KClO_3+6 FeSO_4+3 H_2 SO_4 arrow KCl+3 Fe_2(SO_4)_3+3 H_2 O The above reaction was studied at 300 ~K by monitoring the concentration of FeSO_4 in which initial concentration was 10 M and after half an hour became 8.8 M. The rate of production of Fe_2(SO_4)_3 is \underline\hspace2cm \times 10^-6 ~mol ~L ~s^-1 (Nearest integer)

Chemical Kinetics and Nuclear Chemistry PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q19 with Solution

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Correct answer: 333

The reaction is

\mathrm{KClO_3} + 6\,\mathrm{FeSO_4} + 3\,\mathrm{H_2SO_4} \rightarrow \mathrm{KCl} + 3\,\mathrm{Fe_2(SO_4)_3} + 3\,\mathrm{H_2O}

From stoichiometry,

6\,\mathrm{FeSO_4} \to 3\,\mathrm{Fe_2(SO_4)_3}

So,

-\frac{1}{6}\frac{d[\mathrm{FeSO_4}]}{dt} = \frac{1}{3}\frac{d[\mathrm{Fe_2(SO_4)_3}]}{dt}

Hence,

\frac{d[\mathrm{Fe_2(SO_4)_3}]}{dt} = \frac{3}{6}\left(-\frac{d[\mathrm{FeSO_4}]}{dt}\right) = \frac{1}{2}\left(-\frac{d[\mathrm{FeSO_4}]}{dt}\right)

Initial concentration of $\mathrm{FeSO_4}$ :

[\mathrm{FeSO_4}]_0 = 10\,\mathrm{M}

After half an hour:

[\mathrm{FeSO_4}] = 8.8\,\mathrm{M}

Decrease in concentration:

\Delta[\mathrm{FeSO_4}] = 10 - 8.8 = 1.2\,\mathrm{M}

Time interval:

\Delta t = 0.5\,\mathrm{hour} = 30\,\mathrm{min} = 1800\,\mathrm{s}

Average rate of disappearance of $\mathrm{FeSO_4}$ :

-\frac{\Delta[\mathrm{FeSO_4}]}{\Delta t} = \frac{1.2}{1800} =6.67\times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}

Using the stoichiometric factor:

\text{Rate of formation of } \mathrm{Fe_2(SO_4)_3} = \frac{1}{2}\times 6.67\times 10^{-4}

=3.33\times 10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}

The question asks for

\underline{\hspace{2cm}} \times 10^{-6}\,\mathrm{mol\,L^{-1}\,s^{-1}}

Now,

3.33\times 10^{-4} = 333\times 10^{-6}

So the required nearest integer is

\boxed{333}