Question

The above reaction was studied at 300 K by monitoring the concentration of $\mathrm{FeSO}_{4}$ in which initial concentration was (10 M) and after half an hour became 8.8 M) The rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is _____ x $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$. (Nearest integer)

Answer 1

333

Answer 2

1. Change in concentration of $\mathrm{FeSO}_{4}$: $\Delta[\mathrm{FeSO}_{4}] = 8.8 \mathrm{M} - 10 \mathrm{M} = -1.2 \mathrm{M}$.
2. Time in seconds: $t = 30 \text{ minutes} = 1800 \text{ s}$.
3. Rate of disappearance of $\mathrm{FeSO}_{4}$: $-\frac{\Delta[\mathrm{FeSO}_{4}]}{\Delta t} = -\frac{-1.2 \mathrm{M}}{1800 \mathrm{s}} = \frac{1.2}{1800} \mathrm{M/s}$.
4. Stoichiometric relation: 6 moles of $\mathrm{FeSO}_{4}$ produce 3 moles of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$. Rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ = $\frac{1}{2} \times$ Rate of disappearance of $\mathrm{FeSO}_{4}$.
5. Rate of production of $\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}$: Rate $= \frac{1}{2} \times \frac{1.2}{1800} \mathrm{M/s} = \frac{1.2}{3600} \mathrm{M/s}$.
6. Convert to desired units: $\frac{1.2}{3600} \mathrm{M/s} = \frac{1.2}{3600} \times 10^6 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} = \frac{1000}{3} \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \approx 333.33 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$.
7. Nearest integer: 333.