Question

$R \to P$ is a first-order reaction. The concentration of R changed from 0.04 to 0.03 mol L$^{-1}$ in 40 minutes. What is the average velocity of the reaction in mol L$^{-1}$ s$^{-1}$?

• A. \(2.5 \times 10^{-4}\)
• B. \(4.167 \times 10^{-6}\)
• C. \(4.167 \times 10^{-5}\)
• D. \(2.5 \times 10^{-5}\)

Hint:

Average velocity for first-order reactions can be calculated by dividing the change in concentration by the time interval.

Answer 1

The Correct Option is C

The average velocity of a reaction is: \[ v = \frac{\Delta [R]}{\Delta t} = \frac{0.04 - 0.03}{40 \, \text{min}} = \frac{0.01}{40 \times 60} = 4.167 \times 10^{-5}\, \text{mol L}^{-1} \text{s}^{-1} \]