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nekit [7.7K]
3 years ago
5

If the activation energy for a given compound is found to be 103 kJ/mol, with a frequency factor of 4.0 × 1013 s-1, what is the

rate constant for this reaction at 398 K?2.5 × 107 s-18.2 s-13.9 × 1010 s-11.2 s-11.7 × 1010 s-1
Chemistry
1 answer:
german3 years ago
6 0

Answer:

\large \boxed{\text{1.2 s}^ {-1}}

Explanation:

\begin{array}{rcl}k & = & Ae^{(-E_{a}/{RT)}}\\k & = & 4.0 \times 10^{13}\times e^{-(103000/{(8.314\times398))}}\\& = & 4.0 \times 10^{13}\times e^{-(103000/{3309)}}\\& = & 4.0 \times 10^{13}\times e^{-31.13}\\& = & 4.0 \times 10^{13}\times 3.130  \times 10^{-14}\\& = & \mathbf{1.2} \textbf{ s}^{\mathbf{-1}}\\\end{array}\\\text{The value of the rate constant is $\large \boxed{\textbf{1.2 s}^{\mathbf{-1}}}$}

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hola soy jess, tu respuesta esta aqui

¿cuantos moles de CO2 se requiere para reaccionar 2 moles de Ba(OH)2

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Explanation:

espero que pueda ayudarte

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