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The rate constant for the decomposition of hydrocarbons is \(2.418 \times 10^{-5} \mathrm{~s}^{-1}\) at \(546 \mathrm{~K}\). If the energy of activation is \(179.9 \mathrm{~kJ} / \mathrm{mol}\), what will be the value of pre-exponential factor.
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Given \(k=2.418 \times 10^{-5} \mathrm{~s}^{-1}, E_a=179.9 \mathrm{~kJ} \mathrm{~mol}^{-1}\), \(T=546 \mathrm{~K}\)
Using Arrhenius equation, \(k=A e^{-E_a / R T}\)
\(\ln k=\ln A-\frac{E_a}{R T}\)
\(\log k=\log A-\frac{E_a}{2 \cdot 303 R T}\)
or \(\log A=\log k+\frac{E_a}{2 \cdot 303 R T}\)
\(=\log \left(2.418 \times 10^{-5}\right)+\frac{179.9}{2.303 \times 8.314 \times 546}\)
\(=(-5+0 \cdot 3834)+17 \cdot 2081\)
\(\log A=12 \cdot 5914 \mathrm{~s}^{-1}\)
\(A=\) Antilog \((12 \cdot 5914) \mathrm{s}^{-1}=3.903 \times 10^{12} \mathrm{~s}^{-1}\).
Using Arrhenius equation, \(k=A e^{-E_a / R T}\)
\(\ln k=\ln A-\frac{E_a}{R T}\)
\(\log k=\log A-\frac{E_a}{2 \cdot 303 R T}\)
or \(\log A=\log k+\frac{E_a}{2 \cdot 303 R T}\)
\(=\log \left(2.418 \times 10^{-5}\right)+\frac{179.9}{2.303 \times 8.314 \times 546}\)
\(=(-5+0 \cdot 3834)+17 \cdot 2081\)
\(\log A=12 \cdot 5914 \mathrm{~s}^{-1}\)
\(A=\) Antilog \((12 \cdot 5914) \mathrm{s}^{-1}=3.903 \times 10^{12} \mathrm{~s}^{-1}\).
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