Ethylene is produced by `C_(4)H_(8) overset(Delta)to 2C_(2)H_(4)`
 The rate constant is `2.3 xx 10^(-4) sec^(-1)`. Approximately in what time will the molar ratio of ethylene to cyclobutane in mixture attain the value equal to one?
(log 2=0.3, log 3 =0.47)

Options

A. 1700 sec
B. 1600 sec
C. 2000 sec
D. 1200 sec

Explanation

To solve the problem step by step, we will follow the outlined process in the video transcript.

### Step 1: Write the Reaction
The reaction given is:
\[ C_4H_8 \overset{\Delta}{\rightarrow} 2C_2H_4 \]
This indicates that one mole of cyclobutane (C4H8) produces two moles of ethylene (C2H4).

### Step 2: Define Initial Concentrations
Let the initial concentration of cyclobutane be \( A \) (in moles). At time \( T = 0 \), the concentration of ethylene is 0.

### Step 3: Define Change in Concentration
Let \( x \) be the amount of cyclobutane that has reacted at time \( T \). Therefore, the concentration of cyclobutane at time \( T \) will be \( A - x \), and the concentration of ethylene produced will be \( 2x \).

### Step 4: Set Up the Molar Ratio Equation
We want the molar ratio of ethylene to cyclobutane to be equal to 1:
\[ \frac{2x}{A - x} = 1 \]
Cross-multiplying gives:
\[ 2x = A - x \]
Rearranging this equation gives:
\[ 3x = A \quad \Rightarrow \quad x = \frac{A}{3} \]

### Step 5: Determine the Remaining Concentration of Cyclobutane
The remaining concentration of cyclobutane will be:
\[ A - x = A - \frac{A}{3} = \frac{2A}{3} \]

### Step 6: Identify the Reaction Order
The rate constant \( k \) is given as \( 2.3 \times 10^{-4} \, \text{sec}^{-1} \). Since the unit of the rate constant is \( \text{sec}^{-1} \), this indicates that the reaction is first order.

### Step 7: Use the First Order Kinetics Equation
For a first-order reaction, the time \( T \) can be calculated using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \]
Where:
- \( [A]_0 = A \) (initial concentration)
- \( [A]_t = \frac{2A}{3} \) (concentration at time \( T \))

### Step 8: Substitute Values into the Equation
Substituting the values into the equation:
\[ T = \frac{2.303}{2.3 \times 10^{-4}} \log \left( \frac{A}{\frac{2A}{3}} \right) \]
This simplifies to:
\[ T = \frac{2.303}{2.3 \times 10^{-4}} \log \left( \frac{3}{2} \right) \]

### Step 9: Calculate the Logarithm
Using the logarithm properties:
\[ \log \left( \frac{3}{2} \right) = \log 3 - \log 2 \]
Given \( \log 3 = 0.47 \) and \( \log 2 = 0.3 \):
\[ \log \left( \frac{3}{2} \right) = 0.47 - 0.3 = 0.17 \]

### Step 10: Substitute Logarithm Value and Calculate Time
Now substitute back into the equation:
\[ T = \frac{2.303}{2.3 \times 10^{-4}} \times 0.17 \]
Calculating this gives:
\[ T \approx \frac{2.303 \times 0.17}{2.3 \times 10^{-4}} \approx \frac{0.39151}{2.3 \times 10^{-4}} \approx 1700 \, \text{seconds} \]

### Final Answer
The time required for the molar ratio of ethylene to cyclobutane to attain the value of one is approximately **1700 seconds**.

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