The rate law of a chemical reaction
`2NO(g)+O_(2)(g)=2NO_(2(g)`
is given as rate `= k[NO]^(2) [O_(2)]` .How will the rate of the reaction change if the volume of the reaction vessel is reduced to one - fourth of its initial volume ?

Explanation

To determine how the rate of the reaction changes when the volume of the reaction vessel is reduced to one-fourth of its initial volume, we can follow these steps:

### Step 1: Understand the Rate Law
The rate law for the given reaction is:
\[ \text{Rate} = k[\text{NO}]^2[\text{O}_2] \]
where \( k \) is the rate constant, and \([\text{NO}]\) and \([\text{O}_2]\) are the concentrations of the reactants.

### Step 2: Initial Concentrations
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