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 determine the rate law for the reaction \(2A + B \to \text{Products}\), we will analyze the information given in the question step by step.

### Step 1: Write the general form of the rate law
The rate law for a reaction can be expressed as:
\[ \text{Rate} = k [A]^x [B]^y \]
where \(k\) is the rate constant, \(x\) is the order with respect to reactant \(A\), and \(y\) is the order with respect to reactant \(B\).

### Step 2: Analyze the first condition
The first condition states that doubling the initial concentrations of both reactants increases the rate by a factor of 8.

If we double the concentrations:
\[ [A] \to 2[A], \quad [B] \to 2[B] \]
The new rate can be expressed as:
\[ \text{New Rate} = k (2[A])^x (2[B])^y = k \cdot 2^x [A]^x \cdot 2^y [B]^y = k \cdot 2^{x+y} [A]^x [B]^y \]
According to the problem, this new rate is 8 times the original rate:
\[ k \cdot 2^{x+y} [A]^x [B]^y = 8 \cdot k [A]^x [B]^y \]
Cancelling \(k [A]^x [B]^y\) from both sides gives:
\[ 2^{x+y} = 8 \]
Since \(8 = 2^3\), we have:
\[ x + y = 3 \tag{1} \]

### Step 3: Analyze the second condition
The second condition states that doubling the concentration of \(B\) alone doubles the rate.

If we double \(B\):
\[ [A] \to [A], \quad [B] \to 2[B] \]
The new rate becomes:
\[ \text{New Rate} = k [A]^x (2[B])^y = k [A]^x \cdot 2^y [B]^y = k \cdot 2^y [A]^x [B]^y \]
According to the problem, this new rate is twice the original rate:
\[ k \cdot 2^y [A]^x [B]^y = 2 \cdot k [A]^x [B]^y \]
Cancelling \(k [A]^x [B]^y\) gives:
\[ 2^y = 2 \]
This implies:
\[ y = 1 \tag{2} \]

### Step 4: Solve for \(x\)
Now we can substitute \(y = 1\) into equation (1):
\[ x + 1 = 3 \]
Thus:
\[ x = 2 \tag{3} \]

### Step 5: Write the final rate law
Now that we have both \(x\) and \(y\):
\[ x = 2, \quad y = 1 \]
The rate law can be written as:
\[ \text{Rate} = k [A]^2 [B]^1 \]
or simply:
\[ \text{Rate} = k [A]^2 [B] \]

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

---