JEE Main 2023 — Chemistry Physical Chemistry

$${\mathrm{KClO}}_{3}+6{\mathrm{FeSO}}_{4}+3{H}_{2}{\mathrm{SO}}_{4}\rightarrow \mathrm{KCl}+3{\mathrm{Fe}}_{2}({\mathrm{SO}}_{4}{)}_{3}+3{H}_{2}O$$KClO3​+6FeSO4​+3H2​SO4​→KCl+3Fe2​(SO4​)3​+3H2​O

From the above equation, rate of decomposition ${\mathrm{FeSO}}_{4}$FeSO4​ and rate of production of ${\mathrm{Fe}}_{2}{({\mathrm{SO}}_{4})}_{3}$Fe2​(SO4​)3​ can be written as follows,

$$\frac{1}{6}\times \mathrm{Rate}\mathrm{of}{\mathrm{decomposition}}_{{\mathrm{FeSO}}_{4}}=\frac{1}{3}\times \mathrm{Rate}\mathrm{of}{\mathrm{production}}_{{\mathrm{Fe}}_{2}{({\mathrm{SO}}_{4})}_{3}}$$61​×RateofdecompositionFeSO4​​=31​×RateofproductionFe2​(SO4​)3​​

Rate of decomposition of ${\mathrm{FeSO}}_{4}$FeSO4​

$$\frac{\mathrm{Change}\mathrm{in}\mathrm{concentration}}{\mathrm{time}\mathrm{in}\mathrm{seconds}}=\frac{(10-8.8)}{30\times 60}$$timeinsecondsChangeinconcentration​=30×60(10−8.8)​

$$=\frac{1.2}{1800}\mathrm{mol}{L}^{-1}{s}^{-1}$$=18001.2​molL−1s−1

Rate of production of ${\mathrm{Fe}}_{2}{({\mathrm{SO}}_{4})}_{3}$Fe2​(SO4​)3​

$$=\frac{0.6}{1800}=3.33\times {10}^{-4}$$=18000.6​=3.33×10−4

$$=333.3\times {10}^{-6}{\mathrm{mol}}^{-1}{s}^{-1}$$=333.3×10−6mol−1s−1