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  • In the reaction of formation of sulphur trioxide by contact process <img alt="2SO_{2}+O_{2}\rightleftharpoons 2SO_{3}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%202SO_%7B2

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\left [ O_{2} \right ]}{dt}=-2.5\times 10^{-4}mol\; L^{-1}\; s^{-1}. The rate of reaction in terms of \left [ SO_{2} \right ] in mol L^{-1}\, s^{-1} will be :

Option: 1

-5.00 x 10-4


Option: 2

-2.50 x 10-4


Option: 3

-3.75 x 10-4


Option: 4

-1.25 x 10-4


Answers (1)

best_answer

 \mathrm{Formula = \frac{-d[R]}{dt}= \frac{-d[P]}{dt}}

Now,

 2SO_{2}+O_{2}\rightleftharpoons 2SO_{3}

So,

-\frac{d\left [ SO_{2} \right ]}{2dt}=\frac{-d\left [ O_{2} \right ]}{dt}=\frac{d\left [ SO_{3} \right ]}{2dt}

given that

\frac{d\left [ O_{2} \right ]}{dt}=-2.5\times 10^{-4}

\frac{d\left [ SO_{2} \right ]}{dt}=2\times \frac{d\left [ O_{2} \right ]}{dt}

\frac{d\left [ SO_{2} \right ]}{dt}=2\times\left ( -2.5\times 10^{-4} \right )

\frac{d\left [ SO_{2} \right ]}{dt} = -5\times 10^{-4}

Therefore, option (1) is correct.

Posted by

avinash.dongre

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