Q: Q6. A reaction is first order in A and second order in B.\n(i) Write the differential rate equation.\n(ii) How is the rate affected on increasing the concentration of B three times?\n(

Let the rate of the reaction be 'Rate'.\n\n(i) **Differential rate equation:**\nSince the reaction is first order in A and second order in B, the differential rate equation can be written as:\nRate = k[A]¹[B]² or Rate = k[A][B]²\n\n(ii) **Effect on rate when concentration of B is increased three times:**\nLet the initial concentrations be [A] and [B]. The initial rate is Rate₁ = k[A][B]².\nWhen [B] is increased three times, the new concentration of B becomes 3[B]. The concentration of A remains [A].\nThe new rate, Rate₂ = k[A](3[B])² = k[A]9[B]² = 9 * (k[A][B]²)\nRate₂ = 9 * Rate₁\nThus, the rate of the reaction increases by 9 times.\n\n(iii) **Effect on rate when concentrations of both A and B are doubled:**\nLet the initial concentrations be [A] and [B]. The initial rate is Rate₁ = k[A][B]².\nWhen concentrations of both A and B are doubled, the new concentrations become 2[A] and 2[B].\nThe new rate, Rate₃ = k(2[A])(2[B])² = k(2[A])(4[B]²) = 8 * (k[A][B]²)\nRate₃ = 8 * Rate₁\nThus, the rate of the reaction increases by 8 times.

Question 1

Q1. From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.
(i) 3NO(g) → N₂O(g) Rate = k[NO]²
(ii) H₂O₂(aq)

Accepted Answer

The order of reaction is the sum of the powers of the concentration terms in the rate law. The dimensions (units) of the rate constant 'k' depend on the overall order of the reaction. For a general n-th order reaction, the units of k are (concentration)¹⁻ⁿ (time)⁻¹.

(i) For the reaction 3NO(g) → N₂O(g), Rate = k[NO]²
   - **Order of reaction**: The power of [NO] is 2. So, the order of reaction is 2.
   - **Dimensions of rate constant (k)**:
     Rate = k[NO]²
     mol L⁻¹ s⁻¹ = k (mol L⁻¹)⁻²
     k = (mol L⁻¹ s⁻¹) / (mol L⁻¹)⁻² = mol L⁻¹ s⁻¹ × (mol L⁻¹)⁻² = mol¹⁻² L⁻¹⁻² s⁻¹ = mol⁻¹ L¹ s⁻¹
     Therefore, the units of k are L mol⁻¹ s⁻¹.

(ii) For the reaction H₂O₂(aq) + 3I⁻(aq) + 2H⁺(aq) → 2H₂O(l) + I₃⁻(aq), Rate = k[H₂O₂][I⁻]
   - **Order of reaction**: The sum of the powers of [H₂O₂] (power 1) and [I⁻] (power 1) is 1 + 1 = 2. So, the order of reaction is 2.
   - **Dimensions of rate constant (k)**:
     Since the overall order is 2, the units of k will be the same as for a second-order reaction.
     k = (concentration)¹⁻² (time)⁻¹ = (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹.

(iii) For the reaction CH₃CHO(g) → CH₄(g) + CO(g), Rate = k[CH₃CHO]³/²
   - **Order of reaction**: The power of [CH₃CHO] is 3/2. So, the order of reaction is 3/2.
   - **Dimensions of rate constant (k)**:
     k = (concentration)¹⁻³/² (time)⁻¹ = (mol L⁻¹)⁻¹/² s⁻¹ = L¹/² mol⁻¹/² s⁻¹.

(iv) For the reaction C₂H₅Cl(g) → C₂H₄(g) + HCl(g), Rate = k[C₂H₅Cl]
   - **Order of reaction**: The power of [C₂H₅Cl] is 1. So, the order of reaction is 1.
   - **Dimensions of rate constant (k)**:
     k = (concentration)¹⁻¹ (time)⁻¹ = (mol L⁻¹)⁰ s⁻¹ = s⁻¹.

Question 2

Q2. For a reaction A + B → Products, the rate law is given by Rate = k[A]¹/²[B]². What is the order of the reaction?

Accepted Answer

The order of a reaction is defined as the sum of the powers of the concentration terms of the reactants in the experimentally determined rate law.

Given the rate law: Rate = k[A]¹/²[B]²

Here, the power of the concentration term [A] is 1/2.
And the power of the concentration term [B] is 2.

Therefore, the overall order of the reaction = (power of [A]) + (power of [B]) = 1/2 + 2 = 0.5 + 2 = 2.5.

The order of the reaction is 2.5.

Question 3

Q3. The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?

Accepted Answer

For a reaction that follows second-order kinetics with respect to reactant X, the rate law can be written as:

Rate = k[X]²

Let the initial concentration of X be [X]₀. The initial rate of reaction (Rate₁) is:
Rate₁ = k[X]₀²

If the concentration of X is increased to three times, the new concentration of X will be 3[X]₀. The new rate of reaction (Rate₂) will be:
Rate₂ = k(3[X]₀)²
Rate₂ = k(9[X]₀²)
Rate₂ = 9k[X]₀²

Comparing Rate₂ with Rate₁:
Rate₂ = 9 × (k[X]₀²)
Rate₂ = 9 × Rate₁

Thus, if the concentration of X is increased to three times, the rate of formation of Y (i.e., the rate of reaction) will increase by 9 times.

Question 4

Q4. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant
(i) doubled
(ii) reduced to half?

Accepted Answer

For a reaction that is second order with respect to a reactant, let's denote the reactant as R. The rate law can be written as:

Rate = k[R]²

(i) **If the concentration of the reactant is doubled:**
Let the initial concentration of the reactant be [R]₀. The initial rate (Rate₁) is:
Rate₁ = k[R]₀²

If the concentration of the reactant is doubled, the new concentration will be 2[R]₀. The new rate (Rate₂) will be:
Rate₂ = k(2[R]₀)²
Rate₂ = k(4[R]₀²)
Rate₂ = 4k[R]₀²

Comparing Rate₂ with Rate₁:
Rate₂ = 4 × (k[R]₀²)
Rate₂ = 4 × Rate₁

Thus, if the concentration of the reactant is doubled, the rate of reaction will increase by 4 times.

(ii) **If the concentration of the reactant is reduced to half:**
Let the initial concentration of the reactant be [R]₀. The initial rate (Rate₁) is:
Rate₁ = k[R]₀²

If the concentration of the reactant is reduced to half, the new concentration will be (1/2)[R]₀. The new rate (Rate₃) will be:
Rate₃ = k((1/2)[R]₀)²
Rate₃ = k(1/4)[R]₀²
Rate₃ = (1/4)k[R]₀²

Comparing Rate₃ with Rate₁:
Rate₃ = (1/4) × (k[R]₀²)
Rate₃ = (1/4) × Rate₁

Thus, if the concentration of the reactant is reduced to half, the rate of reaction will be reduced to one-fourth (1/4) of its original rate.

Question 5

Q5. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Accepted Answer

### Effect of Temperature on the Rate Constant of a Reaction

Generally, the rate of a chemical reaction increases with an increase in temperature. Consequently, the rate constant (k) of a reaction also increases significantly with an increase in temperature. For many reactions, the rate constant approximately doubles for every 10°C rise in temperature. This observation is explained by the collision theory, which states that an increase in temperature increases the kinetic energy of reactant molecules, leading to more frequent and more energetic collisions. More importantly, a larger fraction of molecules possess the activation energy (Ea) required to overcome the energy barrier and form products.

### Quantitative Representation of Temperature Effect: Arrhenius Equation

The quantitative relationship between the rate constant (k) and temperature (T) was proposed by Svante Arrhenius and is known as the **Arrhenius equation**:

**k = A e^(-Ea/RT)**

Where:
*   **k** is the rate constant of the reaction.
*   **A** is the Arrhenius factor or pre-exponential factor or frequency factor. It represents the frequency of collisions of reactant molecules in the correct orientation. Its units are the same as the rate constant k.
*   **Ea** is the activation energy of the reaction (in J mol⁻¹ or kJ mol⁻¹). It is the minimum energy that reactant molecules must possess to undergo a chemical reaction.
*   **R** is the universal gas constant (8.314 J K⁻¹ mol⁻¹).
*   **T** is the absolute temperature (in Kelvin).
*   **e^(-Ea/RT)** is the fraction of molecules that have energy equal to or greater than the activation energy.

#### Logarithmic Form of Arrhenius Equation:
Taking the natural logarithm on both sides of the Arrhenius equation:

ln k = ln (A e^(-Ea/RT))
ln k = ln A + ln (e^(-Ea/RT))
**ln k = ln A - Ea/RT**

Converting to base-10 logarithm:

**log k = log A - Ea/(2.303 RT)**

#### Graphical Representation:
If we plot **ln k versus 1/T**, we get a straight line with:
*   **Slope = -Ea/R**
*   **Y-intercept = ln A**

Similarly, a plot of **log k versus 1/T** yields a straight line with:
*   **Slope = -Ea/(2.303 R)**
*   **Y-intercept = log A**

This graphical method allows for the determination of activation energy (Ea) and the pre-exponential factor (A) from experimental data of rate constants at different temperatures.

#### Arrhenius Equation at Two Different Temperatures:
For a reaction at two different temperatures, T₁ and T₂, with corresponding rate constants k₁ and k₂, the Arrhenius equation can be written as:

ln k₁ = ln A - Ea/(RT₁)
ln k₂ = ln A - Ea/(RT₂)

Subtracting the first equation from the second:

ln k₂ - ln k₁ = (ln A - Ea/(RT₂)) - (ln A - Ea/(RT₁))
ln (k₂/k₁) = -Ea/(RT₂) + Ea/(RT₁)
**ln (k₂/k₁) = (Ea/R) (1/T₁ - 1/T₂)**

Or in base-10 logarithm:

**log (k₂/k₁) = (Ea/(2.303 R)) (1/T₁ - 1/T₂)**

This form is particularly useful for calculating the activation energy if rate constants at two different temperatures are known, or for calculating the rate constant at a new temperature if Ea and k at one temperature are known.

Question 6

Q6. A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(

Accepted Answer

Let the rate of the reaction be 'Rate'.

(i) **Differential rate equation:**
Since the reaction is first order in A and second order in B, the differential rate equation can be written as:
Rate = k[A]¹[B]²  or  Rate = k[A][B]²

(ii) **Effect on rate when concentration of B is increased three times:**
Let the initial concentrations be [A] and [B]. The initial rate is Rate₁ = k[A][B]².
When [B] is increased three times, the new concentration of B becomes 3[B]. The concentration of A remains [A].
The new rate, Rate₂ = k[A](3[B])² = k[A]9[B]² = 9 * (k[A][B]²)
Rate₂ = 9 * Rate₁
Thus, the rate of the reaction increases by 9 times.

(iii) **Effect on rate when concentrations of both A and B are doubled:**
Let the initial concentrations be [A] and [B]. The initial rate is Rate₁ = k[A][B]².
When concentrations of both A and B are doubled, the new concentrations become 2[A] and 2[B].
The new rate, Rate₃ = k(2[A])(2[B])² = k(2[A])(4[B]²) = 8 * (k[A][B]²)
Rate₃ = 8 * Rate₁
Thus, the rate of the reaction increases by 8 times.

Question 7

Q7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be quantitatively expressed?

Accepted Answer

### Effect of Temperature on the Rate Constant of a Reaction

Generally, the rate of a chemical reaction increases with an increase in temperature. Consequently, the rate constant (k) of a reaction also increases significantly with an increase in temperature. For most reactions, the rate constant approximately doubles for every 10°C rise in temperature. This is because increasing the temperature increases the kinetic energy of the reactant molecules, leading to:

1.  **Increased Collision Frequency:** Molecules move faster and collide more frequently.
2.  **Increased Fraction of Effective Collisions:** More importantly, a larger fraction of molecules possess kinetic energy equal to or greater than the activation energy (Ea). These molecules can overcome the energy barrier and lead to product formation. Even a small increase in temperature can lead to a substantial increase in the number of molecules with sufficient energy to react.

### Quantitative Expression of Temperature Effect on Rate Constant (Arrhenius Equation)

The quantitative relationship between the rate constant (k) and temperature (T) is given by the **Arrhenius equation**:

**k = A * e^(-Ea / RT)**

Where:
*   **k** is the rate constant of the reaction.
*   **A** is the Arrhenius factor or pre-exponential factor or frequency factor. It represents the frequency of collisions of reactant molecules and the fraction of collisions in which molecules are properly oriented.
*   **Ea** is the activation energy of the reaction (in J mol⁻¹ or kJ mol⁻¹). It is the minimum energy required for reactant molecules to transform into products.
*   **R** is the universal gas constant (8.314 J K⁻¹ mol⁻¹).
*   **T** is the absolute temperature (in Kelvin).
*   **e^(-Ea / RT)** represents the fraction of molecules that have energy equal to or greater than the activation energy.

#### Derivation and Linear Form:
The Arrhenius equation can be expressed in a more convenient linear form by taking the natural logarithm on both sides:

ln k = ln (A * e^(-Ea / RT))
ln k = ln A + ln (e^(-Ea / RT))
**ln k = ln A - (Ea / RT)**

This equation is in the form of a straight line (y = mx + c) if we plot ln k against 1/T:
*   y = ln k
*   x = 1/T
*   m (slope) = -Ea / R
*   c (y-intercept) = ln A

From the slope of such a plot, the activation energy (Ea) can be determined.

#### Arrhenius Equation for Two Different Temperatures:
If the rate constants are k₁ and k₂ at two different temperatures T₁ and T₂ respectively, the Arrhenius equation can be written as:

ln k₁ = ln A - (Ea / RT₁)
ln k₂ = ln A - (Ea / RT₂)

Subtracting the first equation from the second:

ln k₂ - ln k₁ = (ln A - Ea / RT₂) - (ln A - Ea / RT₁)
ln (k₂ / k₁) = -Ea / RT₂ + Ea / RT₁
**ln (k₂ / k₁) = (Ea / R) * (1/T₁ - 1/T₂)**

Or, converting to base-10 logarithm:

**log (k₂ / k₁) = (Ea / 2.303R) * (1/T₁ - 1/T₂)**

This form is particularly useful for calculating activation energy when rate constants at two different temperatures are known, or vice-versa.

Question 8

Q8. The activation energy for the reaction 2HI(g) → I₂(g) + H₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules having energy equal to or greater than activation ener

Accepted Answer

To calculate the fraction of molecules having energy equal to or greater than the activation energy, we use the Boltzmann factor, which is given by e^(-Ea/RT).

Question 9

Q9. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10¹⁰ s⁻¹, calculate Ea f

Accepted Answer

This problem involves using the integrated rate law for a first-order reaction and the Arrhenius equation to find the activation energy (Ea).

Question 10

Q10. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the activation energy of the reaction, assuming that it does not change with temperatu

Accepted Answer

We can use the Arrhenius equation in its integrated form to calculate the activation energy (Ea) when the rate constants at two different temperatures are known.

Question 11

Q11. The activation energy for the reaction 2HI(g) → H₂(g) + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules having energy equal to or greater than activation ener

Accepted Answer

The fraction of molecules having energy equal to or greater than the activation energy (Eₐ) at a given temperature (T) is given by the Boltzmann factor, e^(-Eₐ/RT).

Given:
Activation energy (Eₐ) = 209.5 kJ mol⁻¹ = 209.5 × 10³ J mol⁻¹
Temperature (T) = 581 K
Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

First, calculate the exponent Eₐ/RT:
Eₐ/RT = (209.5 × 10³ J mol⁻¹) / (8.314 J K⁻¹ mol⁻¹ × 581 K)
Eₐ/RT = (209500) / (4831.034)
Eₐ/RT ≈ 43.365

Now, calculate the fraction of molecules:
Fraction = e^(-Eₐ/RT) = e^(-43.365)

Using a calculator, e^(-43.365) ≈ 1.32 × 10⁻¹⁹

Therefore, the fraction of molecules having energy equal to or greater than the activation energy is approximately 1.32 × 10⁻¹⁹.

Question 12

Q12. Explain the terms:
    (i) Rate of a reaction
    (ii) Order of a reaction
    (iii) Molecularity of a reaction
    (iv) Activation energy

Accepted Answer

Let's explain each term in detail:

### (i) Rate of a reaction
**Definition:** The rate of a reaction is defined as the change in concentration of a reactant or product per unit time. It indicates how fast a reaction proceeds.

**Mathematical Expression:**
For a general reaction: R → P

Average rate = - Δ[R] / Δt = + Δ[P] / Δt

Instantaneous rate = - d[R] / dt = + d[P] / dt

For a stoichiometric reaction: aA + bB → cC + dD

Rate = - (1/a) d[A]/dt = - (1/b) d[B]/dt = + (1/c) d[C]/dt = + (1/d) d[D]/dt

**Units:** The units of reaction rate are typically mol L⁻¹ s⁻¹ or M s⁻¹.

### (ii) Order of a reaction
**Definition:** The order of a reaction with respect to a particular reactant is the power to which its concentration term is raised in the experimentally determined rate law. The overall order of a reaction is the sum of the powers of the concentration terms of all reactants in the rate law expression.

**Mathematical Expression:**
For a reaction: aA + bB → Products
If the experimentally determined rate law is: Rate = k[A]ˣ[B]ʸ

*   Order with respect to A = x
*   Order with respect to B = y
*   Overall order of the reaction = x + y

**Key Points:**
*   The order of a reaction can be a whole number, a fraction, or even zero.
*   It is an experimentally determined quantity and cannot be predicted from the stoichiometry of the balanced chemical equation (unless it's an elementary reaction).
*   It can be zero if the reaction rate does not depend on the concentration of a particular reactant.

### (iii) Molecularity of a reaction
**Definition:** Molecularity is defined as the number of reacting species (atoms, ions, or molecules) that collide simultaneously in an elementary reaction to bring about a chemical reaction. It is a theoretical concept applicable only to elementary reactions.

**Types of Molecularity:**
*   **Unimolecular:** Molecularity = 1 (e.g., decomposition of N₂O₅: N₂O₅ → N₂O₄ + ½O₂)
*   **Bimolecular:** Molecularity = 2 (e.g., dissociation of HI: 2HI → H₂ + I₂)
*   **Trimolecular:** Molecularity = 3 (e.g., 2NO + O₂ → 2NO₂)

**Key Points:**
*   Molecularity is always a whole number (1, 2, or 3). It cannot be zero, fractional, or greater than three (as the probability of simultaneous collision of more than three species is very low).
*   It is determined from the stoichiometry of the elementary step.
*   For complex reactions (reactions occurring in multiple steps), molecularity has no meaning for the overall reaction; instead, it refers to the molecularity of each elementary step.
*   For an elementary reaction, molecularity is equal to the order of the reaction.

### (iv) Activation energy
**Definition:** Activation energy (Eₐ) is the minimum amount of extra energy that must be supplied to the reactant molecules so that they can overcome the energy barrier and transform into products. It is the difference between the threshold energy and the average energy of the reactant molecules.

**Concept:** According to collision theory, for a reaction to occur, reactant molecules must collide with sufficient energy (threshold energy) and proper orientation. Molecules possessing energy less than the threshold energy will simply bounce back without reacting. The activation energy represents the energy barrier that must be surmounted for the reaction to proceed.

**Arrhenius Equation:** The relationship between activation energy, temperature, and rate constant is given by the Arrhenius equation:
k = A * e^(-Eₐ/RT)
Where:
*   k = rate constant
*   A = Arrhenius factor or pre-exponential factor (frequency factor)
*   Eₐ = activation energy
*   R = gas constant
*   T = absolute temperature

**Significance:** A higher activation energy implies a slower reaction rate, as fewer molecules will possess the necessary energy to react. Catalysts work by lowering the activation energy, thereby increasing the reaction rate.

Question 13

Q13. What is the effect of temperature on the rate constant of a reaction? How can this effect be quantitatively expressed?

Accepted Answer

### Effect of Temperature on the Rate Constant

Generally, the rate constant (k) of a reaction increases with an increase in temperature. For most reactions, the rate constant approximately doubles or triples for every 10°C rise in temperature. This is because an increase in temperature increases the kinetic energy of the reactant molecules. As a result:

1.  **Increased Collision Frequency:** Molecules move faster, leading to more frequent collisions.
2.  **Increased Fraction of Effective Collisions:** More importantly, a larger fraction of molecules will possess kinetic energy equal to or greater than the activation energy (Eₐ). These molecules can undergo effective collisions, leading to product formation. The increase in the fraction of molecules with sufficient energy is much more significant than the increase in collision frequency.

Therefore, an increase in temperature leads to a higher rate constant and thus a faster reaction rate.

### Quantitative Expression of the Effect of Temperature

The quantitative relationship between the rate constant and temperature is given by the **Arrhenius equation**:

**k = A * e^(-Eₐ/RT)**

Where:
*   **k** = Rate constant of the reaction
*   **A** = Arrhenius factor or pre-exponential factor (also called frequency factor). It represents the frequency of collisions of reactant molecules in the correct orientation.
*   **Eₐ** = Activation energy of the reaction (in J mol⁻¹ or kJ mol⁻¹). It is the minimum energy required for reactant molecules to transform into products.
*   **R** = Gas constant (8.314 J K⁻¹ mol⁻¹)
*   **T** = Absolute temperature (in Kelvin)
*   **e^(-Eₐ/RT)** = Boltzmann factor, representing the fraction of molecules that have energy equal to or greater than the activation energy.

**Derivation and Linear Form:**
Taking the natural logarithm on both sides of the Arrhenius equation:

ln k = ln (A * e^(-Eₐ/RT))
ln k = ln A + ln (e^(-Eₐ/RT))
**ln k = ln A - Eₐ/RT**

This equation is in the form of a straight line (y = mx + c), where:
*   y = ln k
*   x = 1/T
*   m = -Eₐ/R (slope)
*   c = ln A (y-intercept)

A plot of ln k versus 1/T yields a straight line with a negative slope equal to -Eₐ/R, from which the activation energy (Eₐ) can be determined.

**Arrhenius Equation at Two Different Temperatures:**
If k₁ and k₂ are the rate constants at two different temperatures T₁ and T₂ respectively, then:

ln k₁ = ln A - Eₐ/RT₁  (1)
ln k₂ = ln A - Eₐ/RT₂  (2)

Subtracting (1) from (2):

ln k₂ - ln k₁ = (ln A - Eₐ/RT₂) - (ln A - Eₐ/RT₁)
ln (k₂/k₁) = -Eₐ/RT₂ + Eₐ/RT₁
**ln (k₂/k₁) = (Eₐ/R) * (1/T₁ - 1/T₂)**

Converting to base 10 logarithm:

**log (k₂/k₁) = (Eₐ / 2.303R) * (1/T₁ - 1/T₂)**

This equation allows for the calculation of activation energy if rate constants at two different temperatures are known, or vice-versa.

Question 14

Q14. Predict the order of the reaction:
    (i) A reaction whose rate constant has units mol L⁻¹ s⁻¹
    (ii) A reaction whose rate constant has units s⁻¹

Accepted Answer

The general formula for the units of the rate constant (k) for an n-th order reaction is:
Units of k = (concentration)^(1-n) (time)⁻¹
Units of k = (mol L⁻¹)^(1-n) s⁻¹

We can use this formula to determine the order (n) of the reaction.

**(i) A reaction whose rate constant has units mol L⁻¹ s⁻¹**

Given units of k = mol L⁻¹ s⁻¹
Comparing with the general formula: (mol L⁻¹)^(1-n) s⁻¹

We have (mol L⁻¹)^(1-n) = mol L⁻¹
This implies 1-n = 1
So, n = 1 - 1 = 0

Therefore, the reaction is a **zero-order reaction**.

**(ii) A reaction whose rate constant has units s⁻¹**

Given units of k = s⁻¹
Comparing with the general formula: (mol L⁻¹)^(1-n) s⁻¹

We have (mol L⁻¹)^(1-n) = (mol L⁻¹)⁰ (since anything to the power of 0 is 1)
This implies 1-n = 0
So, n = 1

Therefore, the reaction is a **first-order reaction**.

Question 15

Q15. Define the following terms:
    (i) Half-life period of a reaction (t₁/₂)
    (ii) Collision frequency (Z)
    (iii) Activation energy (Eₐ)

Accepted Answer

Let's define each term:

### (i) Half-life period of a reaction (t₁/₂)
**Definition:** The half-life period (t₁/₂) of a reaction is defined as the time required for the concentration of a reactant to be reduced to one-half of its initial concentration. In simpler terms, it is the time taken for half of the reactant to be consumed.

**Significance:**
*   For a **first-order reaction**, the half-life is independent of the initial concentration of the reactant. It is given by the formula: t₁/₂ = 0.693 / k, where k is the rate constant.
*   For a **zero-order reaction**, the half-life is directly proportional to the initial concentration: t₁/₂ = [R]₀ / 2k.
*   For an **n-th order reaction (n ≠ 1)**, the half-life is inversely proportional to the (n-1)th power of the initial concentration.

Half-life is a useful parameter to characterize the speed of a reaction and is often used in radioactive decay (which follows first-order kinetics).

### (ii) Collision frequency (Z)
**Definition:** Collision frequency (Z) is defined as the total number of collisions per unit volume per unit time between the reacting species (molecules, atoms, or ions) in a reaction mixture.

**Concept:** According to the collision theory, for a chemical reaction to occur, reactant molecules must collide with each other. The collision frequency gives an idea of how often these collisions happen. It is a factor in determining the rate of a reaction.

**Key Points:**
*   Collision frequency typically increases with an increase in temperature and concentration of reactants.
*   However, not all collisions lead to a reaction. Only a small fraction of collisions, known as 'effective collisions', result in product formation. Effective collisions require both sufficient energy (activation energy) and proper orientation.
*   In the Arrhenius equation (k = A * e^(-Eₐ/RT)), the pre-exponential factor 'A' is related to the collision frequency and the steric factor (orientation factor).

### (iii) Activation energy (Eₐ)
**Definition:** Activation energy (Eₐ) is the minimum amount of extra energy that must be supplied to the reactant molecules so that they can overcome the energy barrier and transform into products. It is the difference between the threshold energy and the average energy of the reactant molecules.

**Concept:** According to collision theory, for a reaction to occur, reactant molecules must collide with sufficient energy (threshold energy) and proper orientation. Molecules possessing energy less than the threshold energy will simply bounce back without reacting. The activation energy represents the energy barrier that must be surmounted for the reaction to proceed.

**Arrhenius Equation:** The relationship between activation energy, temperature, and rate constant is given by the Arrhenius equation:
k = A * e^(-Eₐ/RT)
Where:
*   k = rate constant
*   A = Arrhenius factor or pre-exponential factor (frequency factor)
*   Eₐ = activation energy
*   R = gas constant
*   T = absolute temperature

**Significance:** A higher activation energy implies a slower reaction rate, as fewer molecules will possess the necessary energy to react. Catalysts work by lowering the activation energy, thereby increasing the reaction rate.

Question 16

Q16. The activation energy for the reaction 2HI(g) → H₂(g) + I₂(g) is 209.5 kJ mol⁻¹ at 581 K. Calculate the fraction of molecules having energy equal to or greater than activation ener

Accepted Answer

The fraction of molecules having energy equal to or greater than the activation energy (Ea) is given by the Boltzmann factor, e^(-Ea/RT).

Given:
Activation energy (Ea) = 209.5 kJ mol⁻¹ = 209.5 × 10³ J mol⁻¹
Temperature (T) = 581 K
Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

Fraction of molecules = e^(-Ea/RT)

First, calculate Ea/RT:
Ea/RT = (209.5 × 10³ J mol⁻¹) / (8.314 J K⁻¹ mol⁻¹ × 581 K)
Ea/RT = 209500 / 4831.234
Ea/RT ≈ 43.363

Now, calculate e^(-Ea/RT):
Fraction = e^(-43.363)

Using a calculator, e^(-43.363) ≈ 1.34 × 10⁻¹⁹

Therefore, the fraction of molecules having energy equal to or greater than the activation energy is approximately 1.34 × 10⁻¹⁹.

Question 17

Q17. The decomposition of A into product has a value of k as 4.5 × 10³ s⁻¹ at 10°C and activation energy 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?

Accepted Answer

We can use the two-point form of the Arrhenius equation to solve this problem:

ln(k₂/k₁) = (Ea/R) * (1/T₁ - 1/T₂)

Given:
k₁ = 4.5 × 10³ s⁻¹ at T₁ = 10°C
k₂ = 1.5 × 10⁴ s⁻¹ at T₂ = ?
Activation energy (Ea) = 60 kJ mol⁻¹ = 60 × 10³ J mol⁻¹
Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

Convert temperatures to Kelvin:
T₁ = 10°C + 273.15 = 283.15 K

Substitute the given values into the Arrhenius equation:
ln((1.5 × 10⁴) / (4.5 × 10³)) = (60 × 10³ J mol⁻¹ / 8.314 J K⁻¹ mol⁻¹) * (1/283.15 K - 1/T₂)

ln(15000 / 4500) = (60000 / 8.314) * (1/283.15 - 1/T₂)

ln(10/3) = 7216.743 * (0.003531 - 1/T₂)

2.3026 * log(10/3) = 7216.743 * (0.003531 - 1/T₂)

1.20397 = 7216.743 * (0.003531 - 1/T₂)

Divide both sides by 7216.743:
1.20397 / 7216.743 = 0.003531 - 1/T₂

0.0001668 = 0.003531 - 1/T₂

Rearrange to solve for 1/T₂:
1/T₂ = 0.003531 - 0.0001668
1/T₂ = 0.0033642

Now, calculate T₂:
T₂ = 1 / 0.0033642
T₂ ≈ 297.23 K

Convert T₂ back to Celsius:
T₂ (°C) = 297.23 - 273.15 = 24.08 °C

Therefore, the temperature at which k would be 1.5 × 10⁴ s⁻¹ is approximately 297.23 K or 24.08 °C.

Question 18

Q18. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s⁻¹, calculate Ea a

Accepted Answer

This problem involves both first-order kinetics and the Arrhenius equation.

**Part 1: Calculate rate constants at 298 K (k₁) and 308 K (k₂)**
For a first-order reaction, the integrated rate law is:
t = (2.303 / k) * log([A]₀ / [A])

Let [A]₀ = 100 (initial concentration)

At 298 K (T₁):
10% completion means [A] = 100 - 10 = 90
t₁ = (2.303 / k₁) * log(100 / 90)
t₁ = (2.303 / k₁) * log(1.1111)
t₁ = (2.303 / k₁) * 0.04576
t₁ = 0.1054 / k₁  --- (Equation 1)

At 308 K (T₂):
25% completion means [A] = 100 - 25 = 75
t₂ = (2.303 / k₂) * log(100 / 75)
t₂ = (2.303 / k₂) * log(1.3333)
t₂ = (2.303 / k₂) * 0.1249
t₂ = 0.2876 / k₂  --- (Equation 2)

Given that t₁ = t₂:
0.1054 / k₁ = 0.2876 / k₂
k₂ / k₁ = 0.2876 / 0.1054
k₂ / k₁ = 2.7286  --- (Equation 3)

**Part 2: Calculate Ea at 308 K**
We use the Arrhenius equation: ln k = ln A - Ea / RT
Or, in terms of log: log k = log A - Ea / (2.303 RT)

Given A = 4 × 10¹⁰ s⁻¹
For T₂ = 308 K:
log k₂ = log(4 × 10¹⁰) - Ea / (2.303 × 8.314 J K⁻¹ mol⁻¹ × 308 K)
log k₂ = 10.6021 - Ea / (5907.64)

For T₁ = 298 K:
log k₁ = log(4 × 10¹⁰) - Ea / (2.303 × 8.314 J K⁻¹ mol⁻¹ × 298 K)
log k₁ = 10.6021 - Ea / (5714.65)

Subtracting the two equations:
log k₂ - log k₁ = (10.6021 - Ea / 5907.64) - (10.6021 - Ea / 5714.65)
log(k₂ / k₁) = Ea / 5714.65 - Ea / 5907.64
log(k₂ / k₁) = Ea * (1/5714.65 - 1/5907.64)

From Equation 3, k₂ / k₁ = 2.7286
log(2.7286) = Ea * (0.0001750 - 0.0001693)
0.4359 = Ea * (0.0000057)

Ea = 0.4359 / 0.0000057
Ea = 76473.68 J mol⁻¹
Ea = 76.47 kJ mol⁻¹

**Part 3: Calculate the rate constant at 298 K (k₁)**
We can use the Arrhenius equation for k₁:
ln k₁ = ln A - Ea / (R T₁)
k₁ = A * e^(-Ea / (R T₁))

k₁ = (4 × 10¹⁰ s⁻¹) * e^(-76473.68 J mol⁻¹ / (8.314 J K⁻¹ mol⁻¹ × 298 K))
k₁ = (4 × 10¹⁰) * e^(-76473.68 / 2477.572)
k₁ = (4 × 10¹⁰) * e^(-30.866)
k₁ = (4 × 10¹⁰) * (1.86 × 10⁻¹⁴)
k₁ = 7.44 × 10⁻⁴ s⁻¹

Alternatively, using log form:
log k₁ = log A - Ea / (2.303 R T₁)
log k₁ = log(4 × 10¹⁰) - 76473.68 / (2.303 × 8.314 × 298)
log k₁ = 10.6021 - 76473.68 / 5714.65
log k₁ = 10.6021 - 13.381
log k₁ = -2.7789
k₁ = 10^(-2.7789)
k₁ = 1.66 × 10⁻³ s⁻¹

Let's recheck the calculation of k₂/k₁ ratio and then use the two-point Arrhenius equation directly for Ea.

Using the two-point Arrhenius equation:
ln(k₂/k₁) = (Ea/R) * (1/T₁ - 1/T₂)
ln(2.7286) = (Ea/8.314) * (1/298 - 1/308)
1.0037 = (Ea/8.314) * (0.0033557 - 0.0032468)
1.0037 = (Ea/8.314) * (0.0001089)
Ea = (1.0037 * 8.314) / 0.0001089
Ea = 8.346 / 0.0001089
Ea = 76639.1 J mol⁻¹ = 76.64 kJ mol⁻¹
This value is consistent with the previous calculation.

Now, calculate k₁ using the Arrhenius equation and A:
ln k₁ = ln A - Ea / (R T₁)
ln k₁ = ln(4 × 10¹⁰) - 76639.1 / (8.314 × 298)
ln k₁ = 24.412 - 76639.1 / 2477.572
ln k₁ = 24.412 - 30.933
ln k₁ = -6.521
k₁ = e^(-6.521)
k₁ = 0.00147 s⁻¹ = 1.47 × 10⁻³ s⁻¹

Final Answer:
Activation energy (Ea) = 76.64 kJ mol⁻¹
Rate constant at 298 K (k₁) = 1.47 × 10⁻³ s⁻¹

Question 19

Q19. The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ mol⁻¹, what will be the value of pre-exponential facto

Accepted Answer

We can use the Arrhenius equation to find the pre-exponential factor (A).

Arrhenius equation: k = A * e^(-Ea/RT)
Taking the natural logarithm on both sides: ln k = ln A - Ea/RT

Given:
Rate constant (k) = 2.418 × 10⁻⁵ s⁻¹
Temperature (T) = 546 K
Activation energy (Ea) = 179.9 kJ mol⁻¹ = 179.9 × 10³ J mol⁻¹
Gas constant (R) = 8.314 J K⁻¹ mol⁻¹

Rearrange the equation to solve for ln A:
ln A = ln k + Ea/RT

First, calculate ln k:
ln k = ln(2.418 × 10⁻⁵) ≈ -10.630

Next, calculate Ea/RT:
Ea/RT = (179.9 × 10³ J mol⁻¹) / (8.314 J K⁻¹ mol⁻¹ × 546 K)
Ea/RT = 179900 / 4540.644
Ea/RT ≈ 39.620

Now, substitute these values into the equation for ln A:
ln A = -10.630 + 39.620
ln A = 28.990

Finally, calculate A:
A = e^(28.990)
A ≈ 3.52 × 10¹² s⁻¹

Therefore, the value of the pre-exponential factor (A) is approximately 3.52 × 10¹² s⁻¹.

Question 20

Q20. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temper

Accepted Answer

The energy of activation (Ea) for the reaction is approximately 52.27 kJ mol⁻¹.

Question 21

Q21. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 x 10^10 s^-1. Calculate Ea

Accepted Answer

The activation energy (Ea) is approximately 76.62 kJ mol⁻¹. The rate constant (k) at 318 K is approximately 3.65 x 10⁻³ s⁻¹.

Question 22

Q22. The rate constant for the decomposition of N2O5 at various temperatures is given below:

| T (°C) | k (s⁻¹) |
|---|---|
| 0 | 7.87 × 10⁻⁸ |
| 20 | 1.70 × 10⁻⁶ |
| 40 | 2.57 ×

Accepted Answer

To calculate the values of A (pre-exponential factor) and Ea (activation energy) from the given data, we use the Arrhenius equation in its logarithmic form:

ln k = ln A - (Ea / R) * (1/T)

This equation is in the form of a straight line y = mx + c, where:
- y = ln k
- x = 1/T
- m (slope) = -Ea / R
- c (y-intercept) = ln A

First, we convert the given temperatures from °C to Kelvin and calculate 1/T and ln k for each data point.

Question 23

Q23. The decomposition of A into product has a value of k as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?

Accepted Answer

This problem involves the Arrhenius equation, which relates the rate constant (k) to temperature (T) and activation energy (Ea). We can use the two-point form of the Arrhenius equation to solve for the unknown temperature.

Question 24

Q24. The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s⁻¹. Calculate Ea a

Accepted Answer

This problem combines first-order reaction kinetics with the Arrhenius equation. We need to first establish a relationship between the rate constants at two different temperatures using the given information about reaction completion times, then use the Arrhenius equation to find Ea and finally calculate the rate constant at a third temperature.

Question 25

Q25. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temper

Accepted Answer

The energy of activation (Eₐ) of the reaction is 52.86 kJ mol⁻¹.

Question 26

Q26. The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with tempera

Accepted Answer

The energy of activation (Eₐ) for the reaction is calculated to be 52.86 kJ/mol. This is determined using the Arrhenius equation, which relates the rate constants at two different temperatures to the activation energy. Given that the reaction rate quadruples, the ratio of the final rate constant (k₂) to the initial rate constant (k₁) is 4.

Question 27

Q27. The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 10⁴ K/T
Calculate Eₐ for this reaction and at what temperat

Accepted Answer

The activation energy (Eₐ) for the reaction is 239.34 kJ/mol. The temperature at which the half-period of the reaction will be 256 minutes is 668.9 K.

The activation energy is found by comparing the given equation with the standard Arrhenius equation. The temperature is calculated by first finding the rate constant (k) from the given half-life and then substituting it back into the given equation to solve for T.

Chapter 4: Chemical Kinetics