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  •  Areaction given below is stredied at some temperature by monitoring the concentration of <img alt="\mathrm{FSOSO}_4" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BFSOS

 Areaction given below is stredied at some temperature by monitoring the concentration of \mathrm{FSOSO}_4 in which initial concentration was \mathrm{12 \mathrm{M}} and after \mathrm{40 \mathrm{~min} } becomes \mathrm{5.2 \mathrm{M}}. Calculate the rate of: production of \mathrm{ \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3} :-
 

Option: 1

2.6 \times 10^{-3}


Option: 2

1.4 \times 10^{+3}


Option: 3

3.6 \times 10^{-2}


Option: 4

1.4 \times 10^{-3}


Answers (1)

best_answer

\mathrm{KClO}_3+6 \mathrm{Fe} \mathrm{SO}+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}
From the above equation, rate of decomposition \mathrm{\mathrm{FeSO}_4} and rate of production of \mathrm{\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3} can be written as:
\mathrm{\frac{1}{6} \times} Rate of decomposition \mathrm{\mathrm{FeSO}_4} = \mathrm{\frac{1}{3} \times} Rate of production \mathrm{ \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3}
 Rate of decomposition of \mathrm{\mathrm{FeSO}_4}
\mathrm{\begin{aligned} \frac{\text { Change in concentration }}{\text { Pto Time in seconds }} & =\frac{(12-5.2)}{40 \times 60} \\ & =\frac{6.8}{2400} \end{aligned} }
Rate of productio of \mathrm{(\left.\mathrm{Fe}_2 \mathrm{SO}_4\right)_3}
\mathrm{\begin{aligned} \frac{3.4}{2400} & =0.00141 \\ & =1.41 \times 10^{-3} \end{aligned} }

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