# Chemistry

At 425°C, a reaction has a rate constant of 7.7 x 10–4 s–1 and has an activation energy of 275 kJ mol–1. What is the frequency factor?

The answer is 2.9 x 1017 s–1 , I need help with working out I used arrhenius equation and rearranged for A = K /e^(-Ea/RT). My answer comes out too big.

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1. As I understand it, the ‘frequency factor’ does not have units. What you are showing in the posted question is ‘rate constant’ calculated from the Arrhenius Equation. The Arrhenius Equation is derived from the frequency factor. From the KM Theory f = e¯∆E/RT and represents the fraction of molecular collisions having energy values greater than the activation energy. In the posted problem, if ∆E = 275 Kj/mole = 2.75 x 10ˢ J/mole, T = 275⁰C = 698 K and R = 8.314 J/mol·K => ∆E/RT = [(2.75 x 10ˢ J/mole)/(8.314 J/ mol·K)(698 K)] = 47.39. All units cancel out. Substituting into f = e¯∆E/RT => f = e^-(47.39) = 2.63 x 10¯²¹. Also, this type problem usually comes with a couple of temperature values which serve to illustrate – after calculation – the significant increase in number of collisions having energy values greater than the activation energy. Compare frequency at 698 K to frequency at 708 K, a 10⁰ increase in temperature. Frequency at 708 K = e^-(46.72) = 5.13 x 10¯²¹. This represents about a 2-times increase in reaction rate. Hence, the generalization ‘For every 10⁰C increase in temperature, the reaction rate doubles … b/c frequency of collisions greater than the activation energy doubles.

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posted by Dr Rebel

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