Collision Theory: What are the two main barriers—energy and orientation—that determine an effective collision

Chapter: Chemical Kinetics

Topic: Collision Theory – Energy and Orientation Barriers

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What determines an Effective Collision?

According to Collision Theory, molecules must overcome two main barriers for a collision to produce a chemical reaction:

1. Energy Barrier

Collision must have sufficient energy.

+

2. Orientation Barrier

Molecules must collide in the correct orientation.

Effective Collision = Sufficient Energy + Proper Orientation

Collision Theory states that a chemical reaction occurs only when reacting particles collide with each other effectively.

However, every collision does not result in a reaction. A collision must satisfy two conditions:

1. The particles must possess sufficient energy.
2. The particles must have proper relative orientation.

Only collisions satisfying both conditions are called Effective Collisions.

Energy Requirement

Reacting molecules must possess a minimum amount of energy to overcome the energy barrier and form the activated complex.

Energy of Collision ≥ Activation Energy (E_a)

If the collision energy is less than the activation energy, the molecules collide but return to their original state.

Insufficient Energy

Collision occurs

❌ No reaction

Sufficient Energy

Collision occurs

✅ Reaction possible

Example: A molecule may collide with another molecule, but if its kinetic energy is below E_a, the collision will not produce products.

Orientation Requirement

Even when molecules have sufficient energy, the collision may fail if the reacting groups are not properly oriented.

Correct Orientation → Effective Collision

Incorrect Orientation → Ineffective Collision

The atoms or functional groups that participate in bond breaking and bond formation must approach one another in a suitable geometrical arrangement.

The orientation requirement is especially important for complex molecules because only certain parts of the molecules can participate in the reaction.

Condition	Requirement	Result
Energy	Energy ≥ Ea	Collision can overcome energy barrier
Orientation	Correct molecular orientation	Reactive atoms can interact

Energy Condition + Orientation Condition = Effective Collision

Case 1: Low Energy

The molecules collide, but their energy is insufficient to cross the activation-energy barrier.

Result: No Reaction

Case 2: Wrong Orientation

The molecules possess sufficient energy but approach each other in an unfavourable orientation.

Result: No Reaction

Case 3: Both Conditions Satisfied

The molecules have sufficient energy and proper orientation.

Result: Effective Collision → Reaction

When temperature increases, the average kinetic energy of molecules increases.

Therefore, a larger fraction of molecules possesses energy equal to or greater than the activation energy.

Temperature ↑ → Kinetic Energy ↑ → Effective Collisions ↑

As the number of effective collisions increases, the reaction rate increases.

Higher Temperature → More Effective Collisions → Faster Reaction

A catalyst provides an alternative reaction pathway with lower activation energy.

E_a(catalysed) < E_a(uncatalysed)

Because the activation-energy barrier is lower, a larger fraction of collisions can become effective.

Catalyst → Lower E_a → More Effective Collisions → Faster Reaction

Q. What are the two main conditions required for an effective collision?

Answer:

1. The colliding molecules must possess energy equal to or greater than the activation energy.
2. The molecules must collide with proper orientation.

Both sufficient energy and proper orientation are necessary for an effective collision.

Q. Explain the role of energy and orientation in an effective collision.

According to collision theory, a reaction occurs only when molecules undergo effective collisions.

1. Energy: The colliding molecules must possess energy equal to or greater than the activation energy.

2. Orientation: The molecules must approach each other in a proper orientation so that the reacting atoms or groups can interact.

Sufficient Energy + Proper Orientation = Effective Collision

Q. Explain collision theory and the two barriers that determine an effective collision.

Collision theory states that molecules must collide to undergo a chemical reaction. However, every collision is not effective.

Energy Barrier

The molecules must possess energy equal to or greater than the activation energy.

E ≥ E_a

Orientation Barrier

The molecules must collide with proper orientation. The reactive parts must face each other in a suitable arrangement.

Correct Orientation → Reaction Possible

Both conditions must be satisfied simultaneously.

Q. Describe the two main barriers involved in an effective collision according to collision theory.

1. Energy Barrier

Reacting molecules must have sufficient kinetic energy to cross the activation-energy barrier.

Collision Energy ≥ E_a

Collisions having insufficient energy are ineffective.

2. Orientation Barrier

Even if the molecules have sufficient energy, they must approach each other in the correct orientation.

Incorrect orientation prevents the appropriate bonds from breaking or forming.

3. Effective Collision

Effective Collision = Sufficient Energy + Correct Orientation

Thus, both requirements are essential for a chemical reaction to occur.

Q. Explain collision theory in detail. Discuss the energy and orientation barriers and explain their effect on the rate of a chemical reaction.

1. Collision Theory

Collision theory states that reactant molecules must collide with each other for a reaction to occur.

But all collisions do not produce products. Only effective collisions lead to chemical transformation.

2. Energy Barrier

The colliding particles must have sufficient energy to overcome the activation-energy barrier.

E_collision ≥ E_a

If the energy is insufficient, the collision is ineffective.

3. Orientation Barrier

The colliding particles must also possess the correct orientation. The reactive atoms or groups must approach each other in a favourable geometrical arrangement.

4. Effective Collision

Sufficient Energy + Correct Orientation = Effective Collision

5. Effect of Temperature

An increase in temperature increases molecular kinetic energy and the fraction of molecules having energy ≥ E_a. Thus, the number of effective collisions increases.

T ↑ → Effective Collisions ↑ → Rate ↑

6. Effect of Catalyst

A catalyst lowers the activation energy by providing an alternative reaction pathway. Hence, more collisions become effective.

Conclusion:

The two essential requirements for an effective collision are sufficient energy and proper orientation. Both must be satisfied for a collision to result in a chemical reaction.

1. According to collision theory, a reaction occurs when molecules:

A) Only come close to each other

B) Collide effectively

C) Have low energy

D) Stop moving

✅ Answer

B) Collide effectively

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2. An effective collision requires:

A) Only high pressure

B) Only correct orientation

C) Sufficient energy and proper orientation

D) Low temperature

✅ Answer

C

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3. The minimum energy required for a reaction is called:

A) Kinetic energy

B) Activation energy

C) Potential energy

D) Ionisation energy

✅ Answer

B) Activation energy

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4. A collision with energy less than E_a is:

A) Effective

B) Ineffective

C) Explosive

D) Catalytic

✅ Answer

B) Ineffective

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5. Correct orientation means:

A) Molecules do not collide

B) Reactive groups approach suitably

C) Molecules have zero energy

D) Molecules move randomly without collision

✅ Answer

B

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6. Which is NOT a condition for an effective collision?

A) Sufficient energy

B) Proper orientation

C) Correct collision

D) Zero kinetic energy

✅ Answer

D) Zero kinetic energy

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7. Increasing temperature generally:

A) Decreases effective collisions

B) Increases effective collisions

C) Stops collisions

D) Eliminates activation energy

✅ Answer

B

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8. A catalyst increases reaction rate mainly by:

A) Increasing reactant mass

B) Lowering activation energy

C) Increasing product mass

D) Removing orientation

✅ Answer

B

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9. If energy is sufficient but orientation is wrong, the collision is:

A) Effective

B) Ineffective

C) Always explosive

D) Catalysed

✅ Answer

B) Ineffective

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10. If orientation is correct but energy is insufficient:

A) Reaction occurs

B) Reaction does not occur

C) Catalyst is formed

D) Temperature becomes zero

✅ Answer

B

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11. Which two factors determine an effective collision?

A) Pressure and volume

B) Energy and orientation

C) Mass and volume

D) Temperature and pressure only

✅ Answer

B

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12. Collision theory is mainly concerned with:

A) Atomic mass

B) Molecular collisions

C) Nuclear stability

D) Periodic trends

✅ Answer

B

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13. The rate of reaction increases when the number of effective collisions:

A) Decreases

B) Increases

C) Becomes zero

D) Remains zero

✅ Answer

B

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14. Activation energy is associated with:

A) Energy barrier

B) Orientation only

C) Volume

D) Pressure only

✅ Answer

A

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15. Proper orientation is especially important for:

A) All collisions being automatically effective

B) Formation and breaking of specific bonds

C) Eliminating kinetic energy

D) Reducing temperature

✅ Answer

B

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16. Higher temperature increases reaction rate because:

A) Molecules stop moving

B) More molecules can overcome E_a

C) E_a becomes zero

D) Orientation becomes unnecessary

✅ Answer

B

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17. A catalyst provides:

A) A higher-energy pathway

B) A lower-energy alternative pathway

C) No reaction pathway

D) Zero collisions

✅ Answer

B

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18. Which statement is correct?

A) Every collision causes reaction

B) Only effective collisions cause reaction

C) Energy is never important

D) Orientation is never important

✅ Answer

B

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19. Effective collision can be represented as:

A) Low energy + wrong orientation

B) High energy + wrong orientation

C) Sufficient energy + proper orientation

D) Zero energy + proper orientation

✅ Answer

C

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20. The two main barriers in collision theory are:

A) Energy and orientation

B) Mass and pressure

C) Volume and temperature

D) Density and mass

✅ Answer

A) Energy and orientation

🎯 Quick Revision

Collision Theory: Reacting molecules must collide effectively.
Energy Barrier: Collision energy ≥ Activation Energy
Orientation Barrier: Reacting molecules must have proper orientation.
Effective Collision: Sufficient Energy + Proper Orientation
Temperature: T ↑ → Effective Collisions ↑ → Rate ↑
Catalyst: E_a ↓ → More Effective Collisions → Rate ↑

Activation Energy: Explain the Temperature Dependence of reaction rates using the Arrhenius equation

Chapter: Chemical Kinetics

Topic: Activation Energy and Temperature Dependence of Reaction Rate

Arrhenius Equation

k = A e^−E_a/RT

k = rate constant   |   A = Arrhenius/frequency factor   |   E_a = activation energy   |   R = gas constant   |   T = absolute temperature

Activation energy (E_a) is the minimum amount of energy that reacting molecules must possess for an effective collision and successful chemical reaction to occur.

Reactants + Activation Energy → Activated Complex → Products

Even when reactants collide, every collision does not produce a reaction. The molecules must have sufficient energy to cross the activation-energy barrier.

Higher E_a → fewer effective collisions → slower reaction

When temperature increases, the kinetic energy of molecules increases. Consequently, a greater fraction of molecules acquires energy equal to or greater than the activation energy.

Lower Temperature

↓ Kinetic Energy

↓ Effective Collisions

↓ Reaction Rate

→

Higher Temperature

↑ Kinetic Energy

↑ Effective Collisions

↑ Reaction Rate

As temperature increases, the rate constant k increases rapidly.

The Arrhenius equation is:

k = A e^−E_a/RT

Taking natural logarithm on both sides:

ln k = ln A − E_a/RT

Therefore:

ln k = ln A − (E_a/R)(1/T)

This equation has the form:

y = c + mx

Hence, a plot of ln k versus 1/T gives a straight line.

Slope = −E_a/R

Therefore, the slope of the Arrhenius plot can be used to determine the activation energy.

For two temperatures T₁ and T₂, the Arrhenius equation can be written as:

k₁ = A e^−E_a/RT₁

k₂ = A e^−E_a/RT₂

Dividing the equations:

k₂/k₁ = e^{(E_a/R)(1/T₁ − 1/T₂)}

Taking logarithm:

ln(k₂/k₁) = (E_a/R) (1/T₁ − 1/T₂)

Using common logarithm:

log(k₂/k₁) = E_a/(2.303R) (1/T₁ − 1/T₂)

1. Increasing temperature increases the average kinetic energy of molecules.
2. The molecules collide more frequently.
3. More importantly, a larger fraction of molecules has energy greater than or equal to E_a.
4. The number of effective collisions increases.
5. Therefore, the rate constant k increases.
6. Consequently, the reaction rate increases.

Important: Temperature does not simply increase the number of collisions. It significantly increases the fraction of molecules that can cross the activation-energy barrier.

Temperature ↑ → Fraction of molecules with E ≥ E_a ↑ → Effective collisions ↑ → k ↑ → Reaction rate ↑

Q. The rate constant of a reaction at 300 K is 2.0 × 10⁻³ s⁻¹. If its activation energy is 50 kJ mol⁻¹, explain what happens to the rate constant when temperature is increased.

According to the Arrhenius equation:

k = A e^−E_a/RT

When T increases, the magnitude of −E_a/RT becomes less negative. Therefore the exponential factor increases.

Temperature increases → k increases → reaction becomes faster.

Thus, even a moderate increase in temperature can produce a significant increase in the rate constant, particularly for reactions having appreciable activation energy.

A catalyst provides an alternative reaction pathway having a lower activation energy.

E_a(catalysed) < E_a(uncatalysed)

From the Arrhenius equation:

k = A e^−E_a/RT

A lower E_a gives a larger value of k at the same temperature. Therefore, the catalysed reaction is faster.

Catalyst → E_a decreases → k increases → Reaction rate increases

Symbol	Meaning	Effect
k	Rate constant	Increases with temperature
A	Frequency/Arrhenius factor	Related to collision frequency and orientation
Ea	Activation energy	Higher Ea generally gives lower k
R	Gas constant	8.314 J mol⁻¹ K⁻¹
T	Absolute temperature	Higher T → higher k

Q. Write the Arrhenius equation and define activation energy.

Answer:

k = A e^−E_a/RT

Activation energy is the minimum energy required by reacting molecules to undergo an effective collision and form products.

Q. Explain the effect of temperature on the rate constant using the Arrhenius equation.

According to Arrhenius:

k = A e^−E_a/RT

When temperature increases, the exponential factor increases. Therefore, the rate constant k increases. Consequently, the reaction rate increases.

T ↑ → k ↑ → Rate ↑

Q. Derive the logarithmic form of the Arrhenius equation.

Starting with:

k = A e^−E_a/RT

Taking natural logarithm:

ln k = ln A − E_a/RT

Rearranging:

ln k = ln A − (E_a/R)(1/T)

This is the equation of a straight line:

y = c + mx

Therefore:

Slope = −E_a/R

Q. Explain temperature dependence of reaction rate using the Arrhenius equation.

Step 1: Arrhenius Equation

k = A e^−E_a/RT

Step 2: Increase in Temperature

When T increases, the exponent −E_a/RT becomes less negative. Therefore, the value of the exponential term increases.

Step 3: Increase in Rate Constant

T ↑ → k ↑

Step 4: Effect on Reaction Rate

Since the rate constant increases, the reaction rate also increases. More molecules acquire energy equal to or greater than the activation energy.

Higher temperature → More effective collisions → Higher k → Faster reaction

Q. Explain in detail the temperature dependence of the rate of a chemical reaction using the Arrhenius equation. Derive its logarithmic form and explain the significance of activation energy.

1. Arrhenius Equation

k = A e^−E_a/RT

where:

• k = rate constant
• A = Arrhenius or frequency factor
• E_a = activation energy
• R = gas constant
• T = absolute temperature

2. Logarithmic Form

Taking natural logarithm:

ln k = ln A − E_a/RT

or:

ln k = ln A − (E_a/R)(1/T)

3. Temperature Dependence

When temperature increases, the value of the exponential term increases. Therefore, the rate constant k increases.

T ↑ → k ↑ → Reaction rate ↑

4. Molecular Explanation

At higher temperature, molecules possess greater kinetic energy. Consequently, a greater fraction of molecules possesses energy greater than or equal to E_a.

This produces more effective collisions and hence a faster reaction.

5. Arrhenius Plot

A plot of ln k against 1/T is a straight line:

Slope = −E_a/R

Therefore, activation energy can be calculated from the slope.

6. Two-Temperature Equation

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)

Conclusion:

The Arrhenius equation quantitatively explains why reaction rates increase with temperature. An increase in temperature increases the rate constant because a larger fraction of molecules can overcome the activation-energy barrier.

1. The Arrhenius equation is:

A) k = A + E_a/RT

B) k = Ae^−E_a/RT

C) k = E_a/RT

D) k = RT/E_a

✅ Answer

B

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2. E_a represents:

A) Equilibrium energy

B) Activation energy

C) Electrical energy

D) Bond energy only

✅ Answer

B) Activation energy

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3. When temperature increases, generally:

A) k decreases

B) k increases

C) k becomes zero

D) k remains unchanged

✅ Answer

B

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4. The unit of activation energy is:

A) mol L⁻¹

B) J mol⁻¹

C) s⁻¹

D) L mol⁻¹

✅ Answer

B) J mol⁻¹

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5. The Arrhenius factor A is also called:

A) Frequency factor

B) Equilibrium factor

C) Pressure factor

D) Energy factor

✅ Answer

A

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6. The value of R used in Arrhenius calculations is:

A) 8.314 J mol⁻¹ K⁻¹

B) 0.0821 J mol⁻¹ K⁻¹

C) 9.8 m s⁻²

D) 6.022 × 10²³

✅ Answer

A

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7. The Arrhenius equation shows dependence of k on:

A) Temperature

B) Activation energy

C) Both A and B

D) Neither

✅ Answer

C

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8. The logarithmic form of Arrhenius equation is:

A) ln k = ln A − E_a/RT

B) ln k = A + E_a

C) ln k = RT/E_a

D) ln k = kT

✅ Answer

A

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9. A plot of ln k versus 1/T gives:

A) Circle

B) Straight line

C) Parabola

D) Hyperbola

✅ Answer

B

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10. The slope of ln k versus 1/T plot is:

A) E_a/R

B) −E_a/R

C) R/E_a

D) −R/E_a

✅ Answer

B

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11. A catalyst generally:

A) Increases E_a

B) Decreases E_a

C) Has no effect on reaction pathway

D) Stops the reaction

✅ Answer

B

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12. Higher E_a generally means:

A) Faster reaction at the same temperature

B) Lower rate constant at the same temperature

C) No effect

D) Zero temperature

✅ Answer

B

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13. Increasing temperature increases the fraction of molecules:

A) Below E_a

B) With energy ≥ E_a

C) With zero energy

D) Without kinetic energy

✅ Answer

B

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14. In the Arrhenius equation, T must be expressed in:

A) °C

B) °F

C) Kelvin

D) Any unit

✅ Answer

C) Kelvin

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15. The two-temperature Arrhenius equation is:

A) ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)

B) k₂ − k₁ = RT

C) k₂/k₁ = T₂/T₁

D) ln k = RT

✅ Answer

A

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16. A reaction with lower activation energy is generally:

A) Faster

B) Slower

C) Impossible

D) Independent of temperature

✅ Answer

A) Faster

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17. The exponential term in Arrhenius equation represents the:

A) Fraction of molecules having sufficient energy

B) Total mass

C) Pressure

D) Volume

✅ Answer

A

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18. If temperature is increased, the value of −E_a/RT:

A) Becomes less negative

B) Becomes more negative

C) Becomes zero always

D) Remains unchanged

✅ Answer

A

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19. The rate constant is most sensitive to temperature when:

A) E_a is appreciable

B) E_a is zero

C) Temperature is zero Kelvin

D) A is zero

✅ Answer

A

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20. Which statement is correct?

A) Temperature increase usually decreases reaction rate

B) Temperature increase usually increases rate constant

C) Activation energy is always zero

D) Catalyst increases activation energy

✅ Answer

B

🎯 Quick Revision

Arrhenius Equation: k = Ae^−E_a/RT
Log Form: ln k = ln A − E_a/RT
Arrhenius Plot: Slope = −E_a/R
Temperature: T ↑ → k ↑ → Reaction rate ↑
Catalyst: E_a ↓ → k ↑ → Reaction rate ↑

Chemical Kinetics: For a first-order reaction, prove that the half-life is independent of the initial concentration

Chapter: Chemical Kinetics

Topic: Half-Life of a First-Order Reaction

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For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant.

Rate ∝ [A]

Therefore:

−d[A]/dt = k[A]

where k is the first-order rate constant.

For a first-order reaction:

−d[A]/dt = k[A]

On integration between the initial concentration [A]₀ and concentration [A] at time t:

ln([A]₀/[A]) = kt

or:

k = (2.303/t) log([A]₀/[A])

At half-life, the concentration of the reactant becomes half of its initial concentration.

[A] = [A]₀/2

Substitute this value into the first-order integrated rate equation:

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

Therefore:

k = (2.303/t_1/2) log 2

Since:

log 2 = 0.3010

we get:

k = (2.303 × 0.3010)/t_1/2

Since 2.303 × 0.3010 ≈ 0.693:

k = 0.693/t_1/2

Hence:

t_1/2 = 0.693/k

The half-life equation obtained for a first-order reaction is:

t_1/2 = 0.693/k

Notice that the expression contains only the rate constant k.

There is no term containing the initial concentration [A]₀.

Therefore, even if the initial concentration is changed, the half-life will remain the same, provided the temperature and other conditions remain unchanged.

Hence proved:

The half-life of a first-order reaction is independent of its initial concentration.

Q. A first-order reaction has a rate constant k = 0.693 min⁻¹. Calculate its half-life.

We know:

t_1/2 = 0.693/k

Substituting:

t_1/2 = 0.693/0.693

t_1/2 = 1 minute

The same half-life would be obtained whether the initial concentration were 1 M, 0.5 M, 0.1 M or any other value, as long as k remains unchanged.

Order	Half-Life	Depends on Initial Concentration?
Zero order	t1/2 = [A]0/2k	Yes
First order	t1/2 = 0.693/k	No
Second order	t1/2 = 1/(k[A]0)	Yes

Special property of first-order reactions:

Half-life is independent of the initial concentration.

Q. Write the expression for the half-life of a first-order reaction.

Answer:

t_1/2 = 0.693/k

Since the expression does not contain the initial concentration, half-life is independent of initial concentration.

Q. Prove that the half-life of a first-order reaction is independent of initial concentration.

For a first-order reaction:

k = (2.303/t) log([A]₀/[A])

At half-life:

[A] = [A]₀/2

Therefore:

k = (2.303/t_1/2) log 2

Since log 2 = 0.301:

t_1/2 = 0.693/k

The initial concentration [A]₀ cancels out. Hence, half-life is independent of initial concentration.

Q. Derive the half-life equation for a first-order reaction and explain its significance.

Integrated first-order rate equation:

k = (2.303/t) log([A]₀/[A])

At t = t_1/2:

[A] = [A]₀/2

Hence:

k = (2.303/t_1/2) log 2

Therefore:

t_1/2 = 0.693/k

The initial concentration is absent from the final expression. Therefore, t_1/2 is independent of initial concentration.

Q. For a first-order reaction, prove mathematically that the half-life is independent of the initial concentration.

Step 1: First-order equation

k = (2.303/t) log([A]₀/[A])

Step 2: Condition at half-life

At t = t_1/2, half of the original reactant remains:

[A] = [A]₀/2

Step 3: Substitute

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

The [A]₀ terms cancel:

k = (2.303/t_1/2) log 2

Step 4: Simplify

k = 0.693/t_1/2

Therefore:

t_1/2 = 0.693/k

Since [A]₀ does not occur in the final equation, changing the initial concentration does not change the half-life.

Q. For a first-order reaction, derive the expression for half-life and prove that it is independent of the initial concentration. Explain the physical significance of this result.

1. Rate Law

For a first-order reaction:

−d[A]/dt = k[A]

2. Integrated Rate Equation

k = (2.303/t) log([A]₀/[A])

3. Apply the Half-Life Condition

At half-life, the concentration becomes half of the initial concentration:

[A] = [A]₀/2

Substituting:

k = (2.303/t_1/2) log([A]₀/([A]₀/2))

4. Cancellation of Initial Concentration

[A]₀/([A]₀/2) = 2

Thus:

k = (2.303/t_1/2) log 2

Using log 2 = 0.3010:

k = (2.303 × 0.3010)/t_1/2

Therefore:

t_1/2 = 0.693/k

5. Proof of Independence

The final equation contains only k. It does not contain [A]₀. Hence, for a fixed temperature and reaction conditions, the half-life remains constant regardless of the initial concentration.

6. Physical Significance

Suppose two samples of the same first-order reactant have initial concentrations of 1.0 M and 0.5 M. Both will take the same time to reduce their respective concentrations by half.

1.0 M → 0.5 M → 0.25 M → 0.125 M

0.5 M → 0.25 M → 0.125 M → 0.0625 M

Each successive half-life is equal because the half-life depends only on the rate constant.

Final Conclusion:

For a first-order reaction, t_1/2 = 0.693/k.

Therefore, the half-life is independent of the initial concentration.

1. The half-life of a first-order reaction is:

A) [A]₀/2k

B) 0.693/k

C) 1/k[A]₀

D) k/0.693

✅ Answer

B) 0.693/k

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2. The half-life of a first-order reaction depends on:

A) Initial concentration only

B) Rate constant

C) Initial volume only

D) Initial mass only

✅ Answer

B) Rate constant

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3. The half-life of a first-order reaction is independent of:

A) Rate constant

B) Temperature

C) Initial concentration

D) Reaction conditions

✅ Answer

C) Initial concentration

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4. At half-life, [A] is equal to:

A) 2[A]₀

B) [A]₀

C) [A]₀/2

D) Zero

✅ Answer

C) [A]₀/2

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5. The value of log 2 is approximately:

A) 0.3010

B) 0.693

C) 2.303

D) 1.301

✅ Answer

A) 0.3010

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6. The numerical value 0.693 is approximately equal to:

A) log 2

B) 2.303 log 2

C) log 10

D) 1/log 2

✅ Answer

B) 2.303 log 2

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7. If k = 0.693 min⁻¹, t_1/2 is:

A) 0.693 min

B) 1 min

C) 2 min

D) 0.5 min

✅ Answer

B) 1 min

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8. If the initial concentration is doubled, the first-order half-life:

A) Doubles

B) Becomes half

C) Remains unchanged

D) Becomes zero

✅ Answer

C) Remains unchanged

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9. For a first-order reaction, the rate law is:

A) Rate = k

B) Rate = k[A]

C) Rate = k[A]²

D) Rate = [A]/k

✅ Answer

B) Rate = k[A]

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10. The unit of k for a first-order reaction is:

A) mol L⁻¹ s⁻¹

B) s⁻¹

C) L mol⁻¹ s⁻¹

D) mol² L⁻² s⁻¹

✅ Answer

B) s⁻¹

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11. Which order has a half-life independent of initial concentration?

A) Zero order

B) First order

C) Second order

D) Third order

✅ Answer

B) First order

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12. If k increases, the half-life:

A) Increases

B) Decreases

C) Remains unchanged

D) Becomes infinite

✅ Answer

B) Decreases

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13. If k is doubled, t_1/2 becomes:

A) Double

B) Half

C) Four times

D) Unchanged

✅ Answer

B) Half

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14. The integrated rate equation for a first-order reaction is:

A) [A] = [A]₀ − kt

B) k = (2.303/t) log([A]₀/[A])

C) 1/[A] = kt

D) Rate = k[A]²

✅ Answer

B

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15. After two half-lives, the fraction remaining is:

A) 1/2

B) 1/3

C) 1/4

D) 1/8

✅ Answer

C) 1/4

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16. After three half-lives, the fraction remaining is:

A) 1/2

B) 1/4

C) 1/6

D) 1/8

✅ Answer

D) 1/8

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17. The first-order half-life equation contains:

A) Initial concentration

B) Final concentration

C) Rate constant

D) Both A and B

✅ Answer

C) Rate constant

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18. A first-order reaction is characterised by a constant:

A) t_1/2 only for all temperatures

B) Fraction decomposed per unit time

C) Initial concentration

D) Volume

✅ Answer

B

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19. If t_1/2 = 10 s, the rate constant is approximately:

A) 0.0693 s⁻¹

B) 6.93 s⁻¹

C) 0.693 s⁻¹

D) 10 s⁻¹

✅ Answer

A) 0.0693 s⁻¹

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20. The statement "half-life is independent of initial concentration" is true for:

A) Zero-order reactions

B) First-order reactions

C) Second-order reactions

D) All reactions

✅ Answer

B) First-order reactions

🎯 Quick Revision

First-order rate equation: k = (2.303/t) log([A]₀/[A])
At half-life: [A] = [A]₀/2
Half-life: t_1/2 = 0.693/k
Key point: [A]₀ cancels out → Half-life is independent of initial concentration.