Question

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

Answer 1

Given,
Initial concentration, $\left[A_1\right]=0.5M$
Final concentration, $\left[A_2\right]=0.4M$
Time interval, $△t=10$ min

Rate of reaction
For a general reaction, $mA\to nB$, rate of
reaction is given by:
Rate $=-\frac{1}{m}\frac{d\left[A\right]}{dt}=\frac{1}{n}\frac{d\left[B\right]}{dt}$
So,
For $2A\to$ Products
Rate $=-\frac{1}{2}\frac{△\left[A\right]}{△t}=-\frac{1}{2}\frac{(0.4-0.5)\text{mol}{L}^{-1}}{10\text{min}}$

$=\frac{0.1\text{mol}{L}^{-1}}{20\text{min}}$

$=5\times {10}^{-3}\text{mol}{L}^{-1}{\text{min}}^{-1}$

Final answer:

Rate $=5\times {10}^{-3}\text{mol}{L}^{-1}{\text{min}}^{-1}$