The order of the gaseous reaction `A(g)rarr2B(g)+C(g)` is found to be one at the intial pressure `P_(0)=90` torr. The total pressure after ten minutes is found to be 180 torr. Find the value of the rate constant?

Options

A. `1.15xx10^(-3)s^(-1)`
B. `2.30xx10^(-3)s^(-1)`
C. `3.45xx10^(-3)s^(-1)`
D. `4.60xx10^(-3)s^(-1)`

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the rate constant for the given reaction.

### Step 1: Understand the Reaction and Initial Conditions
The reaction is given as:
\[ A(g) \rightarrow 2B(g) + C(g) \]

We know:
- The order of the reaction is 1 (first order).
- Initial pressure, \( P_0 = 90 \) torr.
- Total pressure after 10 minutes = 180 torr.

### Step 2: Set Up the Pressure Changes
Let \( x \) be the change in pressure of \( A \) that has reacted after 10 minutes. According to the stoichiometry of the reaction:
- The pressure of \( A \) decreases by \( x \).
- The pressure of \( B \) increases by \( 2x \) (since 2 moles of \( B \) are produced for every mole of \( A \)).
- The pressure of \( C \) increases by \( x \).

### Step 3: Write the Total Pressure Equation
The total pressure after 10 minutes can be expressed as:
\[ P_0 - x + 2x + x = 180 \]
This simplifies to:
\[ 90 - x + 2x + x = 180 \]
\[ 90 + 2x = 180 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ 2x = 180 - 90 \]
\[ 2x = 90 \]
\[ x = 45 \text{ torr} \]

### Step 5: Determine the Final Pressure of A
The final pressure of \( A \) after 10 minutes is:
\[ P_A = P_0 - x = 90 - 45 = 45 \text{ torr} \]

### Step 6: Use the First Order Rate Constant Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{T} \log \left( \frac{P_0}{P_A} \right) \]
Where:
- \( P_0 = 90 \) torr (initial pressure)
- \( P_A = 45 \) torr (pressure of \( A \) after 10 minutes)
- \( T = 10 \text{ minutes} = 10 \times 60 \text{ seconds} = 600 \text{ seconds} \)

### Step 7: Calculate the Logarithmic Term
Now, we calculate the logarithmic term:
\[ \log \left( \frac{90}{45} \right) = \log(2) \]

### Step 8: Substitute Values into the Rate Constant Formula
Now substituting the values into the formula:
\[ k = \frac{2.303}{600} \log(2) \]

Using \( \log(2) \approx 0.301 \):
\[ k = \frac{2.303}{600} \times 0.301 \]
\[ k \approx \frac{0.693}{600} \]
\[ k \approx 0.001155 \text{ torr}^{-1} \text{ s}^{-1} \]

### Step 9: Final Answer
Thus, the value of the rate constant \( k \) is:
\[ k \approx 1.155 \times 10^{-3} \text{ s}^{-1} \]