The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is
`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i)
Half life of `nth` order reaction depends on the initial concentration according to the following relation:
`t_(1//2) prop (1)/(a^(n-1))` ...(ii)
The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is
Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii)
The differential rate law for `nth` order reaction may be given as:
`(dx)/(dt) = k[A]^(n)` ...(iv)
where `A` denotes the reactant.
In a chemical reaction `A rarr B`, it is found that the rate of the reaction doubles when the concentration of `A` is increased four times. The order of the reaction with respect to `A` is:

Options

A. `0`
B. `(1)/(2)`
C. `1`
D. `2`

Explanation

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

### Step 1: Write the rate law expression
The rate of a reaction can be expressed as:
\[ \text{Rate} = k[A]^n \]
where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of reactant A,
- \( n \) is the order of the reaction with respect to A.

### Step 2: Set up the initial and changed conditions
Let:
- \( r_1 \) be the initial rate of the reaction when the concentration of A is \( [A]_1 \).
- When the concentration of A is increased four times, it becomes \( [A]_2 = 4[A]_1 \).
- The new rate of the reaction is \( r_2 = 2r_1 \) (the rate doubles).

### Step 3: Write the rate equations for both conditions
For the initial condition:
\[ r_1 = k[A]_1^n \]

For the new condition:
\[ r_2 = k[A]_2^n = k(4[A]_1)^n = k(4^n[A]_1^n) \]

### Step 4: Set up the equation based on the rate change
Since \( r_2 = 2r_1 \), we can substitute the expressions for \( r_1 \) and \( r_2 \):
\[ k(4^n[A]_1^n) = 2(k[A]_1^n) \]

### Step 5: Cancel out common terms
We can cancel \( k \) and \( [A]_1^n \) from both sides (assuming \( [A]_1 \neq 0 \)):
\[ 4^n = 2 \]

### Step 6: Solve for \( n \)
To solve for \( n \), we can rewrite \( 4^n \) as \( (2^2)^n = 2^{2n} \):
\[ 2^{2n} = 2^1 \]

This implies:
\[ 2n = 1 \]
\[ n = \frac{1}{2} \]

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

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