Verified Solution • 2025–26 Session moderate

QJEEM23Che097: JEE Main 2023 Chemistry - Class 12 JEEM23Che097 Solution: The above reaction was studied at by monitoring the…

KClO3+6FeSO4+3H2SO4→KCl+3Fe2(SO4)3+3H2O\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}

The above reaction was studied at 300 K300 \mathrm{~K} by monitoring the concentration of FeSO4\mathrm{FeSO}_{4} in which initial concentration was 10M10 \mathrm{M} and after half an hour became 8.8 M. The rate of production of Fe2(SO4)3\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} is _________ ×10−6 mol L s−1\times 10^{-6} \mathrm{~mol} \mathrm{~L} \mathrm{~s}^{-1} (Nearest integer)

Step-by-Step Verified Solution

The Rate of Reaction (ROR) for a reaction can be calculated based on the changes in concentrations of the reactants or the products over time. In the given reaction,

KClO3+6FeSO4+3H2SO4→KCl+3Fe2(SO4)3+3H2O\mathrm{KClO}_{3}+6 \mathrm{FeSO}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{KCl}+3 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}+3 \mathrm{H}_{2} \mathrm{O}

The rates of reaction can be written as :

amp;ROR=−Δ[KClO3]Δt=−16Δ[FeSO4]Δtamp;=+13Δ[Fe2(SO4)3]Δt\begin{aligned}& \mathrm{ROR}=-\frac{\Delta\left[\mathrm{KClO}_3\right]}{\Delta \mathrm{t}}=\frac{-1}{6} \frac{\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}} \\\\&=\frac{+1}{3} \frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}} \\\end{aligned}

From this, we can express the rate of formation of Fe2(SO4)3\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 in terms of the change in concentration of FeSO4\mathrm{FeSO}_4 :

Δ[Fe2(SO4)3]Δt=12−Δ[FeSO4]Δt\frac{\Delta\left[\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\right]}{\Delta \mathrm{t}}=\frac{1}{2} \frac{-\Delta\left[\mathrm{FeSO}_4\right]}{\Delta \mathrm{t}}

We know that the initial concentration of FeSO4\mathrm{FeSO}_4 was 10 M and after 30 minutes (or 1800 seconds), it became 8.8 M. Substituting these values in, we get :

=12(10−8.8)30×60=0.333×10−3=\frac{1}{2} \frac{(10-8.8)}{30 \times 60} = 0.333 \times 10^{-3}

To express the rate in terms of 10−6 mol L−1 s−110^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}} , we multiply the rate by 10310^{3} :

=333×10−6 mol L−1 s−1=333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}}

Therefore, the rate of production of Fe2(SO4)3\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} is 333×10−6 mol L−1 s−1333 \times 10^{-6} \, \mathrm{mol \, L^{-1} \, s^{-1}} .

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring