For the reaction `A + B` products, it is observed that:
(1) on doubling the initial concentration of `A` only, the rate of reaction is also doubled and
(2) on doubling te initial concentration of both `A` and `B`, there is a charge by a factor of `8` in the rate of the reaction.
The rate of this reaction is given by

Options

A. `Rate =k[A]^(2)[B]`
B. `rate=k [A][B]^(2)`
C. `Rate =k[A]^(2)[b]^(2)`
D. `rate =k [A][B]`

Explanation

To determine the rate law for the reaction \( A + B \rightarrow \text{products} \), we will analyze the information provided step by step.

### Step 1: Understand the first observation
The first observation states that when the concentration of \( A \) is doubled, the rate of the reaction is also doubled. This implies that the rate of reaction is directly proportional to the concentration of \( A \).

Mathematically, we can express this as:
\[ \text{Rate} = k[A]^m[B]^n \]
where \( k \) is the rate constant, and \( m \) and \( n \) are the orders of the reaction with respect to \( A \) and \( B \) respectively.

From the first observation:
- If we double \( [A] \) (i.e., \( [A] \rightarrow 2[A] \)), the rate also doubles:
\[ 2 \cdot \text{Rate} = k(2[A])^m[B]^n \]
This simplifies to:
\[ 2 \cdot \text{Rate} = 2^m k[A]^m[B]^n \]
Since the original rate is \( k[A]^m[B]^n \), we can equate:
\[ 2 = 2^m \]
This implies \( m = 1 \). Thus, the reaction is first order with respect to \( A \).

### Step 2: Understand the second observation
The second observation states that when the concentrations of both \( A \) and \( B \) are doubled, the rate of reaction increases by a factor of 8.

Using the rate law we established:
\[ \text{Rate} = k[A]^m[B]^n \]
If we double both \( [A] \) and \( [B] \):
\[ \text{Rate}' = k(2[A])^m(2[B])^n = k(2^1[A])(2^n[B]) = k \cdot 2^m \cdot 2^n \cdot [A]^m \cdot [B]^n \]
Substituting \( m = 1 \):
\[ \text{Rate}' = k \cdot 2^1 \cdot 2^n \cdot [A]^1 \cdot [B]^n = 2^{1+n} \cdot k[A][B]^n \]
According to the second observation, this new rate is 8 times the original rate:
\[ \text{Rate}' = 8 \cdot \text{Rate} = 8 \cdot k[A][B]^n \]
Equating the two expressions for \( \text{Rate}' \):
\[ 2^{1+n} \cdot k[A][B]^n = 8 \cdot k[A][B]^n \]
Dividing both sides by \( k[A][B]^n \) (assuming \( [A] \) and \( [B] \) are not zero):
\[ 2^{1+n} = 8 \]
Since \( 8 = 2^3 \), we have:
\[ 1 + n = 3 \implies n = 2 \]
Thus, the reaction is second order with respect to \( B \).

### Conclusion
Combining the orders we found:
- The reaction is first order with respect to \( A \) (i.e., \( m = 1 \))
- The reaction is second order with respect to \( B \) (i.e., \( n = 2 \))

Therefore, the rate law for the reaction \( A + B \rightarrow \text{products} \) is:
\[ \text{Rate} = k[A]^1[B]^2 \]

### Final Answer
\[ \text{Rate} = k[A][B]^2 \]