The reaction `A(g) + 2B(g) rarr C(g) + D(g)` is an elementary process. In an experiment involvig this reaction, the initial partial pressure of `A` and `B` are `p_(A) = 0.60 atm` and `p_(B) = 0.80 atm`, respectively. When `p_(C ) = 0.20 atm`, the rate of reaction relative to the initial rate is

Options

A. `(1)/(6)`
B. `(1)/(48)`
C. `(1)/(4)`
D. `(1)/(24)`

Explanation

To solve the problem, we need to determine the rate of reaction when the partial pressure of C is 0.20 atm and compare it to the initial rate of the reaction.

### Step-by-Step Solution:

1. **Write the rate expression for the reaction:**
The reaction is given as:
\[ A(g) + 2B(g) \rightarrow C(g) + D(g) \]
Since this is an elementary reaction, the rate can be expressed as:
\[ \text{Rate} = k [A]^1 [B]^2 \]
where \( k \) is the rate constant.

2. **Determine the initial concentrations (partial pressures):**
The initial partial pressures are given as:
\[ p_A = 0.60 \, \text{atm}, \quad p_B = 0.80 \, \text{atm} \]

3. **Calculate the initial rate of the reaction (R1):**
Substitute the initial pressures into the rate expression:
\[ R_1 = k \cdot (0.60) \cdot (0.80)^2 \]
\[ R_1 = k \cdot (0.60) \cdot (0.64) = k \cdot 0.384 \]

4. **Determine the change in partial pressures when \( p_C = 0.20 \, \text{atm} \):**
According to the stoichiometry of the reaction:
- For every 1 mole of A consumed, 2 moles of B are consumed, and 1 mole of C is produced.
- If \( p_C = 0.20 \, \text{atm} \), then \( p_A \) decreases by 0.20 atm and \( p_B \) decreases by \( 2 \times 0.20 = 0.40 \, \text{atm} \).

Thus, the new partial pressures will be:
\[ p_A = 0.60 - 0.20 = 0.40 \, \text{atm} \]
\[ p_B = 0.80 - 0.40 = 0.40 \, \text{atm} \]

5. **Calculate the rate of the reaction when \( p_C = 0.20 \, \text{atm} \) (R2):**
Substitute the new pressures into the rate expression:
\[ R_2 = k \cdot (0.40) \cdot (0.40)^2 \]
\[ R_2 = k \cdot (0.40) \cdot (0.16) = k \cdot 0.064 \]

6. **Find the ratio of the final rate to the initial rate:**
To find the relative rate, we compute:
\[ \frac{R_2}{R_1} = \frac{k \cdot 0.064}{k \cdot 0.384} \]
The \( k \) cancels out:
\[ \frac{R_2}{R_1} = \frac{0.064}{0.384} = \frac{1}{6} \]

### Final Answer:
The rate of reaction relative to the initial rate is:
\[ \frac{R_2}{R_1} = \frac{1}{6} \]