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For a certain reaction the values of Arr...

For a certain reaction the values of Arrhenius factor and Activation energy are 4 x `10^13` collision/sec and 98.6KJ/mol at 303K. Calculate the rate constant if reaction is 1st order?( `R=8.341mol^-1K^-1`)

A

6.07 x `10^-3`

B

3.02 x `10^-5`

C

4.07 x `10^-4`

D

7.42 x `10^-3`

Text Solution

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The correct Answer is:
To calculate the rate constant (k) for a first-order reaction using the Arrhenius equation, we will follow these steps: ### Step 1: Identify the given values - Arrhenius factor (A) = \(4 \times 10^{13}\) collision/sec - Activation energy (E_a) = 98.6 kJ/mol - Temperature (T) = 303 K - Gas constant (R) = 8.314 J/(mol·K) ### Step 2: Convert activation energy to Joules Since the gas constant R is in J/(mol·K), we need to convert the activation energy from kJ/mol to J/mol: \[ E_a = 98.6 \, \text{kJ/mol} = 98.6 \times 10^3 \, \text{J/mol} = 98600 \, \text{J/mol} \] ### Step 3: Write the Arrhenius equation for the rate constant The Arrhenius equation in logarithmic form is given by: \[ \log k = \log A - \frac{E_a}{2.303 R T} \] ### Step 4: Substitute the known values into the equation Now we substitute the values into the equation: \[ \log k = \log(4 \times 10^{13}) - \frac{98600}{2.303 \times 8.314 \times 303} \] ### Step 5: Calculate \(\log A\) Calculate \(\log(4 \times 10^{13})\): \[ \log(4 \times 10^{13}) = \log 4 + \log(10^{13}) = 0.6021 + 13 = 13.6021 \] ### Step 6: Calculate the denominator Calculate \(2.303 \times R \times T\): \[ 2.303 \times 8.314 \times 303 = 2.303 \times 8.314 \times 303 \approx 5801.584 \] ### Step 7: Calculate the activation energy term Now calculate the term \(\frac{E_a}{2.303 R T}\): \[ \frac{98600}{5801.584} \approx 17.00 \] ### Step 8: Substitute back to find \(\log k\) Now substitute back to find \(\log k\): \[ \log k = 13.6021 - 17.00 \approx -3.3979 \] ### Step 9: Calculate k To find k, we take the antilog: \[ k = 10^{-3.3979} \approx 4.07 \times 10^{-4} \, \text{sec}^{-1} \] ### Final Answer The rate constant \(k\) for the reaction is approximately: \[ k \approx 4.07 \times 10^{-4} \, \text{sec}^{-1} \] ---
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