For the reaction, \(X + Y \longrightarrow Z\), the rate law expression is rate \(= k[X][Y]^3\). If the volume of the vessel is reduced to \(1/3\)rd of its original volume, then rate of reaction will:

increase 81 times

The correct answer is option 2. increase 81 times.

To determine how the rate of reaction changes when the volume is reduced, we first need to understand the relationship between the volume of the vessel and the concentration of the reactants.

Given:

The rate law for the reaction \( X + Y \longrightarrow Z \) is \(\text{rate} = k[X][Y]^3\).

The volume of the vessel is reduced to \(\frac{1}{3}\)rd of its original volume.

Let the initial concentrations of \( X \) and \( Y \) be \([X]_0\) and \([Y]_0\), respectively. When the volume is reduced to \(\frac{1}{3}\)rd, the concentrations will increase proportionally. Specifically:

New concentration of \( X \) = \(\frac{[X]_0}{\frac{1}{3}} = 3[X]_0\)

New concentration of \( Y \) = \(\frac{[Y]_0}{\frac{1}{3}} = 3[Y]_0\)

Substituting these new concentrations into the rate law:

Initial rate = \( k[X]_0[Y]_0^3 \)

New rate = \( k[3X_0][3Y_0]^3 \)

Now, simplify the expression for the new rate:

\(\text{New rate} = k \cdot 3[X]_0 \cdot (3[Y]_0)^3 = k \cdot 3[X]_0 \cdot 27[Y]_0^3 = 81 \cdot k[X]_0[Y]_0^3\)

Thus, the rate of reaction increases by a factor of 81.

So, the correct answer is: 2. increase 81 times