A reaction in which reactants (R ) are converted into products (P) follows second order kinetics. If concentration of R is increased by four times, what will be the increase in the rate formation of p?

Options

A. 9 times
B. 4 times
C. 16 times
D. 8 times

Explanation

To solve the problem step by step, we will analyze the given information about the reaction and apply the principles of chemical kinetics.

### Step 1: Understand the Reaction Order
The problem states that the reaction follows second-order kinetics. This means that the rate of reaction depends on the square of the concentration of the reactant \( R \).

### Step 2: Write the Rate Law
For a second-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k [R]^2 \]
where \( k \) is the rate constant and \( [R] \) is the concentration of the reactant.

### Step 3: Define the Initial Concentration
Let the initial concentration of \( R \) be \( [R] \).

### Step 4: Increase the Concentration
According to the problem, the concentration of \( R \) is increased by four times. Therefore, the new concentration \( [R'] \) can be expressed as:
\[ [R'] = 4[R] \]

### Step 5: Calculate the New Rate
Now we substitute the new concentration into the rate law:
\[ \text{New Rate} = k [R']^2 = k (4[R])^2 \]
Calculating this gives:
\[ \text{New Rate} = k \cdot 16[R]^2 = 16 \cdot (k [R]^2) \]
This shows that the new rate is 16 times the original rate.

### Step 6: Relate Rate to Formation of Product
The rate of formation of product \( P \) is directly related to the rate of the reaction. Therefore, if the rate of the reaction increases by a factor of 16, the rate of formation of product \( P \) also increases by the same factor.

### Conclusion
Thus, the increase in the rate of formation of \( P \) when the concentration of \( R \) is increased by four times is:
\[ \text{Increase in rate of formation of } P = 16 \text{ times} \]

### Final Answer
The correct option is \( 16 \) times.

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