For a certain decompoistion, the rate is `0.30 M s^(-1)` when the concentration of the reactant is `0.20 M`. If the reaction is order, the rate (in `M s^(-1)`) when concentration is increased theee times is

Options

A. `0.30`
B. `0.90`
C. `0.60`
D. `2.70`

Explanation

To solve the problem step by step, we need to use the information given about the reaction rate and the order of the reaction.

### Step 1: Understand the relationship between rate and concentration
For a second-order reaction, the rate \( R \) can be expressed as:
\[ R = k[A]^2 \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

### Step 2: Set up the equation with the initial conditions
We know from the problem that when the concentration \( [A] = 0.20 \, M \), the rate \( R_1 = 0.30 \, M s^{-1} \). We can write:
\[ 0.30 = k(0.20)^2 \]

### Step 3: Solve for the rate constant \( k \)
Rearranging the equation gives:
\[ k = \frac{0.30}{(0.20)^2} \]
Calculating \( (0.20)^2 \):
\[ (0.20)^2 = 0.04 \]
Now substituting back:
\[ k = \frac{0.30}{0.04} = 7.5 \, M^{-1}s^{-1} \]

### Step 4: Determine the new concentration
The problem states that the concentration is increased three times. Therefore, the new concentration \( [A] \) will be:
\[ [A] = 3 \times 0.20 = 0.60 \, M \]

### Step 5: Calculate the new rate \( R_2 \)
Using the new concentration in the rate equation:
\[ R_2 = k[A]^2 = k(0.60)^2 \]
Calculating \( (0.60)^2 \):
\[ (0.60)^2 = 0.36 \]
Now substituting for \( k \):
\[ R_2 = 7.5 \times 0.36 \]
Calculating the multiplication:
\[ R_2 = 2.7 \, M s^{-1} \]

### Final Answer
The rate when the concentration is increased three times is:
\[ \boxed{2.7 \, M s^{-1}} \]

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