Question

The rate constant for a second-order reaction, A \(\rightarrow\) Product is 1.62 M\(^{-1}\)s\(^{-1}\). What will be the rate of reaction when the concentration of reactant is 2 \(\times\) 10\(^{-3}\) M?

• A. 3.24 \(\times\) 10\(^-3\) M s\(^{-1}\)
• B. 3.24 \(\times\) 10\(^-6\) M s\(^{-1}\)
• C. 6.48 \(\times\) 10\(^-6\) M s\(^{-1}\)
• D. 2 \(\times\) 10\(^-3\) M s\(^{-1}\)

Hint:

For a second-order reaction, the rate is proportional to the square of the concentration of the reactant: \(\text{Rate} = k[A]^2\).

Answer 1

The Correct Option is C

Step 1: Rate Law for a Second-Order Reaction.
The rate of a second-order reaction is given by the equation: \[ \text{Rate} = k[A]^2 \] where \(k\) is the rate constant, and \([A]\) is the concentration of reactant A.

Step 2: Substituting the Given Values.
We are given that the rate constant \(k = 1.62 \, \text{M}^{-1} \text{s}^{-1}\) and the concentration of A is \([A] = 2 \times 10^{-3} \, \text{M}\). Substituting these values into the rate law equation: \[ \text{Rate} = 1.62 \, \text{M}^{-1}\text{s}^{-1} \times (2 \times 10^{-3} \, \text{M})^2 = 1.62 \times 4 \times 10^{-6} = 6.48 \times 10^{-6} \, \text{M/s} \]
Step 3: Conclusion.
The rate of reaction is 6.48 \(\times\) 10\(^-6\) M s\(^{-1}\), corresponding to option (C).