The rate law of a chemical reaction given below:2NO(g)+O2(g)⟶2NO2(g)is given as rate =kNO2⋅O2 How will the rate of reaction change if the volume of reaction vessel is reduced to 14 th of its original value?

If volume becomes 14 th , conc. becomes 4 times⇒ New rate 42 (old rate) = 64 times ↑

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The rate law of a chemical reaction given below:
$2 NO (g) + O_{2} (g) ⟶ 2 {NO}_{2} (g)$
is given as rate $= k [{NO}^{2} \cdot [O_{2}]$ How will the rate of reaction change if the volume of reaction vessel is reduced to $\frac{1}{4} th$ of its original value?

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answer is 64.

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Detailed Solution

If volume becomes $\frac{1}{4} th$ , conc. becomes 4 times

$\Rightarrow$ New rate $4^{2}$ (old rate) = 64 times $↑$

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