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

• A. 0.005
• B. 0.015
• C. 0.001
• D. 0.15
• E. 0.05

Answer 1

The Correct Option is B

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) : \(0.015\)

Answer 2

The rate of decrease in the concentration of A can be calculated using the formula: 
Rate = \(\frac{\Delta [A]}{\Delta t}\)
where: - \(\Delta [A]\) is the change in concentration of A, - \(\Delta t\) is the time interval.

Given: - Initial concentration of A, [A]_initial = 0.4 mol/L, - Final concentration of A, [A]_final = 0.1 mol/L, - Time interval, \(\Delta t = 20\) minutes.

The change in concentration, \(\Delta [A] = [A]_{\text{final}} - [A]_{\text{initial}} = 0.1 - 0.4 = -0.3 \text{ mol/L}\).
The negative sign indicates a decrease in concentration.

Now, calculating the rate:
Rate = \(\frac{-0.3}{20} = -0.015 \text{ mol/L/min}\)
Taking the absolute value (since rate is typically positive), the rate of decrease is 0.015 mol/L/min.