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

Options

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

Answer

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

### Step 1: Understand the Rate Law
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,
- \( y \) is the order of the reaction with respect to B.

### Step 2: Set Up the Initial Rate
Let the initial rate of the reaction be:
\[ \text{Rate}_1 = k[A]^x[B]^y \]

### Step 3: Change the Concentration of B
According to the problem, when the concentration of B is doubled, the new rate becomes:
\[ \text{Rate}_2 = k[A]^x[2B]^y \]
This can be rewritten as:
\[ \text{Rate}_2 = k[A]^x(2^y[B]^y) = 2^y \cdot k[A]^x[B]^y \]

### Step 4: Relate the Two Rates
We know that the rate decreases by a factor of 4 when the concentration of B is doubled:
\[ \text{Rate}_2 = \frac{\text{Rate}_1}{4} \]

### Step 5: Set Up the Equation
Substituting the expressions for the rates, we have:
\[ 2^y \cdot k[A]^x[B]^y = \frac{k[A]^x[B]^y}{4} \]

### Step 6: Cancel Common Terms
Dividing both sides by \( k[A]^x[B]^y \) gives:
\[ 2^y = \frac{1}{4} \]

### Step 7: Rewrite the Equation
We can rewrite \( \frac{1}{4} \) as:
\[ \frac{1}{4} = 2^{-2} \]
Thus, we have:
\[ 2^y = 2^{-2} \]

### Step 8: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ y = -2 \]

### Step 9: Conclusion
The order of the reaction with respect to B is:
\[ \text{Order with respect to B} = y = -2 \]

### Final Answer
The order of the reaction with respect to reactant B is **-2**.

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