Q.A reaction is first order in A and second order in B

The key idea is that the order of a reaction with respect to a reactant tells you how the rate changes when that reactant’s concentration changes.

• Doubling [A] quadruples the rate → rate ∝ [A]² → order with respect to A is 2.
• Doubling [B] has no effect → rate ∝ [B]⁰ → order with respect to B is 0.

So the rate law is simply rate = k[A]²[B]⁰ = k[A]².

1. Interpret the effect of doubling [A]:

   The problem says: “doubling the concentration of A increases the reaction rate by four times.”

   If rate ∝ [A]ⁿ, then doubling [A] gives a factor of 2ⁿ in the rate.

   We have 2ⁿ = 4 → n = 2. So the reaction is second order in A.

2. Interpret the effect of doubling [B]:

   “Doubling the concentration of B there is no apparent effect on the rate.”

   That means rate ∝ [B]ᵐ with 2ᵐ = 1 → m = 0. So the reaction is zero order in B.

3. Write the overall rate equation:

   Combining the orders: rate = k [A]² [B]⁰ = k [A]².

   This matches option (B). …

The rate of a chemical reaction is highly dependent on temperature. This relationship is quantitatively described by the Arrhenius equation, which links the rate constant ($k$) of a reaction to its activation energy ($E_a$) and the absolute temperature ($T$).

Concept and Intuition

For a chemical reaction to occur, reactant molecules must collide with sufficient energy to overcome an energy barrier. This minimum energy required for a successful collision is called the activation energy ($E_a$).

• Temperature's Role: As temperature increases, the average kinetic energy of molecules rises. This means a larger fraction of molecules will possess energy equal to or greater than the activation energy, leading to more effective collisions and, consequently, a faster reaction rate. The rate constant ($k$) is a direct measure of this reaction speed.
• Arrhenius Equation: The Arrhenius equation mathematically expresses this relationship:

where:
*   $k$ is the rate constant
*   $A$ is the pre-exponential factor (or frequency factor), which accounts for the frequency of collisions and their proper orientation.
*   $E_a$ is the activation energy (in J/mol or kJ/mol)
*   $R$ is the universal gas constant ($8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)
*   $T$ is the absolute temperature (in Kelvin)

When we compare the rate constants of a reaction at two different temperatures, we can use a derived form of the Arrhenius equation that eliminates the pre-exponential factor $A$ (assuming it remains constant over the temperature range). This allows us to directly calculate the activation energy.

Step-by-Step Calculation

1. Write the Arrhenius equation for two different temperatures: Let $k_1$ be the rate constant at temperature $T_1$, and $k_2$ be the rate constant at temperature $T_2$.

2. Derive the two-point Arrhenius equation: Divide Equation 2 by Equation 1:

Taking the natural logarithm of both sides gives:
> [!FORMULA]
> $$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
This equation is particularly useful for calculating activation energy when rate constants at two different temperatures are known.

3. Identify the given values:

* Initial temperature, $T_1 = 300 \text{ K}$

* Final temperature, $T_2 = 600 \text{ K}$ …

The key idea here is that in a crystal lattice, atoms at different positions contribute differently to a single unit cell. A corner atom is shared among 8 adjacent cells, so only $\frac{1}{8}$ of it belongs to one cell. A face-center atom is shared between 2 cells, so $\frac{1}{2}$ of it belongs to one cell. This fractional counting is the foundation of determining the composition of an alloy from its lattice arrangement.

Let’s apply this to the given problem.

1. B atoms at corners

   A cube has 8 corners. Each corner atom is shared by 8 unit cells.

   Effective number of B atoms per unit cell = $8 \times \frac{1}{8} = 1$.

2. A atoms at face centers

   A cube has 6 faces. Each face-center atom is shared by 2 unit cells.

   Effective number of A atoms per unit cell = $6 \times \frac{1}{2} = 3$.

3. Ratio of A to B

   In one unit cell, we have 3 A atoms and 1 B atom.

   So the simplest whole-number ratio is $A : B = 3 : 1$, giving the formula $A_3B$. …