Question

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

Answer 1

Step 1: Concentration of reactant and product.

Initial Concentration

$\left[{\mathrm{A}}_1\right]=0.5\mathrm{mol}{\mathrm{L}}^{-1}$

and, Final Concentration

$\left[{\mathrm{A}}_2\right]=0.4\mathrm{mol}{\mathrm{L}}^{-1}$

Step 2: Finding the rate of the reaction

Time = 10 min

$\mathrm{Rate}\mathrm{of}\mathrm{reactions}=\mathrm{Rate}\mathrm{of}\mathrm{disappearance}\mathrm{of}\mathrm{A}$

$\mathrm{Rate}\mathrm{of}\mathrm{reactions}=$

$-\frac{1}{2}\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}$

$=-\frac{1}{2}\frac{(0.4-0.5)\mathrm{mol}{\mathrm{L}}^{-1}}{10\min }$

$=\frac{(0.1)\mathrm{mol}{\mathrm{L}}^{-1}}{20\min }$

$=0.005\mathrm{mol}{\mathrm{L}}^{-1}{\min }^{-1}$

Therefore, the rate during this interval is $0.005\mathrm{mol}{\mathrm{L}}^{-1}{\min }^{-1}$