How will the rate `r=k[SO_(2)][[O_(2)]` of reaction,
`2SO_(2)+O_(2)rarr2SO_(3)` change if the volume of the reaction vessel is tripled?

Options

A. It wil be `(1)/(3)` rd of its initial value.
B. It wil be 3 times of its initial value.
C. It wil be 9 times of its initial value.
D. It wil be `(1)/(9)th` of its initial value.

Answer

To determine how the rate of the reaction changes when the volume of the reaction vessel is tripled, we can follow these steps:

### Step 1: Write the Rate Expression
The rate of the reaction is given by the expression:
\[ r = k [SO_2][O_2] \]
where \( k \) is the rate constant, and \([SO_2]\) and \([O_2]\) are the concentrations of sulfur dioxide and oxygen, respectively.

### Step 2: Understand the Effect of Volume on Concentration
Concentration is defined as the number of moles of a substance per unit volume. If we denote the initial volume as \( V \), the concentration of a substance is given by:
\[ [SO_2] = \frac{n_{SO_2}}{V} \]
\[ [O_2] = \frac{n_{O_2}}{V} \]
where \( n_{SO_2} \) and \( n_{O_2} \) are the number of moles of sulfur dioxide and oxygen, respectively.

### Step 3: Change in Volume
When the volume is tripled, the new volume becomes:
\[ V' = 3V \]
As a result, the new concentrations will be:
\[ [SO_2]' = \frac{n_{SO_2}}{3V} \]
\[ [O_2]' = \frac{n_{O_2}}{3V} \]

### Step 4: Substitute New Concentrations into the Rate Expression
Now, we can substitute the new concentrations into the rate expression:
\[ r' = k [SO_2]' [O_2]' = k \left(\frac{n_{SO_2}}{3V}\right) \left(\frac{n_{O_2}}{3V}\right) \]

### Step 5: Simplify the New Rate Expression
This simplifies to:
\[ r' = k \left(\frac{n_{SO_2} n_{O_2}}{(3V)(3V)}\right) = k \left(\frac{n_{SO_2} n_{O_2}}{9V^2}\right) \]
Thus, we find that:
\[ r' = \frac{1}{9} k [SO_2][O_2] = \frac{1}{9} r \]

### Conclusion
Therefore, when the volume of the reaction vessel is tripled, the rate of the reaction decreases to one-ninth of its initial value:
\[ r' = \frac{1}{9} r \]

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