For a hypothetical reaction, `A + B + C to` Products, the rate law is determined to be rate `= k[A][B]^(2)`. If the concentration of B is doubled without changing the concentration of A and C, the reaction rate:

Options

A. doubles
B. increases by a factor of four
C. increases by a factor of six
D. decreases by a factor of eight

Explanation

To solve the problem, we need to analyze how the reaction rate changes when the concentration of B is doubled. Let's break down the steps:

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

### Step 2: Define the initial concentrations
Let:
- The concentration of A be \( [A] = x \)
- The concentration of B be \( [B] = y \)
- The concentration of C be \( [C] = z \)

### Step 3: Calculate the initial rate
Using the initial concentrations, the initial rate \( r_1 \) can be expressed as:
\[ r_1 = k[x][y]^2 \]

### Step 4: Change the concentration of B
Now, we are told that the concentration of B is doubled. Therefore, the new concentration of B will be:
\[ [B] = 2y \]

### Step 5: Write the new rate expression
With the new concentration of B, the new rate \( r_2 \) can be expressed as:
\[ r_2 = k[x][2y]^2 \]

### Step 6: Simplify the new rate expression
Now, substituting \( 2y \) into the rate expression:
\[ r_2 = k[x][2y]^2 = k[x][4y^2] = 4k[x][y^2] \]

### Step 7: Relate the new rate to the initial rate
We can see that:
\[ r_2 = 4r_1 \]

### Conclusion
Thus, the reaction rate increases by a factor of 4 when the concentration of B is doubled.

### Final Answer
The reaction rate increases by a factor of 4.

---