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 ↑