`2A rarrC+D`
In this reaction, if we double the concentration of A, reaction rate become , two time .. What is the order of this reaction ?

Options

A. 1
B. 3
C. 2
D. 1.5

Answer

To determine the order of the reaction given by the equation \(2A \rightarrow C + D\), we need to analyze how the rate of the reaction changes when the concentration of reactant A is altered.

### Step-by-Step Solution:

1. **Understand the Rate Expression**:
The rate of a reaction can be expressed in terms of the concentration of the reactants raised to a power, which is the order of the reaction. For our reaction, we can write the rate expression as:
\[ \text{Rate} = k[A]^n \]
where \(k\) is the rate constant, \([A]\) is the concentration of reactant A, and \(n\) is the order of the reaction.

2. **Initial Conditions**:
Let’s denote the initial concentration of A as \([A]\) and the initial rate of the reaction as \(R\). Thus, we can write:
\[ R = k[A]^n \quad \text{(Equation 1)} \]

3. **Doubling the Concentration**:
According to the problem, if we double the concentration of A, the new concentration becomes \(2[A]\). The new rate of the reaction can be expressed as:
\[ \text{New Rate} = k[2A]^n = k(2[A])^n = k \cdot 2^n \cdot [A]^n \quad \text{(Equation 2)} \]

4. **Relating the Rates**:
We know from the problem statement that doubling the concentration of A results in the rate becoming 2 times the original rate:
\[ 2R = k \cdot 2^n \cdot [A]^n \]

5. **Setting Up the Equation**:
Now, we can set up the equation by substituting \(R\) from Equation 1 into the new rate equation:
\[ 2R = 2 \cdot k[A]^n = k \cdot 2^n \cdot [A]^n \]

6. **Canceling Common Terms**:
Dividing both sides by \(k[A]^n\) (assuming \(k\) and \([A]\) are not zero), we get:
\[ 2 = 2^n \]

7. **Solving for n**:
To find \(n\), we can rewrite the equation as:
\[ 2^1 = 2^n \]
This implies that \(n = 1\).

8. **Conclusion**:
Therefore, the order of the reaction is \(1\).

### Final Answer:
The order of the reaction is **1**.

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