The reaction,`2NO(g)+O_(2)(g)hArr2NO_(2)(g),` is of first order. If the volume of reaction vessel is reduced to `(1)/(3)`, the rate of reaction would be

Options

A. `(1)/(3) "times"`
B. `(2)/(3) "times"`
C. 3 times
D. 6 times

Explanation

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

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction given is:
\[ 2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g) \]
This reaction is of first order. For first-order reactions, the rate of reaction is directly proportional to the concentration of the reactants.

2. **Rate Expression**:
For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k [NO]^x [O_2]^y \]
where \(k\) is the rate constant, and \(x\) and \(y\) are the orders with respect to each reactant. In this case, since the overall reaction is first order, we can consider the concentration of the limiting reactant.

3. **Effect of Volume Change**:
When the volume of the reaction vessel is reduced to \( \frac{1}{3} \), the concentration of the gases will change. Concentration (\(C\)) is given by:
\[ C = \frac{n}{V} \]
where \(n\) is the number of moles and \(V\) is the volume. If the volume is reduced to \( \frac{1}{3} \), the concentration will increase by a factor of 3:
\[ C' = \frac{n}{\frac{V}{3}} = 3 \cdot \frac{n}{V} = 3C \]

4. **New Rate of Reaction**:
Since the rate of a first-order reaction is directly proportional to the concentration, if the concentration increases by a factor of 3, the new rate (\(R_2\)) will be:
\[ R_2 = k [C'] = k [3C] = 3 \cdot k [C] = 3R_1 \]
where \(R_1\) is the original rate of the reaction.

5. **Conclusion**:
Therefore, when the volume of the reaction vessel is reduced to one-third, the rate of the reaction increases by a factor of 3.

### Final Answer:
The rate of the reaction would be \(3R_1\).

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