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  • For reaction, <img alt="2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{g})" src="https://entrancecorner.oncodecogs.com/gif.latex?2%20%5Cmathrm%7BNO%7D

For reaction,
2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{NO}_2(\mathrm{g})
Rate =\mathrm{K}[\mathrm{NO}]^2\left[\mathrm{O}_2\right]. If the volume of the reaction vessel is doubled, then the rate of the reaction:

Option: 1

will diminish to (1 / 4)^{\text {th }} of initial value.
 


Option: 2

will diminish to (1 / 8)^{\text {th }} of initial value.
 


Option: 3

will increase to 4 times.


Option: 4

will increase to 8 times.


Answers (1)

best_answer

\mathrm{ 2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g) }

\mathrm{ \text { Rate }=K[\mathrm{NO}]^2\left[\mathrm{O}_2\right] \\ }

\mathrm{ R=K[\mathrm{NO}]^2\left[\mathrm{O}_2\right] }

Suppose 'x' moles of NO and 'y' mole of \mathrm{\mathrm{O}_2 } are taken in the vessel of volume 'V' litre, then,
\mathrm{ r_1=K^{\prime}\left[\frac{x}{V}\right]^2\left[\frac{y}{V}\right] }

When the volume of reaction vessel is doubled,
\mathrm{ r_2=K^{\prime}\left[\frac{x}{2 V}\right]^2\left[\frac{y}{2 V}\right] \ldots \ldots \text {. eqn. } 1 \\ }

\mathrm{ r_2=K^{\prime} \frac{1}{8}\left[\frac{x}{V}\right]^2\left[\frac{y}{V}\right] \ldots \ldots \text { eqn. } 2 }

Dividing equation 1 by 2 ,
\mathrm{ \frac{r_1}{r_2}=\frac{8}{1} \\ }

\mathrm{ \Rightarrow r_2=\frac{r_1}{8} }
Thus, the rate of reaction become \mathrm{ {1/8}^{th} } times the intial rate.

Posted by

Gunjita

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