The rate constant of the first order reaction , that is decomposition of ethylene oxide into `CH_(4)` and CO, may be described by the following equation
`logk(s^(-1)) = 14.34 -(1.25 xx10^(4))/T K `
rate constant at `397^(@)C`

Explanation

To find the rate constant \( k \) for the decomposition of ethylene oxide at \( 397^\circ C \), we will use the given equation:

\[ \log k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4}}{T} \]

### Step 1: Convert the temperature from Celsius to Kelvin
To use the equation, we first need to convert the temperature from Celsius to Kelvin. The conversion formula is:

\[ T(K) = T(°C) + 273.15 \]

For \( 397^\circ C \):

\[ T = 397 + 273.15 = 670.15 \, K \]

### Step 2: Substitute the temperature into the equation
Now, we substitute \( T = 670.15 \, K \) into the equation for \( \log k \):

\[ \log k = 14.34 - \frac{1.25 \times 10^{4}}{670.15} \]

### Step 3: Calculate the value of \( \frac{1.25 \times 10^{4}}{670.15} \)
Now we calculate \( \frac{1.25 \times 10^{4}}{670.15} \):

\[ \frac{1.25 \times 10^{4}}{670.15} \approx 18.65 \]

### Step 4: Substitute this value back into the equation
Now we can substitute this value back into the equation for \( \log k \):

\[ \log k = 14.34 - 18.65 \]

### Step 5: Calculate \( \log k \)
Now we perform the subtraction:

\[ \log k \approx 14.34 - 18.65 \approx -4.31 \]

### Step 6: Convert \( \log k \) to \( k \)
To find \( k \), we need to convert from logarithmic form to exponential form:

\[ k = 10^{\log k} = 10^{-4.31} \]

### Step 7: Calculate \( k \)
Now we calculate \( k \):

\[ k \approx 4.88 \times 10^{-5} \, s^{-1} \]

### Final Answer
The rate constant \( k \) at \( 397^\circ C \) is approximately:

\[ k \approx 4.88 \times 10^{-5} \, s^{-1} \]

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