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//12`
C. `1//36`
D. `1//18`

Explanation

To solve the problem step by step, we will analyze the reaction and calculate the rates accordingly.

### Step 1: Write the rate law for the reaction
The reaction given is:
\[ A(g) + 2B(g) \rightarrow C(g) + D(g) \]

Since this is an elementary reaction, the rate law can be expressed directly using the stoichiometric coefficients:
\[ \text{Rate} = k [A]^1 [B]^2 \]

### Step 2: Determine the initial concentrations (partial pressures)
From the problem, the initial partial pressures are:
- \( p_A = 0.60 \, \text{atm} \)
- \( p_B = 0.80 \, \text{atm} \)

### Step 3: Calculate the initial rate of the reaction
Using the initial pressures in the rate law:
\[ \text{Initial Rate} (R_i) = k (0.60)^1 (0.80)^2 \]
\[ R_i = k (0.60) (0.64) = k (0.384) \]

### Step 4: Determine the changes in pressures when \( p_C = 0.20 \, \text{atm} \)
When \( p_C = 0.20 \, \text{atm} \), we can determine how much \( A \) and \( B \) have reacted. Since the stoichiometry of the reaction indicates that 1 mole of \( A \) reacts with 2 moles of \( B \) to produce 1 mole of \( C \), we can set up the following relationships:

Let \( x \) be the change in pressure of \( A \) that has reacted:
- For \( C \): \( p_C = 0.20 \, \text{atm} \) implies \( x = 0.20 \, \text{atm} \)
- For \( A \): \( p_A = 0.60 - x = 0.60 - 0.20 = 0.40 \, \text{atm} \)
- For \( B \): \( p_B = 0.80 - 2x = 0.80 - 2(0.20) = 0.80 - 0.40 = 0.40 \, \text{atm} \)

### Step 5: Calculate the rate of the reaction at this point
Now we can calculate the rate of the reaction when \( p_C = 0.20 \, \text{atm} \):
\[ \text{Rate} (R_f) = k (0.40)^1 (0.40)^2 \]
\[ R_f = k (0.40) (0.16) = k (0.064) \]

### Step 6: Calculate the relative rate of the reaction
Now, we need to find the ratio of the final rate \( R_f \) to the initial rate \( R_i \):
\[ \frac{R_f}{R_i} = \frac{k (0.064)}{k (0.384)} \]
The \( k \) cancels out:
\[ \frac{R_f}{R_i} = \frac{0.064}{0.384} \]

### Step 7: Simplify the ratio
To simplify:
\[ \frac{0.064}{0.384} = \frac{64}{3840} = \frac{1}{60} \]

Thus, the relative rate of the reaction when \( p_C = 0.20 \, \text{atm} \) is:
\[ \frac{R_f}{R_i} = \frac{1}{6} \]

### Final Answer
The rate of reaction relative to the initial rate is \( \frac{1}{6} \).

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