Consider the reaction,
`2A + B rarr C + D` , If the rate expression is `r = K[A]^2[B]^1` and if volume is reduced to `1/3` , the rate of reaction will increases

Options

A. 27 times
B. 9 times
C. 8 time
D. Rate will not get affected

Answer

To solve the problem step by step, we will analyze the reaction and how the change in volume affects the rate of reaction.

### Step 1: Write the rate expression
The given reaction is:
\[ 2A + B \rightarrow C + D \]
The rate expression for this reaction is given as:
\[ r = K[A]^2[B]^1 \]

### Step 2: Understand the relationship between concentration and volume
Concentration (C) is defined as the number of moles (n) of a substance divided by the volume (V) of the solution:
\[ [A] = \frac{n_A}{V} \]
\[ [B] = \frac{n_B}{V} \]
Where \( n_A \) and \( n_B \) are the number of moles of A and B, respectively.

### Step 3: Calculate the new concentrations after reducing the volume
If the volume is reduced to \( \frac{1}{3} \) of the original volume (V), the new volume \( V' \) can be expressed as:
\[ V' = \frac{V}{3} \]
Thus, the new concentrations will be:
\[ [A]' = \frac{n_A}{V'} = \frac{n_A}{\frac{V}{3}} = 3 \cdot \frac{n_A}{V} = 3[A] \]
\[ [B]' = \frac{n_B}{V'} = \frac{n_B}{\frac{V}{3}} = 3 \cdot \frac{n_B}{V} = 3[B] \]

### Step 4: Substitute the new concentrations into the rate expression
Now we substitute the new concentrations into the rate expression:
\[ r' = K[A]'^2[B]'^1 \]
\[ r' = K(3[A])^2(3[B]) \]
\[ r' = K \cdot 9[A]^2 \cdot 3[B] \]
\[ r' = 27K[A]^2[B] \]

### Step 5: Compare the new rate with the initial rate
The initial rate \( r \) is:
\[ r = K[A]^2[B] \]
The new rate \( r' \) is:
\[ r' = 27K[A]^2[B] \]

This shows that:
\[ r' = 27r \]
Thus, the rate of reaction increases by a factor of 27 when the volume is reduced to one-third.

### Final Answer
The rate of reaction will increase by a factor of 27 when the volume is reduced to one-third.

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