The following data is obtained during the first order thermal decomposition of
`2A(g)rarrB(g)+C(s)` at constant volume and temperature
`{:("S.No.","Time","Total pressure"),(1.,"At the end of 10 minutes",300),(2.,"After completion",200):}`
The rate constant in `"min"^(-1)` is

Options

A. 0.0693
B. 0.693
C. 6.93
D. `6.93 xx 10^(–3)`

Explanation

To find the rate constant \( k \) for the first-order thermal decomposition reaction \( 2A(g) \rightarrow B(g) + C(s) \), we will follow these steps:

### Step 1: Understand the Reaction and Initial Conditions
Initially, at \( t = 0 \), the total pressure \( P_0 \) corresponds only to the pressure of the reactant \( A \). Since the stoichiometry of the reaction indicates that 2 moles of \( A \) decompose, we will denote the initial pressure as \( P_0 \).

### Step 2: Analyze the Data Given
From the problem, we have:
- At \( t = 10 \) minutes, the total pressure is 300.
- After completion of the reaction (at \( t = \infty \)), the total pressure is 200.

### Step 3: Relate Pressures at Different Times
1. At \( t = \infty \), all \( A \) has decomposed, and the pressure of the products is \( \frac{P_0}{2} = 200 \). Thus, we can solve for \( P_0 \):
\[ P_0 = 200 \times 2 = 400 \text{ (total pressure at initial time)} \]

2. At \( t = 10 \) minutes, the pressure is given as 300. The pressure at this time can be expressed as:
\[ P(t) = P_0 - 2x + x = P_0 - x \]
where \( x \) is the change in pressure due to the decomposition of \( A \). Since we have \( P(t) = 300 \):
\[ 400 - x = 300 \implies x = 100 \]

### Step 4: Determine the Rate Constant \( k \)
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{P_0}{P(t)}\right) \]
Substituting the values we have:
- \( P_0 = 400 \)
- \( P(t) = P_0 - 2x = 400 - 200 = 200 \)
- \( t = 10 \) minutes

Now substituting these values into the formula:
\[ k = \frac{2.303}{10} \log\left(\frac{400}{200}\right) \]
Calculating the logarithm:
\[ \log(2) \approx 0.3010 \]
Thus:
\[ k = \frac{2.303}{10} \times 0.3010 = 0.0693 \text{ min}^{-1} \]

### Final Answer
The rate constant \( k \) is approximately \( 0.0693 \text{ min}^{-1} \).

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