For an elementary reaction. A + 2B `rightarrow` C if volume of the vessel is reduced to `1/3` of its original volume the rate of the reaction will

Options

A. Increase 27 times
B. Remain unaltered
C. Increase 9 times
D. Decrease 3 times

Answer

To solve the problem, we need to analyze how the rate of the reaction changes when the volume of the vessel is reduced to one-third of its original volume.

### Step-by-Step Solution:

1. **Understand the Reaction**: The given reaction is:
\[ A + 2B \rightarrow C \]
This is an elementary reaction, which means the rate law can be directly written from the stoichiometry of the reaction.

2. **Write the Rate Law**: For the elementary reaction, the rate law can be expressed as:
\[ \text{Rate} = k[A]^1[B]^2 \]
where \( k \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of A and B, respectively.

3. **Relate Concentration to Volume**: Concentration is defined as:
\[ [A] = \frac{n_A}{V} \quad \text{and} \quad [B] = \frac{n_B}{V} \]
where \( n_A \) and \( n_B \) are the number of moles of A and B, and \( V \) is the volume of the vessel. If the volume is reduced to one-third, the new volume \( V' = \frac{V}{3} \).

4. **Calculate New Concentrations**: The new concentrations when the volume is reduced to one-third will be:
\[ [A]' = \frac{n_A}{V/3} = 3 \cdot \frac{n_A}{V} = 3[A] \]
\[ [B]' = \frac{n_B}{V/3} = 3 \cdot \frac{n_B}{V} = 3[B] \]

5. **Substitute New Concentrations into Rate Law**: Now, substituting the new concentrations into the rate law:
\[ \text{Rate}' = k[A]'[B]'^2 = k(3[A])(3[B])^2 \]
\[ = k(3[A])(9[B]^2) = 27k[A][B]^2 \]

6. **Compare New Rate with Original Rate**: The original rate was:
\[ \text{Rate} = k[A][B]^2 \]
Therefore, the new rate is:
\[ \text{Rate}' = 27 \cdot \text{Rate} \]

7. **Conclusion**: The rate of the reaction will increase by a factor of 27 when the volume is reduced to one-third of its original volume.

### Final Answer:
The rate of the reaction will increase by a factor of 27.

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