Question:

In a reaction 3A → Products, the concentration of A decreases from 0.4molL-1 to 0.1molL-1 in 20 minutes at 300K. The rate of decrease in [A] during this interval (in molL-1min-1) at 300K is

Updated On: Apr 3, 2025
  • 0.005
  • 0.015
  • 0.001
  • 0.15
  • 0.05
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The Correct Option is B

Solution and Explanation

The rate of decrease in the concentration of A is given by the change in concentration over time. The change in concentration \( \Delta [A] \) is: \[ \Delta [A] = [A]_{\text{initial}} - [A]_{\text{final}} = 0.4 \, \text{mol L}^{-1} - 0.1 \, \text{mol L}^{-1} = 0.3 \, \text{mol L}^{-1} \] The time interval is given as 20 minutes. Therefore, the rate of decrease in concentration is: \[ \text{Rate} = \frac{\Delta [A]}{\text{time}} = \frac{0.3 \, \text{mol L}^{-1}}{20 \, \text{min}} = 0.015 \, \text{mol L}^{-1} \text{min}^{-1} \] Thus, the rate of decrease in [A] is 0.015 mol L$^{-1}$ min$^{-1}$.

The correct option is (B) : \(0.015\)

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