Question

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

Answer 1

Answer:

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

Explanation:

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