For a first order reaction `A(g)rarr2B(g)+C(g)` at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t ae `P_(0) and P_(t)` respectively. Initially, only A is present with concentration `[A]_(0) and t_(1//3)` is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct options (s) is (are) :
(Assume that all these behave as ideal gases)

Options

A.
B.
C.
D.

Explanation

To solve the problem step by step, we will analyze the first-order reaction given and derive the necessary relationships.

### Step 1: Understand the Reaction
The reaction is given as:
\[ A(g) \rightarrow 2B(g) + C(g) \]
Initially, only A is present. Let's denote:
- Initial pressure of A as \( P_0 \)
- Partial pressure of A at time \( t \) as \( P \)
- Total pressure at time \( t \) as \( P_t \)

### Step 2: Express Total Pressure
At time \( t = 0 \), the total pressure \( P_0 \) is due to A alone. As the reaction proceeds, A is converted into products B and C.

If \( x \) is the decrease in pressure of A at time \( t \):
- The pressure of A at time \( t \) is \( P_0 - x \)
- The pressure of B produced is \( 2x \)
- The pressure of C produced is \( x \)

Thus, the total pressure at time \( t \) can be expressed as:
\[ P_t = (P_0 - x) + 2x + x = P_0 + 2x \]

### Step 3: Relate Change in Pressure to Time
For a first-order reaction, the rate constant \( k \) is related to the concentration (or pressure) of the reactant by the equation:
\[ k = \frac{1}{t} \ln\left(\frac{P_0}{P_0 - x}\right) \]

### Step 4: Determine \( x \) at \( t_{1/3} \)
We are given that \( t_{1/3} \) is the time required for the partial pressure of A to reach \( \frac{1}{3} P_0 \). Thus, at \( t = t_{1/3} \):
\[ P = \frac{1}{3} P_0 \]
This means:
\[ x = P_0 - P = P_0 - \frac{1}{3} P_0 = \frac{2}{3} P_0 \]

### Step 5: Substitute \( x \) into Total Pressure Equation
Now substituting \( x \) back into the total pressure equation:
\[ P_t = P_0 + 2\left(\frac{2}{3} P_0\right) = P_0 + \frac{4}{3} P_0 = \frac{7}{3} P_0 \]

### Step 6: Relate \( P_t \) to the Rate Constant
Now substituting \( x \) into the rate constant equation:
\[ k = \frac{1}{t_{1/3}} \ln\left(\frac{P_0}{\frac{1}{3} P_0}\right) = \frac{1}{t_{1/3}} \ln(3) \]
Thus, we can express \( t_{1/3} \) in terms of \( k \):
\[ t_{1/3} = \frac{1}{k} \ln(3) \]

### Step 7: Analyze Graphs
From the derived relationships, we can analyze the graphs:
1. The graph of \( \ln(3P_0 - P_t) \) vs. time will be a straight line with a negative slope, indicating a first-order reaction.
2. The relationship between \( t_{1/3} \) and concentration does not show a direct correlation, hence a graph showing a negative slope would be incorrect.
3. The relationship between the rate constant \( k \) and initial pressure \( P_0 \) shows no direct correlation, as \( k \) is constant for a given reaction.

### Conclusion
The correct options based on the analysis would be the graphs that depict the relationship of \( \ln(3P_0 - P_t) \) vs. time and the graph showing no relationship between \( k \) and \( P_0 \).

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