Question

In the reaction 2A $\longrightarrow$ Products, concentration of A decreases from 0.5 mol L$^{-1$ to 0.4 mol L$^{-1}$ in 10 minutes. Find the rate of reaction during the time interval.}

Hint:

Always convert time into seconds when calculating rates in kinetics problems.

Answer 1

Step 1: Rate expression.
For the reaction: \[ 2A \longrightarrow \text{Products} \] \[ \text{Rate} = -\frac{1}{2}\frac{\Delta [A]}{\Delta t} \] Step 2: Change in concentration of A.
Initial $[A] = 0.5$ mol L$^{-1}$
Final $[A] = 0.4$ mol L$^{-1}$
\[ \Delta [A] = 0.4 - 0.5 = -0.1 \, \text{mol L}^{-1} \] Step 3: Time interval.
\[ \Delta t = 10 \, \text{min} = 600 \, \text{s} \] Step 4: Calculate rate.
\[ \text{Rate} = -\frac{1}{2}\frac{\Delta [A]}{\Delta t} = -\frac{1}{2}\frac{-0.1}{600} \] \[ \text{Rate} = \frac{0.1}{1200} = 8.33 \times 10^{-5} \, \text{mol L}^{-1}\text{s}^{-1} \] Conclusion:
The rate of reaction during the given time interval is: \[ \boxed{8.33 \times 10^{-5} \, \text{mol L}^{-1}\text{s}^{-1}} \]