In the reaction 2A -> Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. What is the rate of reaction in mol L^-1 min^-1?

1. For $2A \rightarrow$ Products, rate $= -\dfrac{1}{2}\dfrac{\Delta[A]}{\Delta t}$. 2. $\Delta[A] = 0.4 - 0.5 = -0.1\ \text{mol L}^{-1}$ over $10\ \text{min}$. 3. Rate $= -\dfrac{1}{2}\times\dfrac{-0.1}{10} = \dfrac{0.1}{20} = 0.005\ \text{mol L}^{-1}\text{min}^{-1}$. 4. Distractor B (0.010) forgets to divide by the stoichiometric coefficient 2. _Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.5_