Question

In a reaction, \( 3A \rightarrow \text{Products} \), the concentration of \( A \) decreases from 0.6 mol L\(^{-1}\) to 0.3 mol L\(^{-1}\) in 20 minutes. What is the rate of the reaction during this interval?

• A. 0.05 mol L\(^{-1}\) min\(^{-1}\)
• B. 0.005 mol L\(^{-1}\) min\(^{-1}\)
• C. 0.03 mol L\(^{-1}\) min\(^{-1}\)
• D. 0.6 mol L\(^{-1}\) min\(^{-1}\)
• E. 0.003 mol L\(^{-1}\) min\(^{-1}\)

Hint:

To find the rate of a reaction, remember to divide the change in concentration by the time, and consider the stoichiometric coefficients of the reaction when the reactant and product coefficients differ.

Answer 1

The Correct Option is B

The rate of the reaction can be calculated by the change in concentration of the reactant over time. The formula for the rate of reaction is: \[ \text{Rate} = \frac{\Delta[\text{A}]}{\Delta t} \] Where: - \( \Delta[\text{A}] \) is the change in concentration of \( A \), - \( \Delta t \) is the time interval. Given: - Initial concentration of \( A = 0.6 \, \text{mol L}^{-1} \), - Final concentration of \( A = 0.3 \, \text{mol L}^{-1} \), - Time interval \( \Delta t = 20 \, \text{minutes} \). The change in concentration \( \Delta[\text{A}] \) is: \[ \Delta[\text{A}] = 0.6 - 0.3 = 0.3 \, \text{mol L}^{-1} \] Now, calculate the rate: \[ \text{Rate} = \frac{0.3}{20} = 0.015 \, \text{mol L}^{-1} \text{ min}^{-1} \] Since the stoichiometry of the reaction is 3:1 (3 moles of \( A \) react to form products), the rate of disappearance of \( A \) is: \[ \text{Rate of reaction} = \frac{0.015}{3} = 0.005 \, \text{mol L}^{-1} \text{ min}^{-1} \] Thus, the correct answer is (B) 0.005 mol L\(^{-1}\) min\(^{-1}\).