For a chemical reaction ` A to B` it is found that the rate of reaction doubles , when the concentration of A is increased four times . The order in A for this reaction is

Options

A. Two
B. One
C. Half
D. Zero

Explanation

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

### Step 1: Write the rate law expression
For the reaction \( A \rightarrow B \), the rate law can be expressed as:
\[ R = k [A]^x \]
where \( R \) is the rate of the reaction, \( k \) is the rate constant, \( [A] \) is the concentration of A, and \( x \) is the order of the reaction with respect to A.

### Step 2: Analyze the change in concentration and rate
According to the problem, when the concentration of A is increased from \( [A] \) to \( 4[A] \), the rate of reaction doubles. This can be expressed mathematically as:
\[ 2R = k (4[A])^x \]

### Step 3: Substitute the original rate expression
We can also express the original rate \( R \) as:
\[ R = k [A]^x \]
Now, substituting this into the equation for the doubled rate, we have:
\[ 2R = k (4[A])^x = k \cdot 4^x \cdot [A]^x \]

### Step 4: Set the equations equal
Now we can equate the two expressions for \( 2R \):
\[ 2(k [A]^x) = k \cdot 4^x \cdot [A]^x \]

### Step 5: Cancel out common terms
Since \( k \) and \( [A]^x \) are common on both sides, we can cancel them out (assuming \( [A] \neq 0 \)):
\[ 2 = 4^x \]

### Step 6: Express \( 4^x \) in terms of powers of 2
We know that \( 4 = 2^2 \), so we can rewrite \( 4^x \) as:
\[ 4^x = (2^2)^x = 2^{2x} \]
Thus, the equation becomes:
\[ 2 = 2^{2x} \]

### Step 7: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 1 = 2x \]

### Step 8: Solve for \( x \)
Dividing both sides by 2 gives:
\[ x = \frac{1}{2} \]

### Conclusion
Thus, the order of the reaction with respect to A is \( \frac{1}{2} \).