Rate constant calculation Using the Arrhenius equation

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SUMMARY

The discussion focuses on calculating the rate constant at 538 K using the Arrhenius equation. Given rate constants of 0.0117/s at 400.0 K and 0.689/s at 450.0 K, participants derive the activation energy (Ea) and pre-exponential factor (A) through the equations ln k1 and ln k2. By substituting these values into the Arrhenius equation, the rate constant at the desired temperature can be accurately determined.

PREREQUISITES
  • Understanding of the Arrhenius equation
  • Familiarity with natural logarithms
  • Knowledge of activation energy (Ea) and pre-exponential factor (A)
  • Basic principles of thermodynamics related to temperature and reaction rates
NEXT STEPS
  • Calculate activation energy (Ea) using the Arrhenius equation
  • Determine the pre-exponential factor (A) from the rate constants
  • Apply the Arrhenius equation to find rate constants at various temperatures
  • Explore the impact of temperature on reaction rates in chemical kinetics
USEFUL FOR

Chemistry students, chemical engineers, and researchers in the field of kinetics who are looking to deepen their understanding of reaction rates and the Arrhenius equation.

Madelin Pierce
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Homework Statement


A reaction has a rate constant of 0.0117/s at 400.0 K and 0.689/s at 450.0 K. What is the value of the rate constant (to 1 decimal place) at 538 K?

Homework Equations


ln k= ln A-E[a]/RT

The Attempt at a Solution


I'm not sure how to approach this problem, although I know the arrhenius equation is involved. When I submitted it blank on the computer, the computer hinted at finding enthalpy. But I don't know how. [/B]
 
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$$\ln{k_1} = \ln{A} - \frac{E_a}{RT_1} \text{...Eqn. 1}$$ and $$\ln{k_2} = \ln{A} - \frac{E_a}{RT_2} \text{...Eqn. 2}$$ for two different ##k_1## and ##k_2## at two different temperatures ##T_1## and ##T_2##.

Subtract Eqn1 from Eqn2:$$\ln{\frac{k2}{k1}} = - E_a R \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \text{...Eqn. 3}$$ Find ##E_a## by putting in the appropriate values.

Then put ##E_a## in Eqn1 or Eqn2, and find ##A##.

Then use $$\ln{k} = \ln{A} - \frac{E_a}{RT}$$ and put ##T= 538K## and find ##k##, the rate constant at temperature ##T##. You already have ##E_a## and ##A##, and they remain constant.
 

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