In a reaction A + B `to` Products
(i) If the initial concentration of A is doubled and B is kept constant, the rate of the reaction is doubled
(ii) If the initial concentration both A and B are doubled the rate of reaction becomes eight times
The rate law of the reaction is :

Options

A. `k[A]^2`
B. `k[A]^2[B]^2`
C. `k[A][B]^2`
D. None of these

Answer

To determine the rate law for the reaction \( A + B \to \text{Products} \), we will analyze the information given in the problem step by step.

### Step 1: Establish the Rate Law Expression
The rate of a reaction can be expressed in terms of the concentrations of the reactants as follows:
\[ \text{Rate} = k [A]^x [B]^y \]
where \( k \) is the rate constant, \( x \) is the order of the reaction with respect to \( A \), and \( y \) is the order of the reaction with respect to \( B \).

### Step 2: Analyze the First Case
From the problem, we know:
- If the initial concentration of \( A \) is doubled (i.e., \( [A] \) becomes \( 2[A] \)) and \( [B] \) is kept constant, the rate of the reaction is doubled.

This can be expressed mathematically as:
\[ 2r = k (2[A])^x [B]^y \]
Simplifying this gives:
\[ 2r = k \cdot 2^x [A]^x [B]^y \]
Since the original rate \( r = k [A]^x [B]^y \), we can substitute \( r \) in the equation:
\[ 2r = 2^x r \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ 2 = 2^x \]
This implies:
\[ x = 1 \]
Thus, the order with respect to \( A \) is 1.

### Step 3: Analyze the Second Case
Now consider the second case:
- If the initial concentrations of both \( A \) and \( B \) are doubled, the rate of the reaction becomes eight times.

This can be expressed as:
\[ 8r = k (2[A])^x (2[B])^y \]
Simplifying this gives:
\[ 8r = k \cdot 2^x \cdot 2^y [A]^x [B]^y \]
This can be rewritten as:
\[ 8r = k \cdot 2^{x+y} [A]^x [B]^y \]
Substituting \( r \) again:
\[ 8r = 2^{x+y} r \]
Dividing both sides by \( r \):
\[ 8 = 2^{x+y} \]
This implies:
\[ 8 = 2^3 \quad \Rightarrow \quad x + y = 3 \]

### Step 4: Solve for \( y \)
We already found \( x = 1 \) from the first case. Now substituting \( x \) into the equation \( x + y = 3 \):
\[ 1 + y = 3 \quad \Rightarrow \quad y = 2 \]

### Step 5: Write the Rate Law
Now that we have both \( x \) and \( y \):
- \( x = 1 \) (order with respect to \( A \))
- \( y = 2 \) (order with respect to \( B \))

The rate law expression for the reaction is:
\[ \text{Rate} = k [A]^1 [B]^2 \]
or simply:
\[ \text{Rate} = k [A] [B]^2 \]

### Final Answer
The rate law of the reaction is:
\[ \text{Rate} = k [A] [B]^2 \]