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  • For a given reaction: R  P following graph was obtained between concentration of

For a given reaction:
R \rightarrow P
following graph was obtained between concentration of reactant, R w.r.t time.


The value for average rate in \mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1} from \mathrm{t}_1=10 \mathrm{~min}$ to $\mathrm{t}_2=15 \mathrm{~min} is x, then find the value of 100x :

Option: 1

4


Option: 2

3


Option: 3

2


Option: 4

1


Answers (1)

best_answer

For R \rightarrow P
Average rate of reaction:
\mathrm{ r_{a v}=-\frac{[R]_{\text {final }}-[R]_{\text {initial }}}{t_{\text {final }}-t_{\text {initial }}}=-\frac{\Delta R}{\Delta t} }

Here \mathrm{ [R]_{\text {final }},[R]_{\text {initial }} } are the concentration of reactant A at \mathrm{ t_{\text {final }}=15 \mathrm{~min}, t_{\text {initial }}=10 \mathrm{~min} } respectively.

\mathrm{ r_{a v}=-\frac{6-9}{15-10} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ }

\mathrm{ r_{a v}=\frac{3}{5 \times 60} \mathrm{~mol} L^{-1} \mathrm{~s}^{-1}=0.01 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} }

\mathrm{ \text { So, } x=0.01 \mathrm{~mol}^{-1} \mathrm{~s}^{-1}\\ }

\mathrm{ \therefore 100 x=1 }

Posted by

avinash.dongre

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