecules cross the activation barrier. �� Effective collisions increase.
According to Maxwell-Boltzmann distribution and Arrhenius theory, increasing temperature increases the fraction of molecules possessing kinetic energy greater than the activation energy (Ea). These molecules undergo effective collisions, resulting in product formation. Since more molecules can overcome the activation energy barrier, the reaction rate increases significantly. Temperature does not directly lower Ea; instead, it increases the number of molecules capable of crossing the existing energy barrier. Hence, the correct answer is C.
- �� Option A → Catalysts lower activation energy, not temperature.
- �� Option B → Temperature does not increase the number of molecules.
- �� Option D → Reaction rate is not accelerated by changing Gibbs energy directly.
Used: NCERT Recall
- Application
- Recall the effect of temperature on molecular energy distribution.
- Final Logic
- Higher temperature increases the fraction of molecules with energy ≥ Ea.
- More Energy → More Effective Collisions
3 Arrange the logical steps to derive the integrated Arrhenius equation for comparing two temperatures.
1. Rearrange to form log(k₂/k₁) = (Ea/2.303R) × ((T₂ − T₁)/(T₁T₂))
2. Take the natural logarithm of k = Ae^(-Ea/RT) to get ln k = -Ea/RT + ln A
3. Subtract the equation for T₁ from the equation for T₂
4. Substitute T₁, k₁ and T₂, k₂ into the logarithmic equation
�� Start with the Arrhenius equation. �� Convert it into logarithmic form. �� Compare rate constants at two temperatures.
The integrated Arrhenius equation is derived by first taking the natural logarithm of the Arrhenius equation, , to obtain a linear logarithmic form. Next, values corresponding to two different temperatures (T₁ and T₂) and their respective rate constants (k₁ and k₂) are substituted into the logarithmic equation. The resulting equations are then subtracted to eliminate the constant term ln A. Finally, the expression is rearranged and converted into common logarithmic form to obtain: log(k₂/k₁) = (Ea/2.303R) × ((T₂ − T₁)/(T₁T₂)) Thus, the correct sequence of steps is: 2 → 4 → 3 → 1 Hence, the correct answer is D.
- �� Option A → The subtraction step cannot be the first step in the derivation.
- �� Option B → Substitution cannot be performed before obtaining the logarithmic form.
- �� Option C → Subtraction cannot occur before substituting values for both temperatures.
Logical Analysis
- Application
- Identify the chronological order of mathematical operations used in deriving the integrated Arrhenius equation.
- Final Logic
- Logarithm → Substitute → Subtract → Rearrange.
"Log → Plug → Subtract → Rearrange"
4 Match List-I with List-II:
| List I | List II |
|---|---|
| 1. y-axis | a. 1/T |
| 2. x-axis | b. ln A |
| 3. Slope | c. ln k |
| 4. Intercept | d. -Ea/R |
�� Arrhenius plot is ln k versus 1/T. �� Slope equals -Ea/R. �� Intercept equals ln A.
The Arrhenius equation is expressed as: ln k = -Ea/RT + ln A This equation has the same form as the straight-line equation: y = mx + c Comparing the two equations: • y corresponds to ln k • x corresponds to 1/T • Slope (m) corresponds to -Ea/R • Intercept (c) corresponds to ln A Therefore, the correct matching is: 1. y-axis → c. ln k 2. x-axis → a. 1/T 3. Slope → d. -Ea/R 4. Intercept → b. ln A An Arrhenius plot of ln k against 1/T gives a straight line whose slope is negative because the activation energy term carries a negative sign. The intercept provides the frequency factor (A) information. This graphical representation is widely used to determine activation energy experimentally.
- �� Option A → The y-axis and x-axis are interchanged, and slope and intercept are incorrectly matched.
- �� Option C → Intercept is incorrectly matched with 1/T instead of ln A.
- �� Option D → The y-axis is incorrectly assigned to the slope term.
NCERT Recall
- Application
- Compare the Arrhenius equation directly with the linear equation y = mx + c.
- Final Logic
- ln k acts as y, 1/T acts as x, slope is -Ea/R, and intercept is ln A.
- x = 1/T
5 What is the precise mathematical relationship defining Threshold Energy?
�� Threshold energy is the minimum energy required for effective collision. �� It is related to activation energy. �� Molecules already possess some energy before collision.
Threshold energy is defined as the minimum kinetic energy that colliding molecules must possess for a reaction to occur. Mathematically: Threshold Energy = Activation Energy + Energy already possessed by reactant molecules Only molecules having energy equal to or greater than threshold energy can undergo effective collisions and form products. Therefore, threshold energy exceeds activation energy by the amount of energy already possessed by reacting species. Hence, the correct answer is A.
- �� Option B → Threshold energy is not exactly equal to activation energy.
- �� Option C → Gibbs free energy is unrelated to this definition.
- �� Option D → Product potential energy is not part of the threshold energy expression.
Used: NCERT Recall
- Application
- Recall the definition of threshold energy from collision theory.
- Final Logic
- Threshold energy includes activation energy plus existing molecular energy.
- Threshold = Existing Energy + Ea
6 According to the table of rate constant units, what is the unit of a first order rate constant?
�� First order rate constant depends only on time. �� Concentration units cancel during derivation. �� Unit is reciprocal of time.
For a first order reaction: Rate = k[R] Rate has units of mol L⁻¹ s⁻¹ and concentration has units of mol L⁻¹. Therefore, k = (mol L⁻¹ s⁻¹)/(mol L⁻¹) = s⁻¹ Thus, the rate constant of a first order reaction has units of reciprocal time. This is a characteristic feature of first order kinetics and is independent of the concentration unit used. Hence, the correct answer is C. s⁻¹.
- �� Option A → This is the unit of reaction rate, not the first order rate constant.
- �� Option B → This is the unit of a second order rate constant.
- �� Option D → Pressure units are not used for the standard first order rate constant.
Used: Dimensional/Unit Analysis
- Application
- Determine the unit of k from the rate law expression.
- Final Logic
- For first order reactions, concentration units cancel, leaving s⁻¹.
- First Order → First Power of Time⁻¹
7 Identify the thermodynamic reaction condition based on the potential energy diagram if energy is released when the complex decomposes to form products:
�� Products are formed with lower energy. �� Energy is released to surroundings. �� ΔH is negative.
In a potential energy diagram, when the activated complex decomposes and releases energy while forming products, the products lie at a lower energy level than the reactants. This energy difference is released to the surroundings and the enthalpy change (ΔH) becomes negative. Such reactions are known as exothermic reactions. Therefore, if energy is released during product formation, the reaction is exothermic. Hence, the correct answer is D. Exothermic reaction.
- �� Option A → Endothermic reactions absorb energy.
- �� Option B → Reversibility is unrelated to energy release in the diagram.
- �� Option C → Isothermic refers to constant temperature, not energy release.
Used: Concept Application
- Application
- Interpret the energy profile and compare reactant and product energy levels.
- Final Logic
- Products at lower energy indicate an exothermic process.
- Energy Out = Exothermic
8 On an energy profile diagram, what dictates the final enthalpy of the reaction?
�� Enthalpy change depends on initial and final states. �� Catalyst does not alter ΔH. �� Reactants and products determine energy difference.
The final enthalpy change of a reaction depends solely on the energy difference between reactants and products. Since enthalpy is a state function, it depends only on the initial and final states and not on the reaction pathway. Therefore, the nature of the reactants and products determines the final enthalpy of reaction. Catalysts may lower activation energy but do not affect ΔH. Hence, the correct answer is B.
- �� Option A → The peak represents the activated complex, not final enthalpy.
- �� Option C → Threshold energy does not determine ΔH.
- �� Option D → Catalysts do not change enthalpy.
Used: NCERT Recall
- Application
- Recall that enthalpy is a state function.
- Final Logic
- ΔH depends only on reactants and products.
- ΔH = Start vs End
9 The specific statistical framework plotting the fraction of molecules versus kinetic energy was developed by which two scientists?
�� Describes molecular energy distribution. �� Forms the Maxwell-Boltzmann distribution. �� Widely used in kinetic theory.
The Maxwell-Boltzmann distribution describes how molecular kinetic energies are distributed among a large number of molecules. This statistical model was developed through the contributions of James Clerk Maxwell and Ludwig Boltzmann. The distribution curve helps explain reaction rates, activation energy concepts, and the effect of temperature on chemical reactions. Therefore, the correct answer is B. Ludwig Boltzmann and James Clark Maxwell.
- �� Option A → Arrhenius and van't Hoff did not develop the kinetic energy distribution law.
- �� Option C → Trautz and Lewis contributed to collision theory.
- �� Option D → Le Chatelier is known for Le Chatelier's principle.
Used: NCERT Recall
- Application
- Recall the scientists associated with the molecular energy distribution curve.
- Final Logic
- The distribution is named after Maxwell and Boltzmann.
- MB Curve = Maxwell + Boltzmann
10 Statements on the Maxwell-Boltzmann distribution at changing temperatures:
1. The area under the curve must be constant since total probability must be one.
2. The total area under the entire curve doubles when temperature increases by 10 degrees.
3. Broadening to the right indicates a greater proportion of molecules with much higher energies.
�� Total probability remains unity. �� Area under the curve remains constant. �� Higher temperature shifts molecules toward higher energies.
In a Maxwell-Boltzmann distribution, the total area under the curve represents total probability and must always remain equal to one. Therefore, Statement 1 is correct. When temperature increases, the distribution broadens and shifts toward higher kinetic energies. This means a larger fraction of molecules possess high energies, making Statement 3 correct. However, the total area under the curve does not double with temperature increase; it remains constant. Hence Statement 2 is incorrect. Therefore, Statements 1 and 3 are correct. Hence, the correct answer is C. 1 and 3 are correct.
- �� Option A → Statement 2 is incorrect.
- �� Option B → Statement 1 is also correct.
- �� Option D → Includes incorrect Statement 2.
Used: Elimination
- Application
- Check which statements agree with the properties of probability distribution curves.
- Final Logic
- Area remains constant while high-energy fraction increases.
- Area Same, Energy Range Wider
11 According to intermediate complex theory, what happens to the catalyst after the intermediate complex decomposes?
�� Catalyst participates in the reaction temporarily. �� An intermediate complex is formed during catalysis. �� Catalyst is regenerated after decomposition of the intermediate complex.
According to the intermediate complex theory of catalysis, a catalyst first combines with one or more reactant molecules to form an intermediate complex. This intermediate complex is unstable and has a transitory existence. It provides an alternative reaction pathway having lower activation energy than the uncatalysed reaction. After the reaction proceeds through this pathway, the intermediate complex decomposes to form the products. During this decomposition, the catalyst is regenerated in its original form and is yielded along with the products. Thus, the catalyst is not consumed during the reaction and remains chemically unchanged at the end of the process. This is one of the most important characteristics of catalysts discussed in NCERT. The regeneration of the catalyst allows a small quantity of catalyst to convert a large amount of reactants into products. Therefore, the correct statement is that the catalyst is yielded along with the products after the intermediate complex decomposes.
- �� Option A → A catalyst does not become part of the final product because it is regenerated after the reaction.
- �� Option B → A catalyst is not consumed permanently during the reaction.
- �� Option D → Evaporation is not related to the catalytic mechanism described by intermediate complex theory.
NCERT Recall
- Application
- Recall the key feature of catalysts from NCERT: catalysts participate in reactions but are regenerated at the end.
- Final Logic
- The intermediate complex decomposes into products and releases the catalyst unchanged. Therefore, the catalyst is yielded along with the products.
"Complex breaks, catalyst wakes."
12 Match Catalyst Properties (List-I) with their effects (List-II).
| List I | List II |
|---|---|
| 1. Gibbs energy | a. Does not change |
| 2. Non-spontaneous reactions | b. Reduces potential energy barrier |
| 3. Equilibrium Constant | c. Does not catalyse |
| 4. Alternate pathway | d. Remains unaltered |
�� Catalysts do not change Gibbs energy. �� Catalysts cannot make non-spontaneous reactions occur. �� Equilibrium constant remains unchanged. �� Catalysts provide an alternate pathway.
Catalysts affect only the rate at which equilibrium is attained and do not alter the thermodynamic parameters of a reaction. The Gibbs energy change of a reaction remains unaltered in the presence of a catalyst. Similarly, catalysts cannot catalyse non-spontaneous reactions because spontaneity depends on thermodynamic feasibility and not on reaction rate. The equilibrium constant also remains unchanged because it depends only on temperature and the thermodynamic properties of the system. A catalyst merely lowers the activation energy by providing an alternate reaction pathway with a lower energy barrier. This alternate pathway increases the rate of both forward and backward reactions equally. Consequently, equilibrium is reached faster, but the position of equilibrium remains unchanged. These principles are clearly explained in NCERT under the topic of catalysis and catalyst mechanism.
- �� Option B → Gibbs energy is incorrectly matched with reduction of activation energy.
- �� Option C → Non-spontaneous reactions are incorrectly matched with Gibbs energy effect.
- �� Option D → Multiple catalyst properties are assigned incorrect effects.
Concept Application
- Application
- Apply fundamental properties of catalysts and identify the effect associated with each property.
- Final Logic
- Catalysts lower activation energy, do not affect Gibbs energy or equilibrium constant, and cannot catalyse non-spontaneous reactions.
A → Alternate pathway
13
Based on the passage, the collision theory provides deeper insight specifically into which aspects of a reaction?
�� Collision theory explains reaction mechanisms. �� It considers molecular collisions. �� It provides energetic interpretation of reactions.
Collision theory was developed by Max Trautz and William Lewis and is based on the kinetic theory of gases. According to this theory, reactant molecules are treated as hard spheres and reactions occur when molecules collide effectively. The theory provides a deeper understanding of why reactions occur at different rates by considering both the energy possessed by colliding molecules and the orientation during collision. Hence, collision theory offers insight into the energetic aspects, such as activation energy and threshold energy, and mechanistic aspects, such as how molecules collide and rearrange to form products. This makes it more informative than simply using the Arrhenius equation, which primarily describes the temperature dependence of reaction rates. NCERT specifically states that collision theory provides greater insight into the energetic and mechanistic aspects of reactions.
- �� Option A → Collision theory does not primarily explain equilibrium or thermodynamic parameters.
- �� Option B → Nuclear and isotopic processes are unrelated to collision theory.
- �� Option C → Pseudo-order reactions and catalysis are separate concepts.
NCERT Recall
- Application
- Recall the exact NCERT statement regarding collision theory.
- Final Logic
- NCERT explicitly states that collision theory provides insight into energetic and mechanistic aspects of reactions.
"Collision = Energy + Mechanism"
14
Identify the theory type foundational to Trautz and Lewis's collision theory as mentioned in the passage.
�� Collision theory is based on molecular motion. �� Molecules are assumed to behave as hard spheres. �� The foundation is kinetic theory.
The collision theory developed by Max Trautz and William Lewis is based on the kinetic theory of gases. The kinetic theory explains the behaviour of gas molecules in terms of their continuous random motion and collisions. Collision theory extends these ideas to chemical reactions by assuming that reactant molecules behave as hard spheres. According to this theory, reactions occur only when molecules collide with sufficient energy and proper orientation. The frequency and effectiveness of collisions determine the reaction rate. Thus, the kinetic theory of gases provides the theoretical framework upon which collision theory is constructed. NCERT clearly mentions that collision theory is based on the kinetic theory of gases and uses its principles to explain chemical reaction rates.
- �� Option A → Quantum theory is not identified as the basis of collision theory in NCERT.
- �� Option C → Thermodynamics explains feasibility, not collision-based reaction rates.
- �� Option D → Intermediate complex theory explains catalysis, not collision theory.
NCERT Recall
- Application
- Remember the theoretical foundation explicitly mentioned in NCERT.
- Final Logic
- Collision theory directly originates from the kinetic theory of gases.
"Kinetic Motion → Collision Notion"
15 During the hydrolysis of 0.01 mol of ethyl acetate with 10 mol of water, what is the exact amount of water remaining at completion?
�� Hydrolysis consumes water. �� Water is present in large excess. �� Only 0.01 mol water reacts.
The hydrolysis reaction of ethyl acetate can be represented as: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH The stoichiometric ratio between ethyl acetate and water is 1:1. Therefore, 0.01 mol of ethyl acetate requires 0.01 mol of water for complete hydrolysis. Initially, 10 mol of water is present. Since only 0.01 mol water participates in the reaction, the remaining amount of water is: Water remaining = Initial water − Water consumed = 10 − 0.01 = 9.99 mol Because water is present in very large excess, its concentration remains nearly constant throughout the reaction. This is the basis of pseudo-first-order kinetics discussed in NCERT. The reaction appears to follow first-order kinetics even though it is fundamentally a second-order reaction. Therefore, the exact amount of water remaining after completion is 9.99 mol.
- �� Option A → All water is not consumed because water is present in huge excess.
- �� Option B → This represents the amount consumed, not the amount remaining.
- �� Option D → Water cannot increase during the reaction.
Formula Application
- Application
- Use stoichiometric mole relationships and subtraction of consumed reactant from the initial amount.
- Final Logic
- 10 mol − 0.01 mol = 9.99 mol water remaining.
"Excess water hardly changes."
16 Arrange the molecular formulas sequentially as they appear from left to right in the chemical equation for the inversion of cane sugar.
1. C₁₂H₂₂O₁₁
2. H₂O
3. C₆H₁₂O₆ (Glucose)
4. C₆H₁₂O₆ (Fructose)
�� Cane sugar undergoes hydrolysis. �� Water participates as a reactant. �� Glucose and fructose are formed as products.
The inversion of cane sugar refers to the hydrolysis of sucrose in the presence of water. The chemical equation is: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ + C₆H₁₂O₆ Here, sucrose (cane sugar) reacts with water to produce glucose and fructose. When arranged from left to right exactly as they appear in the equation, the sequence is sucrose, water, glucose and fructose. This reaction is commonly used in NCERT to explain pseudo-first-order kinetics because water is generally present in large excess. Due to the excess concentration of water, its concentration remains nearly constant during the reaction, causing the reaction to behave like a first-order reaction even though it is actually bimolecular. Understanding the correct order of reactants and products is important for interpreting the reaction mechanism and kinetic treatment of inversion of cane sugar.
- �� Option B → Begins with products instead of reactants.
- �� Option C → Places water before sucrose, which does not match the equation.
- �� Option D → Places products before completion of reactants.
NCERT Recall
- Application
- Recall the exact hydrolysis equation of sucrose given in NCERT.
- Final Logic
- Reactants appear first, followed by products: sucrose + water → glucose + fructose.
"Sugar + Water Gives Sweet Twins"
17 When a catalyst lowers the activation energy for the forward reaction by providing an alternate pathway, what is its effect on the reverse reaction?
�� Catalysts lower activation energy. �� Forward and reverse reactions are both affected. �� Equilibrium position remains unchanged.
A catalyst provides an alternative reaction pathway with lower activation energy. This reduction in activation energy increases the rate of the forward reaction. However, the catalyst does not selectively act only on the forward reaction. It also lowers the activation energy of the reverse reaction by the same amount. As a result, both forward and backward reactions are accelerated equally. Since both reaction rates increase proportionately, the equilibrium constant and equilibrium position remain unchanged. The only effect is that equilibrium is reached more rapidly. NCERT clearly states that catalysts do not affect the thermodynamic properties of a reaction or alter equilibrium composition. Therefore, when a catalyst lowers the activation energy for the forward reaction, it catalyzes the reverse reaction to the same extent.
- �� Option A → Equilibrium state and equilibrium constant remain unchanged.
- �� Option B → Catalysts do not inhibit reverse reactions.
- �� Option C → Catalysts affect both forward and reverse reactions.
Concept Application
- Application
- Apply the NCERT principle that catalysts accelerate both directions equally.
- Final Logic
- Since activation energy decreases for both pathways, both forward and backward reactions are catalyzed equally.
"Fast Forward, Fast Reverse"
18 Identify the type of complex described as having a "transitory existence".
�� Intermediate complexes are short-lived. �� They form during catalytic reactions. �� They decompose rapidly into products.
According to the intermediate complex theory of catalysis, the catalyst first combines with reactants to form an intermediate complex. This complex is unstable and exists only for a very short duration. Because of its temporary nature, NCERT describes it as having a transitory existence. The intermediate complex provides an alternative reaction pathway with lower activation energy. After formation, it rapidly decomposes into products while regenerating the catalyst. The transitory existence of the intermediate complex is essential because a stable complex would prevent product formation and reduce catalytic efficiency. Therefore, the term "intermediate complex" specifically refers to the short-lived species formed during catalytic reactions.
- �� Option A → Stable products are formed at the end of the reaction and are not transitory.
- �� Option C → Reactants exist before the reaction and are not intermediate species.
- �� Option D → An inhibited complex is not a standard NCERT term in catalyst mechanism.
NCERT Recall
- Application
- Recall the description of intermediate complexes in the catalyst mechanism.
- Final Logic
- The only species described as having a transitory existence is the intermediate complex.
"Intermediate = In Between = Temporary"
19 Identify the correct statements regarding the criteria for effective collisions.
Statements:
1. Proper orientation is required to facilitate breaking of old bonds and formation of new bonds.
2. The steric factor accounts for the fact that molecules must be properly oriented during collision.
3. Molecules must have kinetic energy strictly less than threshold energy.
4. Effective collisions require proper orientation.
�� Orientation affects product formation. �� Steric factor measures orientation effects. �� Threshold energy must be attained or exceeded.
Collision theory states that not all molecular collisions lead to product formation. For a collision to be effective, molecules must possess sufficient kinetic energy equal to or greater than the threshold energy and must collide with proper orientation. Proper orientation allows existing bonds to break and new bonds to form efficiently. The steric factor accounts for the probability that molecules are oriented correctly during collision. Statement 1 is correct because orientation is necessary for bond rearrangement. Statement 2 is correct because steric factor represents orientation requirements. Statement 4 is also correct because proper orientation is one of the conditions for effective collision. Statement 3 is incorrect because molecules must have kinetic energy equal to or greater than the threshold energy, not less than it. Therefore, statements 1, 2 and 4 are correct.
- �� Option B → Includes Statement 3, which is incorrect.
- �� Option C → Includes Statement 3 and excludes Statement 2.
- �� Option D → Incorrect because Statement 3 is false.
Logical Analysis
- Application
- Evaluate each statement using the conditions required for effective molecular collisions.
- Final Logic
- Effective collisions require sufficient energy and proper orientation; therefore Statements 1, 2 and 4 are correct.
"Energy + Angle = Reaction"
20 In the formation of methanol from bromoethane, if molecules bounce back without forming products, it is termed:
�� Orientation determines reaction success. �� Incorrect alignment prevents bond formation. �� Molecules may simply rebound after collision.
According to collision theory, molecules must collide with both sufficient energy and proper orientation for a reaction to occur. In reactions such as the formation of methanol from bromoethane, the reacting molecules must approach each other in a specific arrangement so that old bonds can break and new bonds can form. If the molecules collide with improper orientation, the required atomic interactions do not occur. As a result, no products are formed, and the molecules simply bounce back after collision. Such collisions are termed ineffective collisions due to improper orientation. NCERT explains that the steric factor accounts for this orientation requirement. Therefore, when molecules rebound without product formation because of incorrect alignment, the phenomenon is called improper orientation.
- �� Option A → Proper orientation leads to a greater probability of reaction.
- �� Option B → Steric hindrance refers to crowding effects and is not the term used here.
- �� Option D → Effective collisions produce products and do not result in rebound.
Concept Application
- Application
- Apply the collision theory requirement of proper molecular orientation.
- Final Logic
- No product formation despite collision indicates that the molecules collided with improper orientation.
"Wrong Angle, No Reaction"
