Given $2 N O_{(g)} + O_{2 (g)} \to 2 N O_{2 (g)}$ ; rate =k $[N O]^{2}$ $[O_{2}]^{1}$ . By how many times does the rate of the reaction change when the volume of the reaction vessel is reduced to 1/3rd of its original volume? Will there be any change in the order of the reaction?

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Solution

Suppose let $x$ and $y$ be the moles of $N O$ and $O_{2}$ in the vessel with volume $V$
Rate of the reaction $(R_{1})$ = $k [x / V]^{2} \times [y / V]$
Now the volume $V$ is reduced to $V / 3$
Rate of the reaction $(R_{2})$ = $k [x / (V / 3)]^{2} \times [y / (V / 3)]$
Rate of the reaction $(R_{2})$ = $27 k [x / V]^{2} \times [y / V]$
$\frac{R_{1}}{R_{2}} = \frac{1}{27}$
$R_{2} = 27 R_{1}$
Hence, rate of the reaction becomes 27 times greater than earlier.
There is no change in the order of reaction because the none of the mass of the reactant is increased.

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