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. The reaction given is:

\[ A + 2B \rightarrow C \]

### Step-by-step Solution:

1. **Understanding the Rate of Reaction**:
The rate of an elementary reaction can be expressed as:
\[ \text{Rate} = k [A]^m [B]^n \]
where \( k \) is the rate constant, and \( m \) and \( n \) are the stoichiometric coefficients of the reactants. For our reaction, \( m = 1 \) for \( A \) and \( n = 2 \) for \( B \). Thus, the rate expression becomes:
\[ \text{Rate} = k [A][B]^2 \]

2. **Effect of Volume Change on Concentration**:
When the volume of the vessel is reduced to one-third, the concentration of the reactants will change. Concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the original volume is \( V \), the new volume after reduction is \( \frac{V}{3} \). The concentration of the reactants will increase as follows:
- For \( A \):
\[ [A]_{\text{new}} = \frac{n_A}{\frac{V}{3}} = 3 \cdot [A]_{\text{original}} \]
- For \( B \):
\[ [B]_{\text{new}} = \frac{n_B}{\frac{V}{3}} = 3 \cdot [B]_{\text{original}} \]

3. **Substituting New Concentrations into the Rate Expression**:
Now, substituting the new concentrations into the rate expression:
\[ \text{Rate}_{\text{new}} = k [A]_{\text{new}} [B]_{\text{new}}^2 \]
Substituting the new concentrations:
\[ \text{Rate}_{\text{new}} = k (3[A]) (3[B])^2 \]
Simplifying this:
\[ \text{Rate}_{\text{new}} = k (3[A]) (9[B]^2) = 27 k [A][B]^2 \]

4. **Relating New Rate to Initial Rate**:
The initial rate of the reaction is:
\[ \text{Rate}_{\text{initial}} = k [A][B]^2 \]
Therefore, we can express the new rate in terms of the initial rate:
\[ \text{Rate}_{\text{new}} = 27 \cdot \text{Rate}_{\text{initial}} \]

5. **Conclusion**:
Thus, when the volume of the vessel is reduced to one-third of its original volume, the rate of the reaction increases by a factor of 27.

### Final Answer:
The rate of the reaction will increase by 27 times the initial rate.

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