In the reaction of formation of sulphur triox | vedclass

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(D) For the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$,the rate of reaction is expressed as:Rate $= -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_

Vedclass · Accepted Answer

$-5.00 	imes 10^{-4}$

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(D) For the reaction $2SO_2 + O_2 ightleftharpoons 2SO_3$,the rate of reaction is expressed as:Rate $= -\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt} = \frac{1}{2} \frac{d[SO_3]}{dt}$Given that $\frac{d[O_2]}{dt} = -2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.Equating the terms for $SO_2$ and $O_2$:$-\frac{1}{2} \frac{d[SO_2]}{dt} = -\frac{d[O_2]}{dt}$$\frac{d[SO_2]}{dt} = 2 	imes \frac{d[O_2]}{dt}$$\frac{d[SO_2]}{dt} = 2 	imes (-2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1})$$\frac{d[SO_2]}{dt} = -5.00 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

In the reaction of formation of sulphur trioxide by contact process $2SO_2 + O_2 \rightleftharpoons 2SO_3$,the rate of reaction was measured as $\frac{d[O_2]}{dt} = -2.5 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$. The rate of reaction in terms of $[SO_2]$ in $mol \ L^{-1} \ s^{-1}$ will be:

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