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. 5 times
B. 8 times
C. 27 times
D. 35 times

Answer

To solve the problem, we need to analyze how the change in volume affects the rate of the reaction given by the equation:

\[ 2NO + O_2 \rightarrow 2NO_2 \]

### Step-by-Step Solution:

**Step 1: Understand the initial conditions.**
- Let the initial volume of the vessel be \( V \).
- The initial concentrations of the reactants \( [NO] \) and \( [O_2] \) can be expressed as:
\[ [NO] = \frac{n_{NO}}{V} \quad \text{and} \quad [O_2] = \frac{n_{O_2}}{V} \]
where \( n_{NO} \) and \( n_{O_2} \) are the number of moles of \( NO \) and \( O_2 \), respectively.

**Step 2: Write the rate expression for the reaction.**
- The rate of the reaction can be expressed as:
\[ \text{Rate} = k [NO]^2 [O_2] \]
where \( k \) is the rate constant.

**Step 3: Determine the new volume.**
- The volume of the vessel is diminished to \( \frac{1}{3} \) of its initial volume, so the new volume \( V' \) is:
\[ V' = \frac{V}{3} \]

**Step 4: Calculate the new concentrations.**
- The new concentrations after the volume change will be:
\[ [NO]' = \frac{n_{NO}}{V'} = \frac{n_{NO}}{\frac{V}{3}} = 3 \cdot \frac{n_{NO}}{V} = 3[NO] \]
\[ [O_2]' = \frac{n_{O_2}}{V'} = \frac{n_{O_2}}{\frac{V}{3}} = 3 \cdot \frac{n_{O_2}}{V} = 3[O_2] \]

**Step 5: Write the new rate expression.**
- The new rate of the reaction \( R' \) can be expressed as:
\[ R' = k [NO]'^2 [O_2]' = k (3[NO])^2 (3[O_2]) = k \cdot 9[NO]^2 \cdot 3[O_2] = 27 k [NO]^2 [O_2] \]

**Step 6: Relate the new rate to the old rate.**
- The original rate \( R \) was:
\[ R = k [NO]^2 [O_2] \]
- Therefore, the new rate \( R' \) is:
\[ R' = 27R \]

**Step 7: Conclusion.**
- The rate of the reaction will increase by a factor of 27.

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

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