For a reactions `A + B rarr` product, it was found that rate of reaction increases four times if concentration of `'A'` is doubled, but the rate of reaction remains unaffected. If concentration of `'B'` is doubled. Hence, the rate law for the reaction is

Options

A. Rate = K[A][B]
B. Rate =` k[A]^2`
C. Rate = `K[A]^2[B]^1`
D. Rate = `K[A]^2[B]^2`

Answer

To determine the rate law for the reaction \( A + B \rightarrow \text{products} \), we can analyze the information given in the question step by step.

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

### Step 2: Analyze the effect of doubling the concentration of A
According to the problem, when the concentration of A is doubled, the rate of reaction increases four times. This can be expressed mathematically as:
\[ \text{Rate}_{\text{new}} = k [2A]^x = k (2^x [A]^x) = 4 \cdot \text{Rate}_{\text{initial}} = 4k [A]^x \]
From this, we can equate:
\[ 2^x = 4 \]
This implies:
\[ x = 2 \]
Thus, the order with respect to A is 2.

### Step 3: Analyze the effect of doubling the concentration of B
The problem states that when the concentration of B is doubled, the rate of reaction remains unaffected. This means:
\[ \text{Rate}_{\text{new}} = k [A]^x [2B]^y = k [A]^x (2^y [B]^y) = \text{Rate}_{\text{initial}} = k [A]^x [B]^y \]
From this, we can equate:
\[ 2^y = 1 \]
This implies:
\[ y = 0 \]
Thus, the order with respect to B is 0.

### Step 4: Write the final rate law
Now that we have determined the values of \( x \) and \( y \), we can write the final rate law:
\[ \text{Rate} = k [A]^2 [B]^0 \]
Since \( [B]^0 = 1 \), we can simplify this to:
\[ \text{Rate} = k [A]^2 \]

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

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