Step 1: Identify the given values.

• Initial Concentration $[R]_1 = 0.03\text{ M}$
• Final Concentration $[R]_2 = 0.02\text{ M}$
• Time interval $\Delta t = 25\text{ minutes} = 25 \times 60 = 1500\text{ seconds}$

Step 2: Compute the average rate in minutes.

Step 3: Compute the average rate in seconds.

Answer: The average rate is $4.0 \times 10^{-4}\text{ M min}^{-1}$ or $6.67 \times 10^{-6}\text{ M s}^{-1}$.

1. Differences:

Why Rate Decreases: According to the rate law, the rate of a chemical reaction is directly proportional to the concentration of the reactants. As the reaction proceeds, reactant concentrations decrease because they are consumed. Consequently, the instantaneous rate of reaction continuously decreases over time.

Scientific Reason: For any chemical reaction to occur, reactant molecules must collide and cross a specific potential energy barrier called the **threshold energy**. The difference between threshold energy and the normal energy of the reactant molecules is called the **activation energy ($E_a$)**.

At room temperature, the thermodynamic kinetic energy of wood/coal and oxygen molecules is extremely low compared to their high activation energy barrier. As a result, almost zero molecular collisions possess sufficient energy to form the activated complex. Hence, a spark or external heat source is required to initially provide activation energy, after which the highly exothermic nature of combustion maintains the reaction.

Correct Option: (c)

Explanation: The overall order of a reaction is defined as the sum of the power exponents of concentration terms in the experimental rate law expression.

Definition: A reaction which is not truly of first-order but behaves as first-order under specific chemical conditions (usually when one of the reactants is present in large stoichiometric excess) is called a **pseudo-first-order reaction**.

Example: Acid-catalyzed hydrolysis of ethyl acetate.

The rate law for this reaction is: $\text{Rate} = k [\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O}]$. Because water is the solvent and present in immense excess ($\approx 55.5\text{ mol L}^{-1}$), its concentration remains practically unchanged during the reaction. Hence, $[\text{H}_2\text{O}]$ is constant and combined with the rate constant:

The reaction thus obeys first-order kinetics.

Step 1: Write the rate law expression for the second-order reaction.

Step 2: Express the new rate ($r_2$) with tripled concentration $[X]_2 = 3[X]$.

Answer: The rate of formation of $Y$ will increase by $9$ times.

Correct Option: (d)

Explanation: Molecularity is the physical count of reactant particles colliding to form products in a single elementary reaction. It can never be zero, fractional, or negative (since a fraction of a molecule or zero molecules cannot participate in a collision). Thus, Assertion (A) is false. However, Reason (R) is true as it correctly describes the nature of both properties.

Step 1: Set up the general rate law equation.

Step 2: Solve for $y$ using Run 1 and Run 2 (where $[A]$ is constant).

Thus, the reaction is of first order with respect to $B_2$.

Step 3: Solve for $x$ using Run 2 and Run 3 (where $[B_2]$ is constant).

Thus, the reaction is of zero order with respect to $A$.

Step 4: Express the overall rate law and calculate the rate constant $k$.

Using values from Run 1:

Answer: Order wrt $A$ is $0$, wrt $B_2$ is $1$. Rate Law is $\text{Rate} = k[B_2]$, and $k = \mathbf{3.2 \times 10^{-4}\text{ s}^{-1}}$.

Correct Option: (c)

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

Since the initial concentration parameter $[A]_0$ does not appear in this equation, the half-life of a first-order reaction is completely independent of the starting amount.

General Formula: The unit of rate constant for an $n^{\text{th}}$-order reaction is:

Applying order values:

• (A) Zero-order ($n=0$): $(\text{mol L}^{-1})^{1-0}\text{ s}^{-1} = \mathbf{\text{mol L}^{-1}\text{ s}^{-1}}$
• (B) First-order ($n=1$): $(\text{mol L}^{-1})^{1-1}\text{ s}^{-1} = \mathbf{\text{s}^{-1}}$
• (C) Second-order ($n=2$): $(\text{mol L}^{-1})^{1-2}\text{ s}^{-1} = \mathbf{\text{L mol}^{-1}\text{ s}^{-1}}$

Step 1: Apply the first-order integrated rate law to find $k$.

Let $[A]_0 = 100$. After $30\%$ decomposition, the remaining concentration is:

Step 2: Solve for $k$.

Step 3: Calculate the half-life ($t_{1/2}$).

Answer: The half-life of the reaction is $77.7\text{ minutes}$.

Case I: Time for $99\%$ completion ($t_{99\%}$).

Let $[A]_0 = 100$. For $99\%$ completion, $[A]_t = 100 - 99 = 1$.

Case II: Time for $90\%$ completion ($t_{90\%}$).

Let $[A]_0 = 100$. For $90\%$ completion, $[A]_t = 100 - 90 = 10$.

Step 3: Show the relationship.

Divide Equation 1 by Equation 2:

Hence proved, the time required for $99\%$ completion is exactly twice that of $90\%$ completion.

Step 1: Compute time for $99.9\%$ completion ($t_{99.9\%}$).

Let $[A]_0 = 100$. For $99.9\%$ completion, $[A]_t = 100 - 99.9 = 0.1$.

Step 2: Express half-life period ($t_{1/2}$).

Step 3: Compare both values.

Divide Equation 1 by Equation 2:

Hence proved, the time required for $99.9\%$ completion is $10$ times the half-life.

Step 1: Identify the given values.

• Rate constant $k = 1.15 \times 10^{-3}\text{ s}^{-1}$
• Initial concentration $[A]_0 = 5\text{ g}$
• Final concentration $[A]_t = 3\text{ g}$

Step 2: Apply the first-order integrated rate law.

Step 3: Substitute the log value and solve.

Answer: It will take approximately $444\text{ seconds}$ (or $7.4\text{ minutes}$).

Method 1: Half-life progression logic.

For first-order decay, the amount of substance halved after every half-life interval:

This process takes exactly $2$ half-lives. Therefore:

Method 2: Analytical first-order equation.

Find the rate constant $k$ first:

Now, calculate time for concentration to drop to $25\%$ (where $[A]_0=100$, $[A]_t=25$):

Answer: The time required is $280\text{ days}$.

Activation Energy ($E_a$): It is the minimum extra amount of energy absorbed by reactant molecules to reach the threshold energy level required to form the activated complex and undergo chemical transformation.

Role of a Catalyst:

• A catalyst speeds up a reaction by providing an **alternative reaction pathway** of lower activation energy.
• Because the activation energy barrier is reduced, a significantly larger fraction of reactant molecules possess sufficient kinetic energy to successfully collide and form products per unit time.
• Note that a catalyst does **not** change the Gibbs free energy ($\Delta G$) or enthalpy ($\Delta H$) of the overall reaction.

Correct Option: (c)

Explanation: According to the Maxwell-Boltzmann distribution curve, raising the temperature by $10\text{ K}$ does not significantly increase the total collision frequency (it only increases by about $1$ to $2\%$). Instead, it shifts the curve to the right, nearly **doubling the fraction of molecules** that possess energy equal to or greater than the activation energy. Consequently, the rate of reaction doubles.

Step 1: Set up the Arrhenius equation logarithmic form.

Identify parameters:

• $T_1 = 298\text{ K}$, $T_2 = 308\text{ K}$
• $k_2 / k_1 = 2 \implies \log(k_2/k_1) = \log(2) = 0.3010$
• $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$

Step 2: Substitute values and isolate $E_a$.

Step 3: Calculate $E_a$ in Joules and Kilojoules.

Answer: The activation energy ($E_a$) of the reaction is $52.9\text{ kJ mol}^{-1}$.

Step 1: Recall the Arrhenius equation and identify given values.

• $T_1 = 300\text{ K}$, $T_2 = 320\text{ K}$
• $k_2 / k_1 = 4 \implies \log(k_2/k_1) = \log(4) = 0.6020$
• $2.303 \times R = 19.15\text{ J K}^{-1}\text{ mol}^{-1}$

Step 2: Substitute values to isolate $E_a$.

Step 3: Compute the final value of $E_a$.

Answer: The activation energy ($E_a$) of the reaction is $55.34\text{ kJ mol}^{-1}$.