The rate of reaction between two reactants A and B increases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

Options

A. 2
B. -2
C. 1
D. -1

Answer

To determine the order of the reaction with respect to reactant B, we can follow these steps:

### Step 1: Write the Rate Law Expression
The rate of a reaction can be expressed using the rate law:
\[ \text{Rate} = k [A]^x [B]^y \]
where \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, \( [B] \) is the concentration of reactant B, \( x \) 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 Given Information
We know that if the concentration of B is doubled, the rate of the reaction increases by a factor of 4. This can be mathematically expressed as:
\[ \text{Rate}_2 = 4 \times \text{Rate}_1 \]
When the concentration of B is doubled:
\[ [B]_2 = 2[B]_1 \]

### Step 3: Set Up the Equations
Using the rate law, we can express the rates before and after the change in concentration:
1. For the initial concentration:
\[ \text{Rate}_1 = k [A]^x [B]^y \]
2. For the doubled concentration of B:
\[ \text{Rate}_2 = k [A]^x (2[B])^y = k [A]^x 2^y [B]^y \]

### Step 4: Relate the Two Rates
From the information given:
\[ \text{Rate}_2 = 4 \times \text{Rate}_1 \]
Substituting the expressions for the rates:
\[ k [A]^x 2^y [B]^y = 4 \times k [A]^x [B]^y \]

### Step 5: Cancel Common Terms
We can cancel \( k \), \( [A]^x \), and \( [B]^y \) from both sides:
\[ 2^y = 4 \]

### Step 6: Solve for y
We can express 4 as a power of 2:
\[ 4 = 2^2 \]
Thus, we have:
\[ 2^y = 2^2 \]
Since the bases are the same, we can equate the exponents:
\[ y = 2 \]

### Conclusion
The order of the reaction with respect to reactant B is \( y = 2 \).

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