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In a closed container at a given temperature, the concentration of a product P increases from 0.1 mol/L to 0.5 mol/L in 10 min. Then the average rate of reaction in first 10 min will be:

Option: 1

0.1 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}


Option: 2

0.5 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~min}


Option: 3

25 \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~min}


Option: 4

4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{min}^{-1}


Answers (1)

best_answer

Average rate of reaction:
\mathrm{r_{a v}=\frac{[P]_{\text {final }}-[P]_{\text {initial }}}{t_{\text {final }}-t_{\text {initial }}}=-\frac{\Delta P}{\Delta t}}
Here \mathrm{[P]_{\text {final }},[P]_{\text {initial }}} are the concentrations of product at \mathrm{t_{\text {final }}, t_{\text {initial }}} respectively
\mathrm r_{a v}=\frac{0.5-0.1}{10} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}

\mathrm r_{a v}=4 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}

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