Diazonium salt decomposes as
`C_(6)H_(5)N_(2)^(o+)Cl^(ɵ) rarrC_(6)H_(5)Cl+N_(2)`
at `10^(@)C`, the evolution of `N_(2)` becomes two first faster when the initial concentration of the salt is doubled. Thus, it is

Options

A. A first order reaction
B. A second order reaction
C. Independent of the initial concentration of reactant
D. A zero order reaction.

Explanation

To solve the problem regarding the decomposition of diazonium salt, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The decomposition of diazonium salt can be represented as:
\[ C_6H_5N_2^+Cl^- \rightarrow C_6H_5Cl + N_2 \]
Here, we are interested in the rate of evolution of \( N_2 \).

2. **Understanding Rate of Reaction**:
The rate of a reaction can be expressed as:
\[ \text{Rate} = k [C_6H_5N_2^+]^x \]
where \( k \) is the rate constant, \( [C_6H_5N_2^+] \) is the concentration of the diazonium salt, and \( x \) is the order of the reaction.

3. **Effect of Doubling Concentration**:
According to the problem, when the initial concentration of the diazonium salt is doubled, the rate of evolution of \( N_2 \) becomes two times faster. This can be mathematically expressed as:
\[ \text{Rate}' = 2 \times \text{Rate} \]
If we denote the initial concentration as \( [C_6H_5N_2^+] \), then when it is doubled, we have:
\[ \text{Rate}' = k (2[C_6H_5N_2^+])^x \]

4. **Setting Up the Equation**:
From the above, we can set up the equation:
\[ k (2[C_6H_5N_2^+])^x = 2 \times k [C_6H_5N_2^+]^x \]
Simplifying this gives:
\[ k \cdot 2^x \cdot [C_6H_5N_2^+]^x = 2 \cdot k \cdot [C_6H_5N_2^+]^x \]

5. **Canceling Common Terms**:
We can cancel \( k \) and \( [C_6H_5N_2^+]^x \) from both sides (assuming \( [C_6H_5N_2^+] \neq 0 \)):
\[ 2^x = 2 \]

6. **Solving for x**:
To find \( x \), we can equate:
\[ 2^x = 2^1 \]
This implies:
\[ x = 1 \]

7. **Conclusion**:
The order of the reaction is 1, meaning it is a first-order reaction.

### Final Answer:
The order of the reaction is **1** (first-order reaction).