The rate of a gaseous reaction is given by the expresison `k[A]^(2)[B]^(3)`. The volume of the reaction vessel is suddenly reduced to one-half of the initial volume. The reaction rate relative to the original rate will be

Options

A. `(1)/(8) a`
B. `(1)/(2)a`
C. 2a
D. 32a

Explanation

To solve the problem, we need to analyze how the rate of the reaction changes when the volume of the reaction vessel is halved. The rate of the reaction is given by the expression:

\[ \text{Rate} = k[A]^2[B]^3 \]

### Step 1: Understand the relationship between concentration and volume
Concentration (\([C]\)) is defined as the number of moles of a substance per unit volume. When the volume of the reaction vessel is reduced to half, the concentration of each reactant will double. This is because concentration is inversely proportional to volume.

### Step 2: Define the new concentrations
Let’s denote the original concentrations of \(A\) and \(B\) as \([A]\) and \([B]\). After the volume is halved, the new concentrations will be:

\[ [A'] = 2[A] \]
\[ [B'] = 2[B] \]

### Step 3: Write the new rate expression
Now we can write the new rate of the reaction using the new concentrations:

\[ \text{New Rate} = k[A']^2[B']^3 \]

Substituting the new concentrations into the rate expression:

\[ \text{New Rate} = k(2[A])^2(2[B])^3 \]

### Step 4: Simplify the new rate expression
Now we simplify the expression:

\[ \text{New Rate} = k(4[A]^2)(8[B]^3) \]
\[ \text{New Rate} = 32k[A]^2[B]^3 \]

### Step 5: Relate the new rate to the original rate
The original rate of the reaction is:

\[ \text{Original Rate} = k[A]^2[B]^3 \]

Thus, we can express the new rate in terms of the original rate:

\[ \text{New Rate} = 32 \times \text{Original Rate} \]

### Final Answer
Therefore, the reaction rate relative to the original rate will be:

\[ \text{New Rate} = 32 \times \text{Original Rate} \]

### Conclusion
The correct answer is \(32a\), where \(a\) is the original rate.

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