The rate of simple reaction $2 N O + O_{2} ⟶ 2 {N O}_{2}$ , when the volume of the reaction vessel is doubled.

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Solution

Rate of the reaction depends upon the order of the individual reactants. Since volume is inversly proportional to the concentration, by decreasing the volume the concentration of the reactants should increase resulting in increasing the rate of the reaction. Also here it is evident that the reaction system is gaseous , so we are considering the volume of the reaction vessel same as that of the volume of the reaction speices.

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2 N O + O_{2} \to 2 N O_{2}

2 N O + O_{2} ⟶ 2 {N O}_{2}

\frac{1}{4}

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l^{- 1} s^{- 1}

2 N O + O_{2} \to 2 N O_{2}

= K {[N O]}^{2} [O_{2}]

{1 / 3}^{r d}

2 N O + O_{2} ⟶ {2 N O}_{2}

2 N O (g) + O_{2} (g) \to 2 N O_{2} (g)

R a t e = K [N O]^{2} [O_{2}]

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