If the volume of the vessel in which the reaction `2NO+O_2 rarr2NO_2` is occurring is diminished to 1/3 rd of its initial volume . The rate of the reaction will be increased by

Options

A. One third
B. Three times
C. Nine times
D. Twenty seven times

Answer

To determine how the rate of the reaction `2NO + O2 → 2NO2` changes when the volume of the vessel is reduced to one-third of its initial volume, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction and Rate Law**:
The reaction is `2NO + O2 → 2NO2`. The rate law for this reaction can be expressed as:
\[ \text{Rate} = k [NO]^2 [O_2] \]
where \( k \) is the rate constant, and \([NO]\) and \([O_2]\) are the concentrations of nitrogen monoxide and oxygen, respectively.

2. **Initial Concentrations**:
Let's denote the initial concentrations of \( NO \) and \( O_2 \) as \([NO]_0\) and \([O_2]_0\). The initial rate of the reaction can be expressed as:
\[ \text{Rate}_1 = k [NO]_0^2 [O_2]_0 \]

3. **Effect of Volume Change**:
When the volume of the vessel is reduced to one-third, the concentrations of the reactants will change. The concentration is defined as:
\[ [C] = \frac{n}{V} \]
where \( n \) is the number of moles and \( V \) is the volume. If the volume is reduced to \( \frac{V}{3} \), the new concentrations will be:
\[ [NO] = \frac{n_{NO}}{V/3} = 3 \cdot \frac{n_{NO}}{V} = 3[NO]_0 \]
\[ [O_2] = \frac{n_{O_2}}{V/3} = 3 \cdot \frac{n_{O_2}}{V} = 3[O_2]_0 \]

4. **Calculate New Rate**:
Now, substituting the new concentrations into the rate law gives us the new rate:
\[ \text{Rate}' = k [NO]^2 [O_2] = k (3[NO]_0)^2 (3[O_2]_0) \]
Simplifying this:
\[ \text{Rate}' = k \cdot 9[NO]_0^2 \cdot 3[O_2]_0 = 27 k [NO]_0^2 [O_2]_0 \]

5. **Compare New Rate with Initial Rate**:
We can see that:
\[ \text{Rate}' = 27 \cdot \text{Rate}_1 \]
Therefore, the rate of the reaction increases by a factor of 27.

### Final Answer:
The rate of the reaction will be increased by a factor of **27**.