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Reaction AtoB follows second order kinet...

Reaction `AtoB` follows second order kinetics. Doubling the concentration of A wil increase the rate of formation of B by a factor of :

A

`1//4`

B

`1//2`

C

2

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze how the rate of a second-order reaction changes when the concentration of the reactant is doubled. Here's a step-by-step breakdown of the solution: ### Step 1: Understand the Rate Law for a Second-Order Reaction For a reaction that follows second-order kinetics, the rate law can be expressed as: \[ \text{Rate} = k [A]^2 \] where \( k \) is the rate constant and \([A]\) is the concentration of reactant A. ### Step 2: Determine the Initial Rate Let’s denote the initial concentration of A as \([A]\). Therefore, the initial rate (\(R_i\)) can be written as: \[ R_i = k [A]^2 \] ### Step 3: Calculate the Rate After Doubling the Concentration When the concentration of A is doubled, it becomes \(2[A]\). The new rate (\(R_f\)) can be calculated as: \[ R_f = k (2[A])^2 \] \[ R_f = k \cdot 4[A]^2 \] This simplifies to: \[ R_f = 4k [A]^2 \] ### Step 4: Compare the New Rate to the Initial Rate Now, we can compare the new rate (\(R_f\)) to the initial rate (\(R_i\)): \[ \frac{R_f}{R_i} = \frac{4k [A]^2}{k [A]^2} \] The \(k\) and \([A]^2\) terms cancel out: \[ \frac{R_f}{R_i} = 4 \] ### Step 5: Conclusion This means that doubling the concentration of A increases the rate of formation of B by a factor of 4. Therefore, the answer to the question is: **The rate of formation of B increases by a factor of 4.** ### Final Answer The correct option is **4**. ---
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