An elementary reaction is given as `2P + Q rarr` products. If concentration of `Q` is kept constant and concentration of `P` is doubled then rate of reaction is

Options

A. doubled
B. halved
C. quadrupled
D. remains same

Explanation

To solve the problem, we need to analyze the given elementary reaction and how changes in concentration affect the rate of reaction.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the reaction \(2P + Q \rightarrow \text{products}\) can be expressed as:
\[ \text{Rate} = k [P]^2 [Q]^1 \]
where \(k\) is the rate constant.

2. **Identify the Initial Conditions**:
Let’s denote the initial concentration of \(P\) as \([P]_0\) and the concentration of \(Q\) as \([Q]_0\). The initial rate of reaction (\(R_0\)) can then be expressed as:
\[ R_0 = k [P]_0^2 [Q]_0 \]

3. **Change in Concentration of \(P\)**:
According to the problem, the concentration of \(P\) is doubled while the concentration of \(Q\) is kept constant. Therefore, the new concentration of \(P\) will be:
\[ [P] = 2[P]_0 \]

4. **Calculate the New Rate**:
Now, we can calculate the new rate of reaction (\(R_1\)) with the doubled concentration of \(P\):
\[ R_1 = k [P]^2 [Q]_0 = k (2[P]_0)^2 [Q]_0 \]
Simplifying this gives:
\[ R_1 = k \cdot 4[P]_0^2 [Q]_0 \]

5. **Relate the New Rate to the Initial Rate**:
We can express \(R_1\) in terms of \(R_0\):
\[ R_1 = 4 \cdot k [P]_0^2 [Q]_0 = 4R_0 \]

6. **Conclusion**:
Therefore, when the concentration of \(P\) is doubled, the rate of reaction becomes four times the initial rate:
\[ R_1 = 4R_0 \]

### Final Answer:
The rate of reaction is **four times the initial rate**.

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