For the gaseous reaction `A(g) rarr 4 B(g) + 3C( g)` is found to be first order with respect to A. If at the starting the total pressure was 100mm Hg and after 20 minutes it is found to be 40 mm Hg. The rate constant of the reaction is `:`

Options

A. `20 m i n^(-1)`
B. `1.2 xx 10^(3) sec ^(-1)`
C. `7.6 xx 10^(-4) sec^(-1)`
D. `40 m i n^(-1)`

Answer

To find the rate constant \( k \) for the reaction \( A(g) \rightarrow 4B(g) + 3C(g) \), we will follow these steps:

### Step 1: Understand the reaction and given data
The reaction is first order with respect to \( A \). We are given:
- Initial total pressure \( P_0 = 100 \, \text{mm Hg} \)
- Total pressure after 20 minutes \( P_t = 40 \, \text{mm Hg} \)
- Time \( t = 20 \, \text{minutes} = 20 \times 60 = 1200 \, \text{seconds} \)

### Step 2: Relate pressure change to concentration change
For a first-order reaction, the rate constant can be determined using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{P_0}{P_t}\right) \]
Where:
- \( P_0 \) is the initial pressure
- \( P_t \) is the pressure at time \( t \)

### Step 3: Substitute the values into the equation
Substituting the known values into the equation:
\[ k = \frac{2.303}{1200} \log\left(\frac{100}{40}\right) \]

### Step 4: Calculate the logarithm
First, calculate \( \frac{100}{40} = 2.5 \).
Now, find \( \log(2.5) \):
\[ \log(2.5) \approx 0.3979 \]

### Step 5: Substitute the logarithm value back into the equation
Now substituting back into the equation for \( k \):
\[ k = \frac{2.303}{1200} \times 0.3979 \]

### Step 6: Calculate \( k \)
Calculating \( k \):
\[ k = \frac{2.303 \times 0.3979}{1200} \approx \frac{0.9163}{1200} \approx 0.000763 \]
This can be expressed in scientific notation:
\[ k \approx 7.63 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer
The rate constant \( k \) of the reaction is approximately \( 7.63 \times 10^{-4} \, \text{s}^{-1} \).

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