For a reaction `2A + Bto` Products, doubling the initial concentration of both the reactants increases the rate by a factor of 8, and doubling the concentration of B alone doubles the rate. The rate law for the reaction is:

Options

A. `r = K[A][B]^2`
B. `r = K[A]^2[B]`
C. `r = K[A][B]`
D. `r = [A]^2[B]^2`

Answer

To solve the problem step by step, we need to derive the rate law for the reaction \(2A + B \to \text{Products}\) based on the given conditions.

### Step 1: Understand the given information
We know:
1. Doubling the initial concentration of both reactants \(A\) and \(B\) increases the rate by a factor of 8.
2. Doubling the concentration of \(B\) alone doubles the rate.

### Step 2: Write the general form of the rate law
The rate law can be expressed as:
\[ \text{Rate} = k [A]^m [B]^n \]
where \(m\) and \(n\) are the orders of the reaction with respect to \(A\) and \(B\), respectively.

### Step 3: Analyze the first condition
When we double the concentrations of both reactants:
\[ [A] \to 2[A] \quad \text{and} \quad [B] \to 2[B] \]
The new rate becomes:
\[ \text{Rate}' = k (2[A])^m (2[B])^n = k \cdot 2^m [A]^m \cdot 2^n [B]^n = k \cdot 2^{m+n} [A]^m [B]^n \]
According to the problem, this new rate is 8 times the original rate:
\[ \text{Rate}' = 8 \cdot \text{Rate} \Rightarrow k \cdot 2^{m+n} [A]^m [B]^n = 8 \cdot k [A]^m [B]^n \]
This simplifies to:
\[ 2^{m+n} = 8 \Rightarrow 2^{m+n} = 2^3 \Rightarrow m+n = 3 \]

### Step 4: Analyze the second condition
Now, we double the concentration of \(B\) while keeping \(A\) constant:
\[ [A] \to [A] \quad \text{and} \quad [B] \to 2[B] \]
The new rate becomes:
\[ \text{Rate}'' = k [A]^m (2[B])^n = k [A]^m \cdot 2^n [B]^n = 2^n k [A]^m [B]^n \]
According to the problem, this new rate is double the original rate:
\[ \text{Rate}'' = 2 \cdot \text{Rate} \Rightarrow 2^n k [A]^m [B]^n = 2 \cdot k [A]^m [B]^n \]
This simplifies to:
\[ 2^n = 2 \Rightarrow n = 1 \]

### Step 5: Determine the value of \(m\)
From the previous steps, we have:
1. \(m + n = 3\)
2. \(n = 1\)

Substituting \(n\) into the first equation:
\[ m + 1 = 3 \Rightarrow m = 2 \]

### Step 6: Write the final rate law
The rate law for the reaction is:
\[ \text{Rate} = k [A]^2 [B]^1 \]

### Conclusion
Thus, the rate law for the reaction \(2A + B \to \text{Products}\) is:
\[ \text{Rate} = k [A]^2 [B] \]