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A reaction has an activation energy of 195.0 kJ/mol. When the temperature is increased from 473.0K to 493.0K, what will the rate constant increase by a factor of?

I am confused on this problem because according to the answer key the answer is 7.5, but when I used the arrhenius equation to calculate the rate constant at both temperatures, I got .9999(...) for each one, the numbers after the .9999 were slightly different but I still don't understand how it is a factor of 7.5? Will someone please explain and show work?

User Aleix
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Final answer:

To determine the factor increase of the rate constant when temperature is raised, use the Arrhenius equation, making sure to use the correct activation energy in joules and temperatures in Kelvin.

Step-by-step explanation:

To calculate the factor by which the rate constant increases when the temperature is increased from 473.0K to 493.0K, given an activation energy of 195.0 kJ/mol, the Arrhenius equation is used:

k = Ae-Ea/RT

where:

  • k is the rate constant,
  • A is the frequency factor,
  • Ea is the activation energy,
  • R is the gas constant (8.314 J/mol·K),
  • T is the temperature in Kelvin.

Using the natural logarithm of the Arrhenius equation allows comparison of rate constants at two different temperatures, T1 and T2:

ln(k2/k1) = (-Ea/R)(1/T2 - 1/T1)

To find the factor by which rate constant increases, simply calculate k2/k1 given T2 (493.0K) and T1 (473.0K) and the activation energy (195.0 kJ/mol, which is 195000 J/mol to match the units of R). The calculation will yield a factor significantly different from 1, showing an increase in the rate constant.

If incorrect values are obtained, the calculation needs to be rechecked, ensuring that all units match and that the temperatures are in Kelvin. The correct calculations should give a factor significantly different from 1, suggesting an oversight in the initial attempt.

User Michalh
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