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

Answer

To solve the question regarding the effect of reducing the volume of the reaction vessel on the rate of the reaction `2NO(g) + O2(g) ⇌ 2NO2(g)`, we can follow these steps:

### Step 1: Understand the Reaction Order
The reaction is given as a first-order reaction. This means that the rate of the reaction is directly proportional to the concentration of the reactants.

### Step 2: Write the Rate Expression
For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k \cdot [\text{Reactants}] \]
where \( k \) is the rate constant and \([\text{Reactants}]\) represents the concentration of the reactants.

### Step 3: Determine the Effect of Volume Change on Concentration
When the volume of the reaction vessel is reduced to \( \frac{1}{3} \), the concentration of the reactants will change. The concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the volume is reduced to \( \frac{V}{3} \), the new concentration (\( [\text{Reactants}]' \)) will be:
\[ [\text{Reactants}]' = \frac{\text{Number of moles}}{\frac{V}{3}} = 3 \cdot \frac{\text{Number of moles}}{V} = 3 \cdot [\text{Reactants}] \]
Thus, the new concentration is three times the original concentration.

### Step 4: Calculate the New Rate of Reaction
Now, substituting the new concentration into the rate expression:
\[ \text{New Rate} = k \cdot [\text{Reactants}]' = k \cdot (3 \cdot [\text{Reactants}]) \]
This simplifies to:
\[ \text{New Rate} = 3 \cdot (k \cdot [\text{Reactants}]) = 3 \cdot \text{Original Rate} \]

### Step 5: Conclusion
Therefore, when the volume of the reaction vessel is reduced to \( \frac{1}{3} \), the rate of the reaction becomes three times the original rate.

### Final Answer
The rate of the reaction would be **three times the original rate**.

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