For the reaction,

\(2NO (g) + O_2 (g) \rightarrow 2NO_2 (g)\)

system volume is suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order with respect to \(O_2\) and second order with respect to \(NO_2\), the rate of reaction will:

Increase to eight times of its initial value

The correct answer is option 3. Increase to eight times of its initial value.

When the volume of the reaction system is reduced by half at constant temperature (assuming an ideal gas), the concentration of all the gaseous reactants doubles according to the ideal gas law \((PV = nRT)\).

Since the reaction is:

\(2NO(g) + O_2(g) \longrightarrow 2NO2(g)\)

The rate law for a reaction relates the rate of the reaction \((R)\) to the concentration of the reactants (\([A]\) and \([B]\)) raised to their respective orders (\(n\) and \(m\)) in the reaction:

\(R = k [A]^n [B]^m\)

We are given that the reaction is first order with respect to \(O_2\) (\(m = 1\) for \(O_2\)) and second order with respect to \(NO_2\) (\(n = 2\) for \(NO\)).

Let us denote the initial concentrations of \(NO\) and \(O_2\) as \([NO]\) and \([O_2]\), respectively. After the volume reduction, the concentrations become:

\([NO]’ = 2 \times [NO]\) (due to doubling of concentration)

\([O_2]’ = 2 \times [O_2]\) (due to doubling of concentration)

Now we can plug these new concentrations into the rate law expression:

\(R' = k ([NO]')^2 ([O_2]'\)

Substitute the doubled concentrations:

\(R' = k (2 [NO])^2  (2 [O_2])\)

Expand the squares:

\(R' = k \times 4  [NO]^2 \times 2  [O_2]\)

Simplify:

\(R' = 8 \times k [NO]^2  [O_2]\)

Since the initial rate \((R)\) can be expressed as \(k  [NO]^2  [O_2]\) (without the doubled concentrations), we see that the new rate \((R')\) is eight times the initial rate \((R)\).

Therefore, reducing the volume by half (doubling the concentration) will increase the rate of the reaction eight times for this specific reaction order.