Find the rate constant, given temperature and activation energy.

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SUMMARY

The discussion focuses on calculating the rate constant for a chemical reaction given its activation energy and temperature. The activation energy is specified as 38.0 kJ/mol, and the rate constant at 249 K is 1.60 × 10² M⁻¹s⁻¹. The correct formula used is ln(K₂/K₁) = (Eₐ/R)((1/T₁)-(1/T₂)), where R is the gas constant (8.314 J/mol·K). A common mistake identified was the incorrect unit conversion of R from kJ to J, which led to an erroneous calculation of the rate constant at 436 K.

PREREQUISITES
  • Understanding of the Arrhenius equation and its components.
  • Familiarity with unit conversions between kJ and J.
  • Knowledge of the gas constant (R) and its application in thermodynamic equations.
  • Basic algebra skills for manipulating logarithmic equations.
NEXT STEPS
  • Review the Arrhenius equation and its implications for reaction kinetics.
  • Study unit conversion techniques, specifically between kJ and J.
  • Practice solving problems involving the calculation of rate constants at different temperatures.
  • Explore the impact of activation energy on reaction rates and how to experimentally determine it.
USEFUL FOR

Chemistry students, chemical engineers, and researchers in reaction kinetics who are looking to deepen their understanding of rate constants and activation energy calculations.

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Homework Statement



A reaction is found to have an activation energy of 38.0 kJ/mol. If the rate constant for this reaction is 1.60 × 102 M-1s-1 at 249 K, what is the rate constant at 436 K?

Homework Equations



ln\frac{K_{2}}{K_{1}}=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)

The Attempt at a Solution



Given:

R=8.314
T1=249K
T2=436K
Ea=160

ln(K_{2})=\frac{38}{8.314}\left(\frac{1}{249}-\frac{1}{436}\right)+ln(160)

Which equals 161.257 which is incorrect. Any clues where I went wrong would be greatly appreciated. Thanks in advance.

Joe
 
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Ah ha! I should have noticed that being that R has units of J not Kj. Thank you very much Borek.

Joe
 
kJ, not Kj...
 

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